Explanation:
Okay, so I need to evaluate the sum S = 1/(1 + ω) + 1/(1 + ω²) + ... + 1/(1 + ω¹⁹⁹⁷), where ω is a 1997th root of unity, not equal to 1. Hmm, let me think about how to approach this.

First, I remember that roots of unity have some nice properties. Since ω¹⁹⁹⁷ = 1 and ω ≠ 1, ω is a primitive 1997th root of unity. But maybe I don't need to worry about primitive roots here. The key thing is that ω is a root of the equation x¹⁹⁹⁷ - 1 = 0, so it satisfies that equation. Also, the sum of all the roots of xⁿ - 1 = 0 is 0, right? Because the coefficient of xⁿ⁻¹ is 0. So, for x¹⁹⁹⁷ - 1 = 0, the sum of all roots is 0. That includes 1 and all the other roots ω, ω², ..., ω¹⁹⁹⁶. So, 1 + ω + ω² + ... + ω¹⁹⁹⁶ = 0. Maybe that will come in handy later.

But the problem is about the sum of reciprocals of (1 + ωᵏ) for k from 1 to 1997. Wait, hold on, the original sum is from 1/(1 + ω) up to 1/(1 + ω¹⁹⁹⁷). But ω¹⁹⁹⁷ is 1, right? Because ω¹⁹⁹⁷ = 1. So the last term is 1/(1 + 1) = 1/2. But the original problem says ω ≠ 1, but ω¹⁹⁹⁷ is 1. So maybe the sum is from k=1 to 1997, but since ω¹⁹⁹⁷ = 1, the last term is 1/(1 + 1) = 1/2. However, the problem says "ω ≠ 1", but ω is a root of unity, so maybe the sum is over all the roots except 1? Wait, let me check the original problem again.

Wait, the problem says: "If ω¹⁹⁹⁷ = 1 and ω ≠ 1, then evaluate [sum from k=1 to 1997 of 1/(1 + ωᵏ)]". Wait, but if ω is a root of unity, then ω¹⁹⁹⁷ = 1, but the sum is from k=1 to 1997. However, if ω is a primitive root, then ωᵏ for k from 1 to 1996 are distinct roots, and ω¹⁹⁹⁷ = ω⁰ = 1. So the sum is over all the roots except 1? Wait, no. If we take k from 1 to 1997, then ω¹⁹⁹⁷ is 1, so the terms are 1/(1 + ω), 1/(1 + ω²), ..., 1/(1 + ω¹⁹⁹⁶), and 1/(1 + 1). So the sum is over all the roots except 1? Wait, no. The roots are 1, ω, ω², ..., ω¹⁹⁹⁶. So if we take k from 1 to 1997, then ω¹⁹⁹⁷ is 1, so the terms are 1/(1 + ω), ..., 1/(1 + ω¹⁹⁹⁶), and 1/(1 + 1). Therefore, the sum is over all the roots except 1? No, because the roots are 1, ω, ..., ω¹⁹⁹⁶. So the sum is over k=1 to 1997, which includes ω¹⁹⁹⁷ = 1. Therefore, the sum is over all the roots, but with k=1 to 1997, which includes 1. Wait, but the problem says ω ≠ 1. Wait, maybe there's confusion here.

Wait, hold on. The problem says "If ω¹⁹⁹⁷ = 1 and ω ≠ 1", so ω is a root of unity, not equal to 1. But then the sum is from k=1 to 1997 of 1/(1 + ωᵏ). But if ω is a root of unity, then ωᵏ cycles through the roots. However, if ω is a primitive root, then ω, ω², ..., ω¹⁹⁹⁶ are all distinct roots, and ω¹⁹⁹⁷ = 1. Therefore, the sum is over all the roots of x¹⁹⁹⁷ - 1 = 0, including 1. But the problem says ω ≠ 1, but maybe ω is just a root, not necessarily primitive? Wait, but the problem says "ω¹⁹⁹⁷ = 1 and ω ≠ 1", so ω is a non-trivial root. However, the sum is over k=1 to 1997 of 1/(1 + ωᵏ). Wait, but if we take k from 1 to 1997, then ωᵏ cycles through all the roots, including 1 when k=1997. Therefore, the sum is over all roots of x¹⁹⁹⁷ - 1 = 0, but with each term 1/(1 + ωᵏ). However, the problem says "ω ≠ 1", but maybe the sum is over all roots except 1? Wait, no. The problem says "evaluate [sum from k=1 to 1997 of 1/(1 + ωᵏ)]", so if ω is a root, then ωᵏ for k=1 to 1997 would be all roots, but since ω¹⁹⁹⁷ = 1, then ωᵏ for k=1 to 1997 is the same as ω, ω², ..., ω¹⁹⁹⁶, 1. Therefore, the sum is over all roots of x¹⁹⁹⁷ - 1 = 0. However, the problem says ω ≠ 1, but maybe ω is just a specific root? Wait, maybe the problem is written in a confusing way. Let me check again.

Wait, the problem says: "If ω¹⁹⁹⁷ = 1 and ω ≠ 1, then evaluate [sum from k=1 to 1997 of 1/(1 + ωᵏ)]". Wait, but if ω is a root of unity, then ωᵏ for k=1 to 1997 would be all the roots, but if ω is fixed, then ωᵏ cycles through the roots. Wait, but if ω is fixed, then ω¹, ω², ..., ω¹⁹⁹⁷ are the roots. However, if ω is a primitive root, then ω¹⁹⁹⁷ = 1, so ω¹⁹⁹⁷ is 1. Therefore, the sum is over all roots of x¹⁹⁹⁷ - 1 = 0. However, the problem says "ω ≠ 1", but maybe the sum is over all roots except 1? Wait, no. The problem says "evaluate [sum from k=1 to 1997 of 1/(1 + ωᵏ)]", so if ω is a root, then ωᵏ for k=1 to 1997 would be all roots, but since ω¹⁹⁹⁷ = 1, then the sum is over all roots. However, if ω is a root, then the sum is over all roots. But maybe the problem is written in a way that ω is a specific root, and the sum is over k=1 to 1997, which would be all roots. But maybe there's confusion here. Wait, perhaps the problem is written as sum_{k=1}^{1997} 1/(1 + ω^k), where ω is a primitive 1997th root of unity? But the problem says ω¹⁹⁹⁷ = 1 and ω ≠ 1, so ω is a root of unity, not necessarily primitive. However, if ω is not primitive, then the exponents would cycle through fewer roots. But since 1997 is prime? Wait, 1997 is a prime number? Let me check. 1997 divided by... Well, 1997 is actually a prime number. Yes, 1997 is a prime. So, if ω is a root of x¹⁹⁹⁷ - 1 = 0, then the minimal polynomial is cyclotomic polynomial, but since 1997 is prime, the cyclotomic polynomial is x¹⁹⁹⁶ + x¹⁹⁹⁵ + ... + x + 1. Therefore, if ω ≠ 1, then ω is a primitive 1997th root of unity. Therefore, the sum is over all roots except 1? Wait, no. If we take k from 1 to 1997, then ω^k for k=1 to 1997 would be ω, ω², ..., ω¹⁹⁹⁶, 1. Therefore, the sum is over all roots of x¹⁹⁹⁷ - 1 = 0. Therefore, the sum S is sum_{k=1}^{1997} 1/(1 + ω^k). But since ω^1997 = 1, so the last term is 1/(1 + 1) = 1/2. However, the problem says ω ≠ 1, but maybe the sum is over all roots, including 1. However, if we consider that, then maybe we can find a way to compute this sum.

Alternatively, maybe there's a trick to compute this sum. Let me think. Let me recall that for roots of unity, sometimes you can pair terms or use symmetry. For example, if we consider that ω^k and ω^{1997 - k} are related. Since ω^{1997} = 1, so ω^{1997 - k} = ω^{-k}. Therefore, maybe pairing terms 1/(1 + ω^k) and 1/(1 + ω^{-k}).

Let me try that. Let me consider pairing each term with its reciprocal. Let me note that for each k from 1 to 1996, ω^k and ω^{-k} are distinct roots (since 1997 is prime, so k ≠ -k mod 1997 unless k = 0, which is not in the range). Therefore, for each k from 1 to 1996, we can pair 1/(1 + ω^k) and 1/(1 + ω^{-k}).

Let me compute the sum of these two terms:

1/(1 + ω^k) + 1/(1 + ω^{-k}) = [ (1 + ω^{-k}) + (1 + ω^k) ] / [ (1 + ω^k)(1 + ω^{-k}) ] = [2 + ω^k + ω^{-k}]/[1 + ω^k + ω^{-k} + 1] = [2 + ω^k + ω^{-k}]/[2 + ω^k + ω^{-k}] = 1.

Wait, that's interesting! So each pair sums to 1. Therefore, if we pair all terms from k=1 to 1996, which is 1996 terms, so 998 pairs, each summing to 1, so the total sum would be 998. Then we have the last term, which is when k=1997, which is 1/(1 + ω^{1997}) = 1/(1 + 1) = 1/2. Therefore, total sum S = 998 + 1/2 = 998.5 = 1997/2. Wait, but 998 is 1996/2, so 1996/2 + 1/2 = 1997/2. Therefore, the total sum is 1997/2. But wait, that seems too straightforward. Let me check again.

Wait, let me verify the pairing. Let me take k and 1997 - k. Wait, but since ω^{1997} = 1, ω^{-k} = ω^{1997 - k}. Therefore, pairing k and 1997 - k. However, when k ranges from 1 to 1996, each pair is unique. However, if we pair k and 1997 - k, then for k from 1 to 998, and 1997 - k from 1996 down to 999. Therefore, there are 998 pairs, each contributing 1. Then the term when k=1997 is 1/2. Therefore, total sum is 998 + 1/2 = 1997/2. But 1997 is an odd number, so 1997/2 is 998.5. However, the problem is asking for the sum of 1/(1 + ω^k) from k=1 to 1997. But if that's the case, then the answer is 1997/2. However, let me check with a smaller case. Let me take n=3, which is prime. Let me check if the same logic applies.

Let me take n=3. Let ω be a primitive 3rd root of unity. Then the sum would be 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + ω³). But ω³ = 1, so the last term is 1/2. Now, let's compute the first two terms. Let me compute 1/(1 + ω) + 1/(1 + ω²). Since ω² = ω^{-1}, so pairing them. Let me compute:

1/(1 + ω) + 1/(1 + ω^{-1}) = [ (1 + ω^{-1}) + (1 + ω) ] / [ (1 + ω)(1 + ω^{-1}) ] = [2 + ω + ω^{-1}]/[2 + ω + ω^{-1}] = 1. Therefore, the sum of the first two terms is 1, and the third term is 1/2, so total sum is 1 + 1/2 = 3/2, which is n/2. So for n=3, the answer is 3/2. Similarly, for n=1997, the answer would be 1997/2. Therefore, this seems to hold. Therefore, the answer is 1997/2. But let me check another case. Let me take n=2. Wait, n=2, but 2 is not prime? Wait, n=2, ω²=1, ω ≠ 1, so ω=-1. Then the sum is 1/(1 + ω) + 1/(1 + ω²). But ω²=1, so the sum is 1/(1 + (-1)) + 1/(1 + 1) = 1/0 + 1/2. Wait, division by zero. But n=2 is not prime? Wait, but 2 is prime. However, in this case, the term 1/(1 + ω) is 1/(1 - 1) which is undefined. Therefore, maybe the formula doesn't hold for n=2? But in our problem, n=1997 is prime, so maybe there's no division by zero? Let me check for n=3, which worked. For n=3, the term 1/(1 + ω) is defined because ω ≠ -1. Since ω is a primitive 3rd root of unity, which is e^(2πi/3), which is not equal to -1. Therefore, in our problem, since 1997 is prime, and ω is a primitive root, then ω ≠ -1? Wait, is that necessarily true? For example, if n is an odd prime, then -1 is not a root of unity of order n? Wait, -1 is a root of x² - 1 = 0, so if n is odd, then -1 is not a root of xⁿ - 1 = 0. Because (-1)^n = -1 ≠ 1. Therefore, for odd n, ω ≠ -1. Therefore, in our case, since 1997 is odd, then none of the terms 1/(1 + ω^k) will have denominator zero. Therefore, the pairing works. Therefore, for n odd prime, the sum from k=1 to n of 1/(1 + ω^k) is n/2. Therefore, the answer is 1997/2. Therefore, boxed answer is \boxed{\dfrac{1997}{2}}.

But let me think again. Wait, in the case of n=3, the sum is 3/2. Let me compute it directly. Let me take ω = e^(2πi/3). Then 1 + ω + ω² = 0. Therefore, ω² = -1 - ω. Let me compute 1/(1 + ω) + 1/(1 + ω²) + 1/2. Let me compute 1/(1 + ω) + 1/(1 + ω²). Let me compute 1/(1 + ω) + 1/(1 + ω²). Let me write ω² = -1 - ω. Therefore, 1 + ω² = -ω. Therefore, 1/(1 + ω²) = 1/(-ω) = -1/ω. Similarly, 1/(1 + ω) is 1/(1 + ω). Therefore, the sum is 1/(1 + ω) - 1/ω. Let me compute this. Let me write 1/(1 + ω) - 1/ω = [ω - (1 + ω)] / [ω(1 + ω)] = (-1)/[ω(1 + ω)]. But ω(1 + ω) = ω + ω² = -1, since 1 + ω + ω² = 0. Therefore, this becomes (-1)/(-1) = 1. Therefore, the sum of the first two terms is 1, and adding the third term 1/2 gives 3/2. So that works. Therefore, the formula holds. Therefore, for n=1997, the answer is 1997/2. Therefore, I think that is the answer.

But let me think again if there's another way to approach this problem, maybe using complex analysis or generating functions or something else. Let me try another approach.

Let me consider the sum S = sum_{k=1}^{n} 1/(1 + ω^k), where ω is a primitive n-th root of unity, and n is prime. Wait, but in our problem, n=1997. Let me consider that. Let me note that 1/(1 + ω^k) can be rewritten as [1 - ω^k]/[1 - ω^{2k}]. Wait, no. Wait, 1/(1 + x) = (1 - x)/(1 - x²). So if I let x = ω^k, then 1/(1 + ω^k) = (1 - ω^k)/(1 - ω^{2k}). But maybe this is not helpful. Alternatively, maybe consider multiplying numerator and denominator by something. Alternatively, consider that sum_{k=1}^n 1/(1 + ω^k) = sum_{k=1}^n [1/(1 + ω^k)].

Alternatively, consider that for each term, 1/(1 + ω^k) = 1/(1 + ω^k) * [ (1 - ω^k)/(1 - ω^k) ] = (1 - ω^k)/(1 - ω^{2k}). But this seems complicated. Alternatively, maybe consider that sum_{k=1}^n 1/(1 + ω^k) = sum_{k=1}^n [ (1 - ω^k) / (1 - ω^{2k}) ]? Not sure.

Alternatively, think about the sum S = sum_{k=1}^n 1/(1 + ω^k). Let me note that 1 + ω^k = 0 implies ω^k = -1. But since ω is a primitive n-th root of unity, and n is odd (1997 is odd), then -1 is not a root of unity of order n. Therefore, none of the denominators are zero, so all terms are defined.

Alternatively, consider that sum_{k=1}^n 1/(1 + ω^k) = sum_{k=1}^n [1/(1 + ω^k)].

Let me consider multiplying numerator and denominator by something. Let me think about the sum over all roots. Let me consider that for each root ω^k, we have 1/(1 + ω^k). Let me consider that if I have a function f(x) = 1/(1 + x), then the sum is sum_{k=1}^n f(ω^k). But how can I compute this sum?

Alternatively, consider that the sum over all roots of unity of f(ω^k). There is a formula for sums over roots of unity. For example, if we have a function f(x) that is rational, then we can use residues or other methods. But maybe there's a simpler way.

Alternatively, consider that sum_{k=0}^n 1/(1 + ω^k) = sum_{k=0}^n 1/(1 + ω^k). Since ω^0 = 1, so this sum would be 1/(1 + 1) + sum_{k=1}^{n-1} 1/(1 + ω^k) + 1/(1 + ω^n). But ω^n = 1, so the last term is 1/2. Therefore, sum_{k=0}^n 1/(1 + ω^k) = 1/2 + sum_{k=1}^{n-1} 1/(1 + ω^k) + 1/2 = 1 + sum_{k=1}^{n-1} 1/(1 + ω^k). But our original sum S is sum_{k=1}^n 1/(1 + ω^k) = sum_{k=1}^{n-1} 1/(1 + ω^k) + 1/2. Therefore, if I can compute sum_{k=0}^n 1/(1 + ω^k), then subtract 1/(1 + ω^0) = 1/2, then I get S. But how to compute sum_{k=0}^n 1/(1 + ω^k)?

Alternatively, consider that sum_{k=0}^n 1/(1 + ω^k) = sum_{k=0}^n [1/(1 + ω^k)]. Let me note that for each root ω^k, we can pair it with ω^{-k} as before. However, when k=0, ω^0 = 1, so 1/(1 + 1) = 1/2. For k from 1 to n-1, pairing k and n - k. But since n is odd, n - k ≠ k for any k from 1 to n-1. Therefore, similar to before, each pair sums to 1, and there are (n - 1)/2 pairs. Therefore, sum_{k=0}^n 1/(1 + ω^k) = 1/2 + (n - 1)/2 * 1 = (1 + n - 1)/2 = n/2. Therefore, sum_{k=0}^n 1/(1 + ω^k) = n/2. Therefore, our original sum S = sum_{k=1}^n 1/(1 + ω^k) = sum_{k=0}^n 1/(1 + ω^k) - 1/(1 + ω^0) = n/2 - 1/2 = (n - 1)/2. Wait, but this contradicts our previous result. Wait, what's going on here?

Wait, earlier, when I considered pairing terms from k=1 to n, I thought that pairing k and n - k gives (n - 1)/2 pairs, each summing to 1, and then adding the term for k = n, which is 1/2. But according to this new approach, sum_{k=0}^n 1/(1 + ω^k) = n/2, so sum_{k=1}^n 1/(1 + ω^k) = n/2 - 1/2 = (n - 1)/2. But in the case of n=3, this would give (3 - 1)/2 = 1, but earlier we saw that the sum was 3/2. Therefore, there's a contradiction here. Therefore, I must have made a mistake in one of the reasonings.

Wait, let me check again. For n=3, sum_{k=0}^3 1/(1 + ω^k) = 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + ω³). But ω³ = 1, so that's 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2. Which is 1 + [1/(1 + ω) + 1/(1 + ω²)]. Earlier, we saw that [1/(1 + ω) + 1/(1 + ω²)] = 1, so total sum is 1 + 1 = 2. Which is n/2 = 3/2? No, 2 ≠ 3/2. Wait, so this contradicts the previous conclusion. Therefore, my previous reasoning was wrong. Therefore, there's a mistake here.

Wait, so let me recast. For n=3, sum_{k=0}^n 1/(1 + ω^k) = 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. Which is equal to n/2? No, n=3, n/2 is 1.5, but the sum is 2. Therefore, my previous assertion that sum_{k=0}^n 1/(1 + ω^k) = n/2 is wrong. Therefore, my mistake was there. Therefore, need to re-examine.

So, going back. When I considered pairing terms for k from 1 to n-1, but in the case of n=3, pairing k=1 and k=2 (since n-1=2). Then, as before, 1/(1 + ω) + 1/(1 + ω²) = 1. Then adding the term for k=0: 1/2 and k=3: 1/2. Therefore, total sum is 1 + 1/2 + 1/2 = 2. Which is equal to n. Wait, n=3, sum is 2. So maybe sum_{k=0}^n 1/(1 + ω^k) = n? But for n=3, it's 2. Not sure. Wait, maybe there's another way.

Alternatively, let me think about the sum over all roots of unity. Let me consider that for each root ω^k, we have 1/(1 + ω^k). Let me consider that the sum over all roots of unity of 1/(1 + x) where x^n = 1. Let me consider that sum_{x^n=1} 1/(1 + x). Let me note that if I let f(x) = 1/(1 + x), then the sum is sum_{x^n=1} f(x). There is a formula for such sums. Let me recall that for a polynomial P(x) with roots r_i, sum_{i} 1/(a - r_i) = P'(a)/P(a). But here, we have sum_{x^n=1} 1/(1 + x). Let me write this as sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} 1/(1 - (-x)). Therefore, if I consider the polynomial Q(x) = x^n - 1, which has roots x^n = 1. Then, the sum over roots of 1/(1 - (-x)) = sum_{x^n=1} 1/(1 - (-x)) = sum_{x^n=1} 1/(1 - (-x)).

Using the formula sum_{i} 1/(a - r_i) = P'(a)/P(a), where P(x) = product_{i} (x - r_i). In our case, P(x) = x^n - 1, and we have sum_{i} 1/(a - r_i) = P'(a)/P(a). Therefore, if we set a = -1, then sum_{x^n=1} 1/( -1 - x ) = P'(-1)/P(-1). Therefore, sum_{x^n=1} 1/( -1 - x ) = P'(-1)/P(-1). Therefore, sum_{x^n=1} 1/(1 + x) = - sum_{x^n=1} 1/( -1 - x ) = - P'(-1)/P(-1).

Let me compute this. First, P(x) = x^n - 1. Therefore, P(-1) = (-1)^n - 1. Since n=1997 is odd, P(-1) = -1 - 1 = -2. P'(x) = n x^{n - 1}, so P'(-1) = n (-1)^{n - 1} = n (-1)^{even} = n * 1 = n. Therefore, sum_{x^n=1} 1/(1 + x) = - P'(-1)/P(-1) = -n / (-2) = n/2. Therefore, the sum over all roots x^n=1 of 1/(1 + x) is n/2. Therefore, in our case, sum_{k=0}^n 1/(1 + ω^k) = n/2. Therefore, for n=3, this would be 3/2, but earlier when I calculated manually, I got 2. Wait, there's a contradiction here. What's wrong?

Wait, for n=3, sum_{x^3=1} 1/(1 + x) = 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1/(1 + ω) + 1/(1 + ω²). Earlier, I calculated that 1/(1 + ω) + 1/(1 + ω²) = 1, so total sum is 1/2 + 1 = 3/2, which matches the formula. But earlier, when I thought I calculated it as 2, I must have made a mistake. Let me check again. For n=3, the roots are 1, ω, ω². So sum_{x^3=1} 1/(1 + x) = 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + [1/(1 + ω) + 1/(1 + ω²)]. As before, 1/(1 + ω) + 1/(1 + ω²) = 1, so total sum is 1/2 + 1 = 3/2, which is n/2 = 3/2. Therefore, my previous mistake was when I thought that the sum was 2. Therefore, the formula holds. Therefore, sum_{x^n=1} 1/(1 + x) = n/2. Therefore, in our problem, the sum from k=0 to n of 1/(1 + ω^k) is n/2. Therefore, the original sum S = sum_{k=1}^n 1/(1 + ω^k) = sum_{x^n=1} 1/(1 + x) - 1/(1 + 1) = n/2 - 1/2 = (n - 1)/2. Wait, but for n=3, this would be (3 - 1)/2 = 1, but earlier we saw that sum_{k=1}^3 1/(1 + ω^k) = 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2. Wait, this contradicts again. What is happening here?

Wait, no. Wait, sum_{x^n=1} 1/(1 + x) = n/2. This includes x=1, so sum_{k=0}^n 1/(1 + ω^k) = n/2. Therefore, sum_{k=1}^n 1/(1 + ω^k) = n/2 - 1/(1 + 1) = n/2 - 1/2 = (n - 1)/2. But for n=3, this would be (3 - 1)/2 = 1, but earlier calculation shows that sum_{k=1}^3 1/(1 + ω^k) = 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2. Therefore, there's a contradiction. Therefore, my mistake must be in the formula. Wait, let me check again.

Wait, the formula says sum_{x^n=1} 1/(1 + x) = n/2. For n=3, this is 3/2. But sum_{x^3=1} 1/(1 + x) = 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1 = 3/2. Therefore, that is correct. Therefore, sum_{k=0}^n 1/(1 + ω^k) = n/2. Therefore, sum_{k=1}^n 1/(1 + ω^k) = n/2 - 1/(1 + 1) = n/2 - 1/2 = (n - 1)/2. But for n=3, this would be (3 - 1)/2 = 1, but according to direct calculation, it's 3/2. Therefore, there's a contradiction. Wait, why?

Wait, because when I write sum_{k=0}^n 1/(1 + ω^k), for n=3, that is sum_{k=0}^3 1/(1 + ω^k). But ω^3 = 1, so the terms are k=0: 1/(1 + 1), k=1: 1/(1 + ω), k=2: 1/(1 + ω²), k=3: 1/(1 + ω³) = 1/(1 + 1). Therefore, sum_{k=0}^3 1/(1 + ω^k) = 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. But according to the formula, it should be n/2 = 3/2. Therefore, there's a contradiction. Therefore, my previous application of the formula was wrong. Where is the mistake?

Wait, the formula says sum_{x^n=1} 1/(1 + x) = n/2. But in reality, for n=3, the sum is 2, not 3/2. Therefore, my mistake was in the application of the formula. Let me re-examine the derivation.

Earlier, I said that sum_{x^n=1} 1/(1 + x) = -P'(-1)/P(-1). Let me recalculate this. P(x) = x^n - 1. Then P(-1) = (-1)^n - 1. For n odd, P(-1) = -1 - 1 = -2. P'(x) = n x^{n - 1}, so P'(-1) = n (-1)^{n - 1} = n (-1)^{even} = n * 1 = n. Therefore, sum_{x^n=1} 1/(1 + x) = -P'(-1)/P(-1) = -n / (-2) = n/2. But for n=3, this gives 3/2, but actual calculation gives 2. Therefore, there's a mistake in the formula. Where is the error?

Wait, the formula says sum_{i} 1/(a - r_i) = P'(a)/P(a). But in our case, we have sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} 1/(1 - (-x)) = sum_{x^n=1} 1/(a - (-x)) where a = 1. Wait, no. Wait, if I set a = -1, then sum_{x^n=1} 1/(a - x) = P'(a)/P(a). But we have sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} 1/(1 - (-x)) = sum_{x^n=1} 1/(a - x) where a = 1 and x is replaced by -x. Wait, maybe I need to adjust variables.

Let me think again. Let me denote y = -x. Then, if x^n = 1, then y^n = (-x)^n = (-1)^n x^n = (-1)^n * 1. For n odd, y^n = -1. Therefore, the roots y = -x satisfy y^n = -1. Therefore, the sum sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} 1/(1 - (-x)) = sum_{y^n=-1} 1/(1 - y), where y = -x. But this seems more complicated. Alternatively, maybe I need to use a different approach.

Alternatively, consider that sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} [1/(1 + x)]. Let me consider multiplying numerator and denominator by (1 - x), but not sure. Alternatively, consider that 1/(1 + x) = (1 - x)/(1 - x²). Therefore, sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} (1 - x)/(1 - x²). But since x ≠ 1, because if x=1, then denominator is 0. Wait, but x=1 is a root, so we have to be careful. However, for x ≠ 1, we can write 1/(1 + x) = (1 - x)/(1 - x²). Therefore, sum_{x^n=1, x ≠ 1} 1/(1 + x) = sum_{x^n=1, x ≠ 1} (1 - x)/(1 - x²). But 1 - x² = (1 - x)(1 + x), so this is sum_{x^n=1, x ≠ 1} [1/(1 + x)] = sum_{x^n=1, x ≠ 1} [ (1 - x)/(1 - x²) ] = sum_{x^n=1, x ≠ 1} [1/(1 + x)] = same thing. Not helpful.

Alternatively, consider that sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} [1/(1 + x)]. Let me consider that for each root x ≠ -1, which is true since x^n = 1 and n is odd, so -1 is not a root. Therefore, all denominators are non-zero. Let me consider that sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} [1/(1 + x)]. Let me consider that this is equal to sum_{x^n=1} [ (1 - x)/(1 - x²) ] = sum_{x^n=1} [ (1 - x)/(1 - x²) ].

But since x² ≠ 1 for x ≠ ±1, but since n is odd, x ≠ -1, so x² ≠ 1? No, for example, if x = -1, but x^n = 1, which would require (-1)^n = 1, which for odd n is -1 ≠ 1. Therefore, x ≠ -1, so x² ≠ 1? No, for example, if x is a root of x² = 1, then x = ±1, but since n is odd, x = 1 is a root, but x = -1 is not. Therefore, for x ≠ 1, x² ≠ 1? No, for example, if n=3, x=ω, then x² = ω² ≠ 1. So, for x ≠ 1, x² ≠ 1. Therefore, 1 - x² ≠ 0. Therefore, we can write sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} (1 - x)/(1 - x²). Now, note that 1 - x² = (1 - x)(1 + x), so this is sum_{x^n=1} [ (1 - x) / ( (1 - x)(1 + x) ) ] = sum_{x^n=1} 1/(1 + x), which is the same as before. Not helpful.

Alternatively, consider that sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} [1/(1 + x)]. Let me consider that this is equal to sum_{x^n=1} [ (1 - x) / (1 - x²) ] = sum_{x^n=1} [ (1 - x) / (1 - x²) ]. Now, note that 1 - x² = 1 - x², and x² is another root of unity. But not sure.

Alternatively, consider that the sum over all roots x^n=1 of 1/(1 + x) can be related to the derivative. Let me recall that sum_{x^n=1} 1/(a - x) = P'(a)/P(a), where P(x) = x^n - 1. So if I set a = -1, then sum_{x^n=1} 1/(-1 - x) = P'(-1)/P(-1). Therefore, sum_{x^n=1} 1/(1 + x) = - sum_{x^n=1} 1/(-1 - x) = -P'(-1)/P(-1). As before, for n odd, P(-1) = -2, P'(-1) = n. Therefore, sum_{x^n=1} 1/(1 + x) = -n / (-2) = n/2. But for n=3, this gives 3/2, but when I calculate manually, I get 2. What is the issue?

Wait, let me calculate for n=3. P(x) = x³ - 1. P(-1) = (-1)^3 - 1 = -1 - 1 = -2. P'(x) = 3x², so P'(-1) = 3(1) = 3. Therefore, sum_{x³=1} 1/(1 + x) = -P'(-1)/P(-1) = -3 / (-2) = 3/2. But when I calculate manually, I get 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. Wait, why is there a discrepancy? Because in the formula, sum_{x^n=1} 1/(1 + x) = n/2, but manual calculation gives 2 for n=3. Therefore, there must be a mistake in the formula. Wait, but the formula says sum_{x^n=1} 1/(a - x) = P'(a)/P(a). Let me check this formula.

Let me recall that if P(x) = ∏_{i=1}^n (x - r_i), then P'(x) = ∑_{i=1}^n ∏_{j ≠ i} (x - r_j). Therefore, P'(a)/P(a) = ∑_{i=1}^n 1/(a - r_i). Therefore, this is correct. Therefore, for a = -1, sum_{i=1}^n 1/(-1 - r_i) = P'(-1)/P(-1). Therefore, sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} 1/(1 - (-x)) = sum_{x^n=1} 1/(a - x) where a = 1 and x is replaced by -x. Wait, no. Let me clarify.

Let me define y = -x. Then, if x^n = 1, then y^n = (-x)^n = (-1)^n x^n = (-1)^n * 1. For n odd, y^n = -1. Therefore, the roots y satisfy y^n = -1. Therefore, the sum sum_{x^n=1} 1/(1 + x) = sum_{y^n=-1} 1/(1 - y). Because if y = -x, then x = -y, so 1 + x = 1 - y. Therefore, sum_{x^n=1} 1/(1 + x) = sum_{y^n=-1} 1/(1 - y). Now, let me consider the polynomial Q(y) = y^n + 1, which has roots y^n = -1. Therefore, sum_{y^n=-1} 1/(1 - y) = Q'(1)/Q(1). Let me compute this. Q(y) = y^n + 1, so Q(1) = 1 + 1 = 2. Q'(y) = n y^{n - 1}, so Q'(1) = n * 1^{n - 1} = n. Therefore, sum_{y^n=-1} 1/(1 - y) = Q'(1)/Q(1) = n/2. Therefore, sum_{x^n=1} 1/(1 + x) = n/2. But for n=3, this gives 3/2, but manual calculation gives 2. What is the issue?

Wait, for n=3, the roots y^n = -1 are the cube roots of -1, which are -1, and two other roots. Wait, but y^n = -1 for n=3 is y³ = -1, which has roots y = -1, and y = e^{iπ/3}, y = e^{i5π/3}. But the original sum is over x^n=1, which for n=3, x=1, ω, ω². So y = -x, so y = -1, -ω, -ω². Therefore, the sum sum_{x³=1} 1/(1 + x) = sum_{y³=-1} 1/(1 - y) = 1/(1 - (-1)) + 1/(1 - (-ω)) + 1/(1 - (-ω²)) = 1/2 + 1/(1 + ω) + 1/(1 + ω²). Which is exactly what we calculated before, and it equals 1/2 + 1 = 3/2. But earlier, when I thought I calculated it as 2, I was including the term for x=1 and x=3, but x=3 is the same as x=1. Wait, no. For n=3, the sum over x³=1 is x=1, ω, ω². So the sum is 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1 = 3/2. Therefore, the formula is correct. Earlier, when I thought I calculated it as 2, I must have made a mistake. Therefore, the formula holds. Therefore, sum_{x^n=1} 1/(1 + x) = n/2. Therefore, for our original problem, sum_{k=0}^n 1/(1 + ω^k) = n/2. Therefore, the sum from k=1 to n is n/2 - 1/(1 + 1) = n/2 - 1/2 = (n - 1)/2. But for n=3, this would be (3 - 1)/2 = 1, but the actual sum from k=1 to 3 is 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2. Wait, this is still a contradiction. What is going on?

Wait, no! Wait, the sum from k=0 to n of 1/(1 + ω^k) is n/2. For n=3, this is 3/2. Therefore, the sum from k=1 to n is 3/2 - 1/(1 + 1) = 3/2 - 1/2 = 1. But earlier, I calculated the sum from k=1 to 3 as 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2. Wait, this is a contradiction. Where is the mistake?

Wait, let me recheck. For n=3, the sum from k=0 to 3 of 1/(1 + ω^k) is 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + ω³). But ω³ = 1, so the last term is 1/(1 + 1) = 1/2. Therefore, the sum is 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. But according to the formula, it should be n/2 = 3/2. This is a contradiction. Therefore, there must be a mistake in the formula. What is the error?

Wait, the formula says sum_{x^n=1} 1/(1 + x) = n/2. But for n=3, this sum is 2, not 3/2. Therefore, the formula is wrong. Where is the mistake in the derivation?

Let me go back. I said that sum_{x^n=1} 1/(1 + x) = -P'(-1)/P(-1) = n/2. But for n=3, this gives 3/2, but actual calculation gives 2. Therefore, the mistake must be in the derivation. Let me re-derive it.

Let me recall that if P(x) = ∏_{i=1}^n (x - r_i), then P'(a)/P(a) = ∑_{i=1}^n 1/(a - r_i). Therefore, if I want to compute sum_{i=1}^n 1/(a - r_i), it is P'(a)/P(a). In our case, we want to compute sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} 1/(1 - (-x)) = sum_{x^n=1} 1/(a - x) where a = 1 and x is replaced by -x. Wait, no. Let me define y = -x. Then, if x^n = 1, then y^n = (-1)^n x^n = (-1)^n. For n odd, y^n = -1. Therefore, the sum becomes sum_{y^n=-1} 1/(1 - y). Let me define Q(y) = y^n + 1, which has roots y^n = -1. Then, sum_{y^n=-1} 1/(1 - y) = Q'(1)/Q(1). Q(1) = 1^n + 1 = 2. Q'(y) = n y^{n-1}, so Q'(1) = n. Therefore, sum_{y^n=-1} 1/(1 - y) = n/2. Therefore, sum_{x^n=1} 1/(1 + x) = n/2. But for n=3, this gives 3/2, but manual calculation gives 2. What is the issue?

Wait, for n=3, Q(y) = y³ + 1, which factors as (y + 1)(y² - y + 1). The roots are y = -1, y = e^{iπ/3}, y = e^{i5π/3}. Therefore, sum_{y³=-1} 1/(1 - y) = 1/(1 - (-1)) + 1/(1 - e^{iπ/3}) + 1/(1 - e^{i5π/3}). Let me compute this. First term: 1/(1 + 1) = 1/2. Second term: 1/(1 - e^{iπ/3}). Let me compute this. Let me recall that e^{iπ/3} = cos(π/3) + i sin(π/3) = 1/2 + i√3/2. Therefore, 1 - e^{iπ/3} = 1 - 1/2 - i√3/2 = 1/2 - i√3/2. The magnitude is √[(1/2)^2 + (√3/2)^2] = √[1/4 + 3/4] = √1 = 1. Therefore, 1/(1 - e^{iπ/3}) = conjugate(1 - e^{iπ/3}) / |1 - e^{iπ/3}|² = (1/2 + i√3/2)/1 = e^{iπ/3}. Similarly, 1/(1 - e^{i5π/3}) = e^{i5π/3}. Therefore, the sum is 1/2 + e^{iπ/3} + e^{i5π/3}. Now, e^{iπ/3} + e^{i5π/3} = 2 cos(π/3) = 2*(1/2) = 1. Therefore, total sum is 1/2 + 1 = 3/2. Which matches the formula. Therefore, the manual calculation was wrong earlier. Earlier, I thought that sum_{x³=1} 1/(1 + x) = 2, but it is actually 3/2. Where was the mistake?

Ah! Earlier, I thought that the sum from k=0 to 3 of 1/(1 + ω^k) is 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. But this is incorrect. Because when n=3, the sum from k=0 to n of 1/(1 + ω^k) is sum_{x³=1} 1/(1 + x) = 3/2, not 2. Therefore, my mistake was in the manual calculation. Let me recompute it.

For n=3, the roots are x=1, ω, ω². So the sum is 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1/(1 + ω) + 1/(1 + ω²). As calculated earlier, 1/(1 + ω) + 1/(1 + ω²) = 1, so total sum is 1/2 + 1 = 3/2, which matches the formula. Therefore, my previous mistake was in thinking that there was a fourth term, but for n=3, the sum from k=0 to n is over x³=1, which has 3 roots, not 4. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is n/2, and the sum from k=1 to n is n/2 - 1/2 = (n - 1)/2. But for n=3, this would be (3 - 1)/2 = 1, but the actual sum from k=1 to 3 is 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2. Wait, this is still a contradiction. What is going on?

Wait, no! For n=3, the sum from k=0 to n is sum_{k=0}^3 1/(1 + ω^k). But ω^3 = 1, so the terms are k=0: 1/(1 + 1), k=1: 1/(1 + ω), k=2: 1/(1 + ω²), k=3: 1/(1 + ω³) = 1/(1 + 1). Therefore, the sum is 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. But according to the formula, it should be n/2 = 3/2. This is a contradiction. What is the issue?

Wait, but earlier, when I calculated using the Q(y) approach, I found that sum_{x³=1} 1/(1 + x) = 3/2. But here, I'm getting 2. Where is the mistake?

Wait, the confusion is between the sum over all roots x^n=1 of 1/(1 + x) and the sum from k=0 to n of 1/(1 + ω^k). For n=3, the roots x³=1 are x=1, ω, ω². Therefore, the sum over x³=1 of 1/(1 + x) is 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1 = 3/2. But when I write the sum from k=0 to n of 1/(1 + ω^k), for n=3, it is k=0: 1/(1 + ω^0) = 1/2, k=1: 1/(1 + ω^1), k=2: 1/(1 + ω^2), k=3: 1/(1 + ω^3) = 1/(1 + 1) = 1/2. So the sum is 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. But this contradicts the previous result. Why?

Because in the formula, we are summing over all roots x^n=1, which for n=3, are x=1, ω, ω². Therefore, the sum should be 3/2, but when I include k=3, which is ω^3 = 1, it's adding an extra 1/2. Therefore, the mistake is that when I write the sum from k=0 to n of 1/(1 + ω^k), for n=3, it is actually sum_{k=0}^3 1/(1 + ω^k) = sum_{k=0}^2 1/(1 + ω^k) + 1/(1 + ω^3) = sum_{x³=1} 1/(1 + x) + 1/(1 + 1). Wait, no. Because ω^3 = 1, so the term for k=3 is 1/(1 + 1), which is already included in the sum over x³=1. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is not the same as the sum over x^n=1 of 1/(1 + x). It is the same, because for k=0 to n, ω^k cycles through the roots x^n=1, but since ω^n = 1, the term for k=n is 1/(1 + 1), which is already included in the sum over x^n=1. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1). Wait, no. For example, for n=3, the sum from k=0 to 3 of 1/(1 + ω^k) is 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + 1) = 2 * 1/2 + 1/(1 + ω) + 1/(1 + ω²) = 1 + 1 = 2. But the sum over x³=1 of 1/(1 + x) is 1/2 + 1/(1 + ω) + 1/(1 + ω²) = 3/2. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1). Wait, no, it's equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1) - 1/(1 + 1)? No, this is getting too confusing.

The key point is that when we use the formula, we are summing over all roots x^n=1, which for n=3, are x=1, ω, ω². Therefore, the sum is 3/2. But when we write the sum from k=0 to n of 1/(1 + ω^k), for n=3, it is adding an extra term 1/(1 + ω^3) = 1/2, which is already included in the sum over x³=1. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1). Wait, no, because ω^3 = 1 is already in the sum. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1) - 1/(1 + 1)? No, this is not making sense. I think the confusion is arising from the indexing. Let me clarify:

For the sum over x^n=1 of 1/(1 + x), it includes all roots x=1, ω, ω², ..., ω^{n-1}. For n=3, this is x=1, ω, ω². The sum from k=0 to n of 1/(1 + ω^k) includes k=0 to n, which for n=3 is k=0, 1, 2, 3. But ω^3 = 1, so the term for k=3 is 1/(1 + 1), which is already included in the sum over x^n=1. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1). Wait, no, because the term for k=3 is already in the sum over x^n=1. Therefore, it's double-counting. Therefore, the correct sum from k=0 to n of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1) - 1/(1 + 1) = sum over x^n=1 of 1/(1 + x). No, this is not helpful.

I think the confusion is arising from the fact that when n is the number of roots, but the sum from k=0 to n includes an extra term. However, in reality, for the roots of x^n=1, there are n roots, so k=0 to n-1. Therefore, the sum from k=0 to n-1 of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x). Therefore, in the original problem, the sum from k=1 to n of 1/(1 + ω^k) is equal to the sum from k=1 to n-1 of 1/(1 + ω^k) + 1/(1 + ω^n) = sum from k=1 to n-1 of 1/(1 + ω^k) + 1/2. And the sum from k=0 to n-1 of 1/(1 + ω^k) = n/2. Therefore, the sum from k=1 to n-1 of 1/(1 + ω^k) = n/2 - 1/(1 + 1) = n/2 - 1/2. Therefore, the sum from k=1 to n of 1/(1 + ω^k) = (n/2 - 1/2) + 1/2 = n/2. Wait, this is different from before. Let me clarify:

Let me define S_total = sum_{k=0}^{n-1} 1/(1 + ω^k) = n/2. Then, the sum from k=1 to n-1 is S_total - 1/(1 + 1) = n/2 - 1/2. Then, the sum from k=1 to n is this plus 1/(1 + ω^n) = 1/(1 + 1) = 1/2. Therefore, total sum is (n/2 - 1/2) + 1/2 = n/2. Therefore, the sum from k=1 to n of 1/(1 + ω^k) = n/2. But for n=3, this would be 3/2, which matches the manual calculation. Earlier, when I thought it was (n - 1)/2, I was mistaken. The correct result is n/2. Therefore, the answer is n/2, which for n=1997 is 1997/2.

But let me verify with n=3. Sum from k=1 to 3 of 1/(1 + ω^k) = 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2, which is n/2. Yes, this is correct. Earlier, I had a mistake in thinking that the sum from k=1 to n was (n - 1)/2, but it is actually n/2. Therefore, the correct answer is n/2, which for the given problem is 1997/2.

Therefore, the initial pairing approach was correct, and the confusion was due to miscalculating the sum from k=1 to n. Therefore, the answer is 1997/2.

**Final Answer**
\boxed{\dfrac{1997}{2}}

To evaluate the sum

$$
S = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} + \dots + \frac{1}{1 + \omega^{1997}},
$$

where $\omega^{1997} = 1$ and $\omega \neq 1$, we begin by recognizing that $\omega$ is a **primitive 1997th root of unity**. This is because 1997 is a prime number, and the only roots of $x^{1997} - 1 = 0$ are the 1997th roots of unity, which include 1 and the other 1996 roots $\omega, \omega^2, \dots, \omega^{1996}$.

---

### Step 1: Understanding the Structure of the Sum

We are summing over all the 1997th roots of unity, including $\omega^0 = 1$, and the sum is:

$$
S = \sum_{k=1}^{1997} \frac{1}{1 + \omega^k}.
$$

Note that $\omega^{1997} = 1$, so the last term is $\frac{1}{1 + \omega^{1997}} = \frac{1}{1 + 1} = \frac{1}{2}$.

---

### Step 2: Using Symmetry and Pairing

We can pair terms in the sum using the property of roots of unity:

$$
\omega^k \cdot \omega^{1997 - k} = \omega^{1997} = 1 \Rightarrow \omega^{1997 - k} = \omega^{-k}.
$$

So, for each $k = 1$ to $998$, we can pair:

$$
\frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{-k}} = \frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{1997 - k}}.
$$

Let’s compute this sum:

$$
\frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{-k}} = \frac{(1 + \omega^{-k}) + (1 + \omega^k)}{(1 + \omega^k)(1 + \omega^{-k})} = \frac{2 + \omega^k + \omega^{-k}}{2 + \omega^k + \omega^{-k}} = 1.
$$

So, each such pair sums to 1. There are $998$ such pairs (since $1996$ terms from $k = 1$ to $1996$), and the last term is $\frac{1}{2}$.

Thus, the total sum is:

$$
S = 998 \cdot 1 + \frac{1}{2} = 998 + \frac{1}{2} = \frac{1997}{2}.
$$

---

### Step 3: Verification via Complex Analysis

Alternatively, we can use a more general approach. Let $P(x) = x^{1997} - 1$, which has roots $\omega^k$ for $k = 0$ to $1996$. Then, the sum over all roots of $1/(1 + x)$ is:

$$
\sum_{k=0}^{1996} \frac{1}{1 + \omega^k}.
$$

Using the identity:

$$
\sum_{x^n = 1} \frac{1}{1 + x} = \frac{n}{2},
$$

which holds for any odd $n$, we get:

$$
\sum_{k=0}^{1996} \frac{1}{1 + \omega^k} = \frac{1997}{2}.
$$

Subtracting the term for $k = 0$, which is $\frac{1}{1 + 1} = \frac{1}{2}$, we get:

$$
\sum_{k=1}^{1996} \frac{1}{1 + \omega^k} = \frac{1997}{2} - \frac{1}{2} = \frac{1996}{2} = 998.
$$

Adding the last term $\frac{1}{2}$, we get:

$$
S = 998 + \frac{1}{2} = \frac{1997}{2}.
$$

---

### Final Answer

$$
\boxed{\dfrac{1997}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to evaluate the sum S = 1/(1 + ω) + 1/(1 + ω²) + ... + 1/(1 + ω¹⁹⁹⁷), where ω is a 1997th root of unity, not equal to 1. Hmm, let me think about how to approach this.

First, I remember that roots of unity have some nice properties. Since ω¹⁹⁹⁷ = 1 and ω ≠ 1, ω is a primitive 1997th root of unity. But maybe I don't need to worry about primitive roots here. The key thing is that ω is a root of the equation x¹⁹⁹⁷ - 1 = 0, so it satisfies that equation. Also, the sum of all the roots of xⁿ - 1 = 0 is 0, right? Because the coefficient of xⁿ⁻¹ is 0. So, for x¹⁹⁹⁷ - 1 = 0, the sum of all roots is 0. That includes 1 and all the other roots ω, ω², ..., ω¹⁹⁹⁶. So, 1 + ω + ω² + ... + ω¹⁹⁹⁶ = 0. Maybe that will come in handy later.

But the problem is about the sum of reciprocals of (1 + ωᵏ) for k from 1 to 1997. Wait, hold on, the original sum is from 1/(1 + ω) up to 1/(1 + ω¹⁹⁹⁷). But ω¹⁹⁹⁷ is 1, right? Because ω¹⁹⁹⁷ = 1. So the last term is 1/(1 + 1) = 1/2. But the original problem says ω ≠ 1, but ω¹⁹⁹⁷ is 1. So maybe the sum is from k=1 to 1997, but since ω¹⁹⁹⁷ = 1, the last term is 1/(1 + 1) = 1/2. However, the problem says "ω ≠ 1", but ω is a root of unity, so maybe the sum is over all the roots except 1? Wait, let me check the original problem again.

Wait, the problem says: "If ω¹⁹⁹⁷ = 1 and ω ≠ 1, then evaluate [sum from k=1 to 1997 of 1/(1 + ωᵏ)]". Wait, but if ω is a root of unity, then ω¹⁹⁹⁷ = 1, but the sum is from k=1 to 1997. However, if ω is a primitive root, then ωᵏ for k from 1 to 1996 are distinct roots, and ω¹⁹⁹⁷ = ω⁰ = 1. So the sum is over all the roots except 1? Wait, no. If we take k from 1 to 1997, then ω¹⁹⁹⁷ is 1, so the terms are 1/(1 + ω), 1/(1 + ω²), ..., 1/(1 + ω¹⁹⁹⁶), and 1/(1 + 1). So the sum is over all the roots except 1? Wait, no. The roots are 1, ω, ω², ..., ω¹⁹⁹⁶. So if we take k from 1 to 1997, then ω¹⁹⁹⁷ is 1, so the terms are 1/(1 + ω), ..., 1/(1 + ω¹⁹⁹⁶), and 1/(1 + 1). Therefore, the sum is over all the roots except 1? No, because the roots are 1, ω, ..., ω¹⁹⁹⁶. So the sum is over k=1 to 1997, which includes ω¹⁹⁹⁷ = 1. Therefore, the sum is over all the roots, but with k=1 to 1997, which includes 1. Wait, but the problem says ω ≠ 1. Wait, maybe there's confusion here.

Wait, hold on. The problem says "If ω¹⁹⁹⁷ = 1 and ω ≠ 1", so ω is a root of unity, not equal to 1. But then the sum is from k=1 to 1997 of 1/(1 + ωᵏ). But if ω is a root of unity, then ωᵏ cycles through the roots. However, if ω is a primitive root, then ω, ω², ..., ω¹⁹⁹⁶ are all distinct roots, and ω¹⁹⁹⁷ = 1. Therefore, the sum is over all the roots of x¹⁹⁹⁷ - 1 = 0, including 1. But the problem says ω ≠ 1, but maybe ω is just a root, not necessarily primitive? Wait, but the problem says "ω¹⁹⁹⁷ = 1 and ω ≠ 1", so ω is a non-trivial root. However, the sum is over k=1 to 1997 of 1/(1 + ωᵏ). Wait, but if we take k from 1 to 1997, then ωᵏ cycles through all the roots, including 1 when k=1997. Therefore, the sum is over all roots of x¹⁹⁹⁷ - 1 = 0, but with each term 1/(1 + ωᵏ). However, the problem says "ω ≠ 1", but maybe the sum is over all roots except 1? Wait, no. The problem says "evaluate [sum from k=1 to 1997 of 1/(1 + ωᵏ)]", so if ω is a root, then ωᵏ for k=1 to 1997 would be all roots, but since ω¹⁹⁹⁷ = 1, then ωᵏ for k=1 to 1997 is the same as ω, ω², ..., ω¹⁹⁹⁶, 1. Therefore, the sum is over all roots of x¹⁹⁹⁷ - 1 = 0. However, the problem says ω ≠ 1, but maybe ω is just a specific root? Wait, maybe the problem is written in a confusing way. Let me check again.

Wait, the problem says: "If ω¹⁹⁹⁷ = 1 and ω ≠ 1, then evaluate [sum from k=1 to 1997 of 1/(1 + ωᵏ)]". Wait, but if ω is a root of unity, then ωᵏ for k=1 to 1997 would be all the roots, but if ω is fixed, then ωᵏ cycles through the roots. Wait, but if ω is fixed, then ω¹, ω², ..., ω¹⁹⁹⁷ are the roots. However, if ω is a primitive root, then ω¹⁹⁹⁷ = 1, so ω¹⁹⁹⁷ is 1. Therefore, the sum is over all roots of x¹⁹⁹⁷ - 1 = 0. However, the problem says "ω ≠ 1", but maybe the sum is over all roots except 1? Wait, no. The problem says "evaluate [sum from k=1 to 1997 of 1/(1 + ωᵏ)]", so if ω is a root, then ωᵏ for k=1 to 1997 would be all roots, but since ω¹⁹⁹⁷ = 1, then the sum is over all roots. However, if ω is a root, then the sum is over all roots. But maybe the problem is written in a way that ω is a specific root, and the sum is over k=1 to 1997, which would be all roots. But maybe there's confusion here. Wait, perhaps the problem is written as sum_{k=1}^{1997} 1/(1 + ω^k), where ω is a primitive 1997th root of unity? But the problem says ω¹⁹⁹⁷ = 1 and ω ≠ 1, so ω is a root of unity, not necessarily primitive. However, if ω is not primitive, then the exponents would cycle through fewer roots. But since 1997 is prime? Wait, 1997 is a prime number? Let me check. 1997 divided by... Well, 1997 is actually a prime number. Yes, 1997 is a prime. So, if ω is a root of x¹⁹⁹⁷ - 1 = 0, then the minimal polynomial is cyclotomic polynomial, but since 1997 is prime, the cyclotomic polynomial is x¹⁹⁹⁶ + x¹⁹⁹⁵ + ... + x + 1. Therefore, if ω ≠ 1, then ω is a primitive 1997th root of unity. Therefore, the sum is over all roots except 1? Wait, no. If we take k from 1 to 1997, then ω^k for k=1 to 1997 would be ω, ω², ..., ω¹⁹⁹⁶, 1. Therefore, the sum is over all roots of x¹⁹⁹⁷ - 1 = 0. Therefore, the sum S is sum_{k=1}^{1997} 1/(1 + ω^k). But since ω^1997 = 1, so the last term is 1/(1 + 1) = 1/2. However, the problem says ω ≠ 1, but maybe the sum is over all roots, including 1. However, if we consider that, then maybe we can find a way to compute this sum.

Alternatively, maybe there's a trick to compute this sum. Let me think. Let me recall that for roots of unity, sometimes you can pair terms or use symmetry. For example, if we consider that ω^k and ω^{1997 - k} are related. Since ω^{1997} = 1, so ω^{1997 - k} = ω^{-k}. Therefore, maybe pairing terms 1/(1 + ω^k) and 1/(1 + ω^{-k}).

Let me try that. Let me consider pairing each term with its reciprocal. Let me note that for each k from 1 to 1996, ω^k and ω^{-k} are distinct roots (since 1997 is prime, so k ≠ -k mod 1997 unless k = 0, which is not in the range). Therefore, for each k from 1 to 1996, we can pair 1/(1 + ω^k) and 1/(1 + ω^{-k}).

Let me compute the sum of these two terms:

1/(1 + ω^k) + 1/(1 + ω^{-k}) = [ (1 + ω^{-k}) + (1 + ω^k) ] / [ (1 + ω^k)(1 + ω^{-k}) ] = [2 + ω^k + ω^{-k}]/[1 + ω^k + ω^{-k} + 1] = [2 + ω^k + ω^{-k}]/[2 + ω^k + ω^{-k}] = 1.

Wait, that's interesting! So each pair sums to 1. Therefore, if we pair all terms from k=1 to 1996, which is 1996 terms, so 998 pairs, each summing to 1, so the total sum would be 998. Then we have the last term, which is when k=1997, which is 1/(1 + ω^{1997}) = 1/(1 + 1) = 1/2. Therefore, total sum S = 998 + 1/2 = 998.5 = 1997/2. Wait, but 998 is 1996/2, so 1996/2 + 1/2 = 1997/2. Therefore, the total sum is 1997/2. But wait, that seems too straightforward. Let me check again.

Wait, let me verify the pairing. Let me take k and 1997 - k. Wait, but since ω^{1997} = 1, ω^{-k} = ω^{1997 - k}. Therefore, pairing k and 1997 - k. However, when k ranges from 1 to 1996, each pair is unique. However, if we pair k and 1997 - k, then for k from 1 to 998, and 1997 - k from 1996 down to 999. Therefore, there are 998 pairs, each contributing 1. Then the term when k=1997 is 1/2. Therefore, total sum is 998 + 1/2 = 1997/2. But 1997 is an odd number, so 1997/2 is 998.5. However, the problem is asking for the sum of 1/(1 + ω^k) from k=1 to 1997. But if that's the case, then the answer is 1997/2. However, let me check with a smaller case. Let me take n=3, which is prime. Let me check if the same logic applies.

Let me take n=3. Let ω be a primitive 3rd root of unity. Then the sum would be 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + ω³). But ω³ = 1, so the last term is 1/2. Now, let's compute the first two terms. Let me compute 1/(1 + ω) + 1/(1 + ω²). Since ω² = ω^{-1}, so pairing them. Let me compute:

1/(1 + ω) + 1/(1 + ω^{-1}) = [ (1 + ω^{-1}) + (1 + ω) ] / [ (1 + ω)(1 + ω^{-1}) ] = [2 + ω + ω^{-1}]/[2 + ω + ω^{-1}] = 1. Therefore, the sum of the first two terms is 1, and the third term is 1/2, so total sum is 1 + 1/2 = 3/2, which is n/2. So for n=3, the answer is 3/2. Similarly, for n=1997, the answer would be 1997/2. Therefore, this seems to hold. Therefore, the answer is 1997/2. But let me check another case. Let me take n=2. Wait, n=2, but 2 is not prime? Wait, n=2, ω²=1, ω ≠ 1, so ω=-1. Then the sum is 1/(1 + ω) + 1/(1 + ω²). But ω²=1, so the sum is 1/(1 + (-1)) + 1/(1 + 1) = 1/0 + 1/2. Wait, division by zero. But n=2 is not prime? Wait, but 2 is prime. However, in this case, the term 1/(1 + ω) is 1/(1 - 1) which is undefined. Therefore, maybe the formula doesn't hold for n=2? But in our problem, n=1997 is prime, so maybe there's no division by zero? Let me check for n=3, which worked. For n=3, the term 1/(1 + ω) is defined because ω ≠ -1. Since ω is a primitive 3rd root of unity, which is e^(2πi/3), which is not equal to -1. Therefore, in our problem, since 1997 is prime, and ω is a primitive root, then ω ≠ -1? Wait, is that necessarily true? For example, if n is an odd prime, then -1 is not a root of unity of order n? Wait, -1 is a root of x² - 1 = 0, so if n is odd, then -1 is not a root of xⁿ - 1 = 0. Because (-1)^n = -1 ≠ 1. Therefore, for odd n, ω ≠ -1. Therefore, in our case, since 1997 is odd, then none of the terms 1/(1 + ω^k) will have denominator zero. Therefore, the pairing works. Therefore, for n odd prime, the sum from k=1 to n of 1/(1 + ω^k) is n/2. Therefore, the answer is 1997/2. Therefore, boxed answer is \boxed{\dfrac{1997}{2}}.

But let me think again. Wait, in the case of n=3, the sum is 3/2. Let me compute it directly. Let me take ω = e^(2πi/3). Then 1 + ω + ω² = 0. Therefore, ω² = -1 - ω. Let me compute 1/(1 + ω) + 1/(1 + ω²) + 1/2. Let me compute 1/(1 + ω) + 1/(1 + ω²). Let me compute 1/(1 + ω) + 1/(1 + ω²). Let me write ω² = -1 - ω. Therefore, 1 + ω² = -ω. Therefore, 1/(1 + ω²) = 1/(-ω) = -1/ω. Similarly, 1/(1 + ω) is 1/(1 + ω). Therefore, the sum is 1/(1 + ω) - 1/ω. Let me compute this. Let me write 1/(1 + ω) - 1/ω = [ω - (1 + ω)] / [ω(1 + ω)] = (-1)/[ω(1 + ω)]. But ω(1 + ω) = ω + ω² = -1, since 1 + ω + ω² = 0. Therefore, this becomes (-1)/(-1) = 1. Therefore, the sum of the first two terms is 1, and adding the third term 1/2 gives 3/2. So that works. Therefore, the formula holds. Therefore, for n=1997, the answer is 1997/2. Therefore, I think that is the answer.

But let me think again if there's another way to approach this problem, maybe using complex analysis or generating functions or something else. Let me try another approach.

Let me consider the sum S = sum_{k=1}^{n} 1/(1 + ω^k), where ω is a primitive n-th root of unity, and n is prime. Wait, but in our problem, n=1997. Let me consider that. Let me note that 1/(1 + ω^k) can be rewritten as [1 - ω^k]/[1 - ω^{2k}]. Wait, no. Wait, 1/(1 + x) = (1 - x)/(1 - x²). So if I let x = ω^k, then 1/(1 + ω^k) = (1 - ω^k)/(1 - ω^{2k}). But maybe this is not helpful. Alternatively, maybe consider multiplying numerator and denominator by something. Alternatively, consider that sum_{k=1}^n 1/(1 + ω^k) = sum_{k=1}^n [1/(1 + ω^k)].

Alternatively, consider that for each term, 1/(1 + ω^k) = 1/(1 + ω^k) * [ (1 - ω^k)/(1 - ω^k) ] = (1 - ω^k)/(1 - ω^{2k}). But this seems complicated. Alternatively, maybe consider that sum_{k=1}^n 1/(1 + ω^k) = sum_{k=1}^n [ (1 - ω^k) / (1 - ω^{2k}) ]? Not sure.

Alternatively, think about the sum S = sum_{k=1}^n 1/(1 + ω^k). Let me note that 1 + ω^k = 0 implies ω^k = -1. But since ω is a primitive n-th root of unity, and n is odd (1997 is odd), then -1 is not a root of unity of order n. Therefore, none of the denominators are zero, so all terms are defined.

Alternatively, consider that sum_{k=1}^n 1/(1 + ω^k) = sum_{k=1}^n [1/(1 + ω^k)].

Let me consider multiplying numerator and denominator by something. Let me think about the sum over all roots. Let me consider that for each root ω^k, we have 1/(1 + ω^k). Let me consider that if I have a function f(x) = 1/(1 + x), then the sum is sum_{k=1}^n f(ω^k). But how can I compute this sum?

Alternatively, consider that the sum over all roots of unity of f(ω^k). There is a formula for sums over roots of unity. For example, if we have a function f(x) that is rational, then we can use residues or other methods. But maybe there's a simpler way.

Alternatively, consider that sum_{k=0}^n 1/(1 + ω^k) = sum_{k=0}^n 1/(1 + ω^k). Since ω^0 = 1, so this sum would be 1/(1 + 1) + sum_{k=1}^{n-1} 1/(1 + ω^k) + 1/(1 + ω^n). But ω^n = 1, so the last term is 1/2. Therefore, sum_{k=0}^n 1/(1 + ω^k) = 1/2 + sum_{k=1}^{n-1} 1/(1 + ω^k) + 1/2 = 1 + sum_{k=1}^{n-1} 1/(1 + ω^k). But our original sum S is sum_{k=1}^n 1/(1 + ω^k) = sum_{k=1}^{n-1} 1/(1 + ω^k) + 1/2. Therefore, if I can compute sum_{k=0}^n 1/(1 + ω^k), then subtract 1/(1 + ω^0) = 1/2, then I get S. But how to compute sum_{k=0}^n 1/(1 + ω^k)?

Alternatively, consider that sum_{k=0}^n 1/(1 + ω^k) = sum_{k=0}^n [1/(1 + ω^k)]. Let me note that for each root ω^k, we can pair it with ω^{-k} as before. However, when k=0, ω^0 = 1, so 1/(1 + 1) = 1/2. For k from 1 to n-1, pairing k and n - k. But since n is odd, n - k ≠ k for any k from 1 to n-1. Therefore, similar to before, each pair sums to 1, and there are (n - 1)/2 pairs. Therefore, sum_{k=0}^n 1/(1 + ω^k) = 1/2 + (n - 1)/2 * 1 = (1 + n - 1)/2 = n/2. Therefore, sum_{k=0}^n 1/(1 + ω^k) = n/2. Therefore, our original sum S = sum_{k=1}^n 1/(1 + ω^k) = sum_{k=0}^n 1/(1 + ω^k) - 1/(1 + ω^0) = n/2 - 1/2 = (n - 1)/2. Wait, but this contradicts our previous result. Wait, what's going on here?

Wait, earlier, when I considered pairing terms from k=1 to n, I thought that pairing k and n - k gives (n - 1)/2 pairs, each summing to 1, and then adding the term for k = n, which is 1/2. But according to this new approach, sum_{k=0}^n 1/(1 + ω^k) = n/2, so sum_{k=1}^n 1/(1 + ω^k) = n/2 - 1/2 = (n - 1)/2. But in the case of n=3, this would give (3 - 1)/2 = 1, but earlier we saw that the sum was 3/2. Therefore, there's a contradiction here. Therefore, I must have made a mistake in one of the reasonings.

Wait, let me check again. For n=3, sum_{k=0}^3 1/(1 + ω^k) = 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + ω³). But ω³ = 1, so that's 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2. Which is 1 + [1/(1 + ω) + 1/(1 + ω²)]. Earlier, we saw that [1/(1 + ω) + 1/(1 + ω²)] = 1, so total sum is 1 + 1 = 2. Which is n/2 = 3/2? No, 2 ≠ 3/2. Wait, so this contradicts the previous conclusion. Therefore, my previous reasoning was wrong. Therefore, there's a mistake here.

Wait, so let me recast. For n=3, sum_{k=0}^n 1/(1 + ω^k) = 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. Which is equal to n/2? No, n=3, n/2 is 1.5, but the sum is 2. Therefore, my previous assertion that sum_{k=0}^n 1/(1 + ω^k) = n/2 is wrong. Therefore, my mistake was there. Therefore, need to re-examine.

So, going back. When I considered pairing terms for k from 1 to n-1, but in the case of n=3, pairing k=1 and k=2 (since n-1=2). Then, as before, 1/(1 + ω) + 1/(1 + ω²) = 1. Then adding the term for k=0: 1/2 and k=3: 1/2. Therefore, total sum is 1 + 1/2 + 1/2 = 2. Which is equal to n. Wait, n=3, sum is 2. So maybe sum_{k=0}^n 1/(1 + ω^k) = n? But for n=3, it's 2. Not sure. Wait, maybe there's another way.

Alternatively, let me think about the sum over all roots of unity. Let me consider that for each root ω^k, we have 1/(1 + ω^k). Let me consider that the sum over all roots of unity of 1/(1 + x) where x^n = 1. Let me consider that sum_{x^n=1} 1/(1 + x). Let me note that if I let f(x) = 1/(1 + x), then the sum is sum_{x^n=1} f(x). There is a formula for such sums. Let me recall that for a polynomial P(x) with roots r_i, sum_{i} 1/(a - r_i) = P'(a)/P(a). But here, we have sum_{x^n=1} 1/(1 + x). Let me write this as sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} 1/(1 - (-x)). Therefore, if I consider the polynomial Q(x) = x^n - 1, which has roots x^n = 1. Then, the sum over roots of 1/(1 - (-x)) = sum_{x^n=1} 1/(1 - (-x)) = sum_{x^n=1} 1/(1 - (-x)).

Using the formula sum_{i} 1/(a - r_i) = P'(a)/P(a), where P(x) = product_{i} (x - r_i). In our case, P(x) = x^n - 1, and we have sum_{i} 1/(a - r_i) = P'(a)/P(a). Therefore, if we set a = -1, then sum_{x^n=1} 1/( -1 - x ) = P'(-1)/P(-1). Therefore, sum_{x^n=1} 1/( -1 - x ) = P'(-1)/P(-1). Therefore, sum_{x^n=1} 1/(1 + x) = - sum_{x^n=1} 1/( -1 - x ) = - P'(-1)/P(-1).

Let me compute this. First, P(x) = x^n - 1. Therefore, P(-1) = (-1)^n - 1. Since n=1997 is odd, P(-1) = -1 - 1 = -2. P'(x) = n x^{n - 1}, so P'(-1) = n (-1)^{n - 1} = n (-1)^{even} = n * 1 = n. Therefore, sum_{x^n=1} 1/(1 + x) = - P'(-1)/P(-1) = -n / (-2) = n/2. Therefore, the sum over all roots x^n=1 of 1/(1 + x) is n/2. Therefore, in our case, sum_{k=0}^n 1/(1 + ω^k) = n/2. Therefore, for n=3, this would be 3/2, but earlier when I calculated manually, I got 2. Wait, there's a contradiction here. What's wrong?

Wait, for n=3, sum_{x^3=1} 1/(1 + x) = 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1/(1 + ω) + 1/(1 + ω²). Earlier, I calculated that 1/(1 + ω) + 1/(1 + ω²) = 1, so total sum is 1/2 + 1 = 3/2, which matches the formula. But earlier, when I thought I calculated it as 2, I must have made a mistake. Let me check again. For n=3, the roots are 1, ω, ω². So sum_{x^3=1} 1/(1 + x) = 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + [1/(1 + ω) + 1/(1 + ω²)]. As before, 1/(1 + ω) + 1/(1 + ω²) = 1, so total sum is 1/2 + 1 = 3/2, which is n/2 = 3/2. Therefore, my previous mistake was when I thought that the sum was 2. Therefore, the formula holds. Therefore, sum_{x^n=1} 1/(1 + x) = n/2. Therefore, in our problem, the sum from k=0 to n of 1/(1 + ω^k) is n/2. Therefore, the original sum S = sum_{k=1}^n 1/(1 + ω^k) = sum_{x^n=1} 1/(1 + x) - 1/(1 + 1) = n/2 - 1/2 = (n - 1)/2. Wait, but for n=3, this would be (3 - 1)/2 = 1, but earlier we saw that sum_{k=1}^3 1/(1 + ω^k) = 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2. Wait, this contradicts again. What is happening here?

Wait, no. Wait, sum_{x^n=1} 1/(1 + x) = n/2. This includes x=1, so sum_{k=0}^n 1/(1 + ω^k) = n/2. Therefore, sum_{k=1}^n 1/(1 + ω^k) = n/2 - 1/(1 + 1) = n/2 - 1/2 = (n - 1)/2. But for n=3, this would be (3 - 1)/2 = 1, but earlier calculation shows that sum_{k=1}^3 1/(1 + ω^k) = 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2. Therefore, there's a contradiction. Therefore, my mistake must be in the formula. Wait, let me check again.

Wait, the formula says sum_{x^n=1} 1/(1 + x) = n/2. For n=3, this is 3/2. But sum_{x^3=1} 1/(1 + x) = 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1 = 3/2. Therefore, that is correct. Therefore, sum_{k=0}^n 1/(1 + ω^k) = n/2. Therefore, sum_{k=1}^n 1/(1 + ω^k) = n/2 - 1/(1 + 1) = n/2 - 1/2 = (n - 1)/2. But for n=3, this would be (3 - 1)/2 = 1, but according to direct calculation, it's 3/2. Therefore, there's a contradiction. Wait, why?

Wait, because when I write sum_{k=0}^n 1/(1 + ω^k), for n=3, that is sum_{k=0}^3 1/(1 + ω^k). But ω^3 = 1, so the terms are k=0: 1/(1 + 1), k=1: 1/(1 + ω), k=2: 1/(1 + ω²), k=3: 1/(1 + ω³) = 1/(1 + 1). Therefore, sum_{k=0}^3 1/(1 + ω^k) = 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. But according to the formula, it should be n/2 = 3/2. Therefore, there's a contradiction. Therefore, my previous application of the formula was wrong. Where is the mistake?

Wait, the formula says sum_{x^n=1} 1/(1 + x) = n/2. But in reality, for n=3, the sum is 2, not 3/2. Therefore, my mistake was in the application of the formula. Let me re-examine the derivation.

Earlier, I said that sum_{x^n=1} 1/(1 + x) = -P'(-1)/P(-1). Let me recalculate this. P(x) = x^n - 1. Then P(-1) = (-1)^n - 1. For n odd, P(-1) = -1 - 1 = -2. P'(x) = n x^{n - 1}, so P'(-1) = n (-1)^{n - 1} = n (-1)^{even} = n * 1 = n. Therefore, sum_{x^n=1} 1/(1 + x) = -P'(-1)/P(-1) = -n / (-2) = n/2. But for n=3, this gives 3/2, but actual calculation gives 2. Therefore, there's a mistake in the formula. Where is the error?

Wait, the formula says sum_{i} 1/(a - r_i) = P'(a)/P(a). But in our case, we have sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} 1/(1 - (-x)) = sum_{x^n=1} 1/(a - (-x)) where a = 1. Wait, no. Wait, if I set a = -1, then sum_{x^n=1} 1/(a - x) = P'(a)/P(a). But we have sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} 1/(1 - (-x)) = sum_{x^n=1} 1/(a - x) where a = 1 and x is replaced by -x. Wait, maybe I need to adjust variables.

Let me think again. Let me denote y = -x. Then, if x^n = 1, then y^n = (-x)^n = (-1)^n x^n = (-1)^n * 1. For n odd, y^n = -1. Therefore, the roots y = -x satisfy y^n = -1. Therefore, the sum sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} 1/(1 - (-x)) = sum_{y^n=-1} 1/(1 - y), where y = -x. But this seems more complicated. Alternatively, maybe I need to use a different approach.

Alternatively, consider that sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} [1/(1 + x)]. Let me consider multiplying numerator and denominator by (1 - x), but not sure. Alternatively, consider that 1/(1 + x) = (1 - x)/(1 - x²). Therefore, sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} (1 - x)/(1 - x²). But since x ≠ 1, because if x=1, then denominator is 0. Wait, but x=1 is a root, so we have to be careful. However, for x ≠ 1, we can write 1/(1 + x) = (1 - x)/(1 - x²). Therefore, sum_{x^n=1, x ≠ 1} 1/(1 + x) = sum_{x^n=1, x ≠ 1} (1 - x)/(1 - x²). But 1 - x² = (1 - x)(1 + x), so this is sum_{x^n=1, x ≠ 1} [1/(1 + x)] = sum_{x^n=1, x ≠ 1} [ (1 - x)/(1 - x²) ] = sum_{x^n=1, x ≠ 1} [1/(1 + x)] = same thing. Not helpful.

Alternatively, consider that sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} [1/(1 + x)]. Let me consider that for each root x ≠ -1, which is true since x^n = 1 and n is odd, so -1 is not a root. Therefore, all denominators are non-zero. Let me consider that sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} [1/(1 + x)]. Let me consider that this is equal to sum_{x^n=1} [ (1 - x)/(1 - x²) ] = sum_{x^n=1} [ (1 - x)/(1 - x²) ].

But since x² ≠ 1 for x ≠ ±1, but since n is odd, x ≠ -1, so x² ≠ 1? No, for example, if x = -1, but x^n = 1, which would require (-1)^n = 1, which for odd n is -1 ≠ 1. Therefore, x ≠ -1, so x² ≠ 1? No, for example, if x is a root of x² = 1, then x = ±1, but since n is odd, x = 1 is a root, but x = -1 is not. Therefore, for x ≠ 1, x² ≠ 1? No, for example, if n=3, x=ω, then x² = ω² ≠ 1. So, for x ≠ 1, x² ≠ 1. Therefore, 1 - x² ≠ 0. Therefore, we can write sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} (1 - x)/(1 - x²). Now, note that 1 - x² = (1 - x)(1 + x), so this is sum_{x^n=1} [ (1 - x) / ( (1 - x)(1 + x) ) ] = sum_{x^n=1} 1/(1 + x), which is the same as before. Not helpful.

Alternatively, consider that sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} [1/(1 + x)]. Let me consider that this is equal to sum_{x^n=1} [ (1 - x) / (1 - x²) ] = sum_{x^n=1} [ (1 - x) / (1 - x²) ]. Now, note that 1 - x² = 1 - x², and x² is another root of unity. But not sure.

Alternatively, consider that the sum over all roots x^n=1 of 1/(1 + x) can be related to the derivative. Let me recall that sum_{x^n=1} 1/(a - x) = P'(a)/P(a), where P(x) = x^n - 1. So if I set a = -1, then sum_{x^n=1} 1/(-1 - x) = P'(-1)/P(-1). Therefore, sum_{x^n=1} 1/(1 + x) = - sum_{x^n=1} 1/(-1 - x) = -P'(-1)/P(-1). As before, for n odd, P(-1) = -2, P'(-1) = n. Therefore, sum_{x^n=1} 1/(1 + x) = -n / (-2) = n/2. But for n=3, this gives 3/2, but when I calculate manually, I get 2. What is the issue?

Wait, let me calculate for n=3. P(x) = x³ - 1. P(-1) = (-1)^3 - 1 = -1 - 1 = -2. P'(x) = 3x², so P'(-1) = 3(1) = 3. Therefore, sum_{x³=1} 1/(1 + x) = -P'(-1)/P(-1) = -3 / (-2) = 3/2. But when I calculate manually, I get 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. Wait, why is there a discrepancy? Because in the formula, sum_{x^n=1} 1/(1 + x) = n/2, but manual calculation gives 2 for n=3. Therefore, there must be a mistake in the formula. Wait, but the formula says sum_{x^n=1} 1/(a - x) = P'(a)/P(a). Let me check this formula.

Let me recall that if P(x) = ∏_{i=1}^n (x - r_i), then P'(x) = ∑_{i=1}^n ∏_{j ≠ i} (x - r_j). Therefore, P'(a)/P(a) = ∑_{i=1}^n 1/(a - r_i). Therefore, this is correct. Therefore, for a = -1, sum_{i=1}^n 1/(-1 - r_i) = P'(-1)/P(-1). Therefore, sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} 1/(1 - (-x)) = sum_{x^n=1} 1/(a - x) where a = 1 and x is replaced by -x. Wait, no. Let me clarify.

Let me define y = -x. Then, if x^n = 1, then y^n = (-x)^n = (-1)^n x^n = (-1)^n * 1. For n odd, y^n = -1. Therefore, the roots y satisfy y^n = -1. Therefore, the sum sum_{x^n=1} 1/(1 + x) = sum_{y^n=-1} 1/(1 - y). Because if y = -x, then x = -y, so 1 + x = 1 - y. Therefore, sum_{x^n=1} 1/(1 + x) = sum_{y^n=-1} 1/(1 - y). Now, let me consider the polynomial Q(y) = y^n + 1, which has roots y^n = -1. Therefore, sum_{y^n=-1} 1/(1 - y) = Q'(1)/Q(1). Let me compute this. Q(y) = y^n + 1, so Q(1) = 1 + 1 = 2. Q'(y) = n y^{n - 1}, so Q'(1) = n * 1^{n - 1} = n. Therefore, sum_{y^n=-1} 1/(1 - y) = Q'(1)/Q(1) = n/2. Therefore, sum_{x^n=1} 1/(1 + x) = n/2. But for n=3, this gives 3/2, but manual calculation gives 2. What is the issue?

Wait, for n=3, the roots y^n = -1 are the cube roots of -1, which are -1, and two other roots. Wait, but y^n = -1 for n=3 is y³ = -1, which has roots y = -1, and y = e^{iπ/3}, y = e^{i5π/3}. But the original sum is over x^n=1, which for n=3, x=1, ω, ω². So y = -x, so y = -1, -ω, -ω². Therefore, the sum sum_{x³=1} 1/(1 + x) = sum_{y³=-1} 1/(1 - y) = 1/(1 - (-1)) + 1/(1 - (-ω)) + 1/(1 - (-ω²)) = 1/2 + 1/(1 + ω) + 1/(1 + ω²). Which is exactly what we calculated before, and it equals 1/2 + 1 = 3/2. But earlier, when I thought I calculated it as 2, I was including the term for x=1 and x=3, but x=3 is the same as x=1. Wait, no. For n=3, the sum over x³=1 is x=1, ω, ω². So the sum is 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1 = 3/2. Therefore, the formula is correct. Earlier, when I thought I calculated it as 2, I must have made a mistake. Therefore, the formula holds. Therefore, sum_{x^n=1} 1/(1 + x) = n/2. Therefore, for our original problem, sum_{k=0}^n 1/(1 + ω^k) = n/2. Therefore, the sum from k=1 to n is n/2 - 1/(1 + 1) = n/2 - 1/2 = (n - 1)/2. But for n=3, this would be (3 - 1)/2 = 1, but the actual sum from k=1 to 3 is 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2. Wait, this is still a contradiction. What is going on?

Wait, no! Wait, the sum from k=0 to n of 1/(1 + ω^k) is n/2. For n=3, this is 3/2. Therefore, the sum from k=1 to n is 3/2 - 1/(1 + 1) = 3/2 - 1/2 = 1. But earlier, I calculated the sum from k=1 to 3 as 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2. Wait, this is a contradiction. Where is the mistake?

Wait, let me recheck. For n=3, the sum from k=0 to 3 of 1/(1 + ω^k) is 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + ω³). But ω³ = 1, so the last term is 1/(1 + 1) = 1/2. Therefore, the sum is 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. But according to the formula, it should be n/2 = 3/2. This is a contradiction. Therefore, there must be a mistake in the formula. What is the error?

Wait, the formula says sum_{x^n=1} 1/(1 + x) = n/2. But for n=3, this sum is 2, not 3/2. Therefore, the formula is wrong. Where is the mistake in the derivation?

Let me go back. I said that sum_{x^n=1} 1/(1 + x) = -P'(-1)/P(-1) = n/2. But for n=3, this gives 3/2, but actual calculation gives 2. Therefore, the mistake must be in the derivation. Let me re-derive it.

Let me recall that if P(x) = ∏_{i=1}^n (x - r_i), then P'(a)/P(a) = ∑_{i=1}^n 1/(a - r_i). Therefore, if I want to compute sum_{i=1}^n 1/(a - r_i), it is P'(a)/P(a). In our case, we want to compute sum_{x^n=1} 1/(1 + x) = sum_{x^n=1} 1/(1 - (-x)) = sum_{x^n=1} 1/(a - x) where a = 1 and x is replaced by -x. Wait, no. Let me define y = -x. Then, if x^n = 1, then y^n = (-1)^n x^n = (-1)^n. For n odd, y^n = -1. Therefore, the sum becomes sum_{y^n=-1} 1/(1 - y). Let me define Q(y) = y^n + 1, which has roots y^n = -1. Then, sum_{y^n=-1} 1/(1 - y) = Q'(1)/Q(1). Q(1) = 1^n + 1 = 2. Q'(y) = n y^{n-1}, so Q'(1) = n. Therefore, sum_{y^n=-1} 1/(1 - y) = n/2. Therefore, sum_{x^n=1} 1/(1 + x) = n/2. But for n=3, this gives 3/2, but manual calculation gives 2. What is the issue?

Wait, for n=3, Q(y) = y³ + 1, which factors as (y + 1)(y² - y + 1). The roots are y = -1, y = e^{iπ/3}, y = e^{i5π/3}. Therefore, sum_{y³=-1} 1/(1 - y) = 1/(1 - (-1)) + 1/(1 - e^{iπ/3}) + 1/(1 - e^{i5π/3}). Let me compute this. First term: 1/(1 + 1) = 1/2. Second term: 1/(1 - e^{iπ/3}). Let me compute this. Let me recall that e^{iπ/3} = cos(π/3) + i sin(π/3) = 1/2 + i√3/2. Therefore, 1 - e^{iπ/3} = 1 - 1/2 - i√3/2 = 1/2 - i√3/2. The magnitude is √[(1/2)^2 + (√3/2)^2] = √[1/4 + 3/4] = √1 = 1. Therefore, 1/(1 - e^{iπ/3}) = conjugate(1 - e^{iπ/3}) / |1 - e^{iπ/3}|² = (1/2 + i√3/2)/1 = e^{iπ/3}. Similarly, 1/(1 - e^{i5π/3}) = e^{i5π/3}. Therefore, the sum is 1/2 + e^{iπ/3} + e^{i5π/3}. Now, e^{iπ/3} + e^{i5π/3} = 2 cos(π/3) = 2*(1/2) = 1. Therefore, total sum is 1/2 + 1 = 3/2. Which matches the formula. Therefore, the manual calculation was wrong earlier. Earlier, I thought that sum_{x³=1} 1/(1 + x) = 2, but it is actually 3/2. Where was the mistake?

Ah! Earlier, I thought that the sum from k=0 to 3 of 1/(1 + ω^k) is 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. But this is incorrect. Because when n=3, the sum from k=0 to n of 1/(1 + ω^k) is sum_{x³=1} 1/(1 + x) = 3/2, not 2. Therefore, my mistake was in the manual calculation. Let me recompute it.

For n=3, the roots are x=1, ω, ω². So the sum is 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1/(1 + ω) + 1/(1 + ω²). As calculated earlier, 1/(1 + ω) + 1/(1 + ω²) = 1, so total sum is 1/2 + 1 = 3/2, which matches the formula. Therefore, my previous mistake was in thinking that there was a fourth term, but for n=3, the sum from k=0 to n is over x³=1, which has 3 roots, not 4. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is n/2, and the sum from k=1 to n is n/2 - 1/2 = (n - 1)/2. But for n=3, this would be (3 - 1)/2 = 1, but the actual sum from k=1 to 3 is 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2. Wait, this is still a contradiction. What is going on?

Wait, no! For n=3, the sum from k=0 to n is sum_{k=0}^3 1/(1 + ω^k). But ω^3 = 1, so the terms are k=0: 1/(1 + 1), k=1: 1/(1 + ω), k=2: 1/(1 + ω²), k=3: 1/(1 + ω³) = 1/(1 + 1). Therefore, the sum is 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. But according to the formula, it should be n/2 = 3/2. This is a contradiction. What is the issue?

Wait, but earlier, when I calculated using the Q(y) approach, I found that sum_{x³=1} 1/(1 + x) = 3/2. But here, I'm getting 2. Where is the mistake?

Wait, the confusion is between the sum over all roots x^n=1 of 1/(1 + x) and the sum from k=0 to n of 1/(1 + ω^k). For n=3, the roots x³=1 are x=1, ω, ω². Therefore, the sum over x³=1 of 1/(1 + x) is 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1 = 3/2. But when I write the sum from k=0 to n of 1/(1 + ω^k), for n=3, it is k=0: 1/(1 + ω^0) = 1/2, k=1: 1/(1 + ω^1), k=2: 1/(1 + ω^2), k=3: 1/(1 + ω^3) = 1/(1 + 1) = 1/2. So the sum is 1/2 + 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + [1/(1 + ω) + 1/(1 + ω²)] = 1 + 1 = 2. But this contradicts the previous result. Why?

Because in the formula, we are summing over all roots x^n=1, which for n=3, are x=1, ω, ω². Therefore, the sum should be 3/2, but when I include k=3, which is ω^3 = 1, it's adding an extra 1/2. Therefore, the mistake is that when I write the sum from k=0 to n of 1/(1 + ω^k), for n=3, it is actually sum_{k=0}^3 1/(1 + ω^k) = sum_{k=0}^2 1/(1 + ω^k) + 1/(1 + ω^3) = sum_{x³=1} 1/(1 + x) + 1/(1 + 1). Wait, no. Because ω^3 = 1, so the term for k=3 is 1/(1 + 1), which is already included in the sum over x³=1. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is not the same as the sum over x^n=1 of 1/(1 + x). It is the same, because for k=0 to n, ω^k cycles through the roots x^n=1, but since ω^n = 1, the term for k=n is 1/(1 + 1), which is already included in the sum over x^n=1. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1). Wait, no. For example, for n=3, the sum from k=0 to 3 of 1/(1 + ω^k) is 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + 1) = 2 * 1/2 + 1/(1 + ω) + 1/(1 + ω²) = 1 + 1 = 2. But the sum over x³=1 of 1/(1 + x) is 1/2 + 1/(1 + ω) + 1/(1 + ω²) = 3/2. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1). Wait, no, it's equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1) - 1/(1 + 1)? No, this is getting too confusing.

The key point is that when we use the formula, we are summing over all roots x^n=1, which for n=3, are x=1, ω, ω². Therefore, the sum is 3/2. But when we write the sum from k=0 to n of 1/(1 + ω^k), for n=3, it is adding an extra term 1/(1 + ω^3) = 1/2, which is already included in the sum over x³=1. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1). Wait, no, because ω^3 = 1 is already in the sum. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1) - 1/(1 + 1)? No, this is not making sense. I think the confusion is arising from the indexing. Let me clarify:

For the sum over x^n=1 of 1/(1 + x), it includes all roots x=1, ω, ω², ..., ω^{n-1}. For n=3, this is x=1, ω, ω². The sum from k=0 to n of 1/(1 + ω^k) includes k=0 to n, which for n=3 is k=0, 1, 2, 3. But ω^3 = 1, so the term for k=3 is 1/(1 + 1), which is already included in the sum over x^n=1. Therefore, the sum from k=0 to n of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1). Wait, no, because the term for k=3 is already in the sum over x^n=1. Therefore, it's double-counting. Therefore, the correct sum from k=0 to n of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x) + 1/(1 + 1) - 1/(1 + 1) = sum over x^n=1 of 1/(1 + x). No, this is not helpful.

I think the confusion is arising from the fact that when n is the number of roots, but the sum from k=0 to n includes an extra term. However, in reality, for the roots of x^n=1, there are n roots, so k=0 to n-1. Therefore, the sum from k=0 to n-1 of 1/(1 + ω^k) is equal to the sum over x^n=1 of 1/(1 + x). Therefore, in the original problem, the sum from k=1 to n of 1/(1 + ω^k) is equal to the sum from k=1 to n-1 of 1/(1 + ω^k) + 1/(1 + ω^n) = sum from k=1 to n-1 of 1/(1 + ω^k) + 1/2. And the sum from k=0 to n-1 of 1/(1 + ω^k) = n/2. Therefore, the sum from k=1 to n-1 of 1/(1 + ω^k) = n/2 - 1/(1 + 1) = n/2 - 1/2. Therefore, the sum from k=1 to n of 1/(1 + ω^k) = (n/2 - 1/2) + 1/2 = n/2. Wait, this is different from before. Let me clarify:

Let me define S_total = sum_{k=0}^{n-1} 1/(1 + ω^k) = n/2. Then, the sum from k=1 to n-1 is S_total - 1/(1 + 1) = n/2 - 1/2. Then, the sum from k=1 to n is this plus 1/(1 + ω^n) = 1/(1 + 1) = 1/2. Therefore, total sum is (n/2 - 1/2) + 1/2 = n/2. Therefore, the sum from k=1 to n of 1/(1 + ω^k) = n/2. But for n=3, this would be 3/2, which matches the manual calculation. Earlier, when I thought it was (n - 1)/2, I was mistaken. The correct result is n/2. Therefore, the answer is n/2, which for n=1997 is 1997/2.

But let me verify with n=3. Sum from k=1 to 3 of 1/(1 + ω^k) = 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2, which is n/2. Yes, this is correct. Earlier, I had a mistake in thinking that the sum from k=1 to n was (n - 1)/2, but it is actually n/2. Therefore, the correct answer is n/2, which for the given problem is 1997/2.

Therefore, the initial pairing approach was correct, and the confusion was due to miscalculating the sum from k=1 to n. Therefore, the answer is 1997/2.

**Final Answer**
\boxed{\dfrac{1997}{2}}

To evaluate the sum

$$
S = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} + \dots + \frac{1}{1 + \omega^{1997}},
$$

where $\omega^{1997} = 1$ and $\omega \neq 1$, we begin by recognizing that $\omega$ is a **primitive 1997th root of unity**. This is because 1997 is a prime number, and the only roots of $x^{1997} - 1 = 0$ are the 1997th roots of unity, which include 1 and the other 1996 roots $\omega, \omega^2, \dots, \omega^{1996}$.

---

### Step 1: Understanding the Structure of the Sum

We are summing over all the 1997th roots of unity, including $\omega^0 = 1$, and the sum is:

$$
S = \sum_{k=1}^{1997} \frac{1}{1 + \omega^k}.
$$

Note that $\omega^{1997} = 1$, so the last term is $\frac{1}{1 + \omega^{1997}} = \frac{1}{1 + 1} = \frac{1}{2}$.

---

### Step 2: Using Symmetry and Pairing

We can pair terms in the sum using the property of roots of unity:

$$
\omega^k \cdot \omega^{1997 - k} = \omega^{1997} = 1 \Rightarrow \omega^{1997 - k} = \omega^{-k}.
$$

So, for each $k = 1$ to $998$, we can pair:

$$
\frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{-k}} = \frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{1997 - k}}.
$$

Let’s compute this sum:

$$
\frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{-k}} = \frac{(1 + \omega^{-k}) + (1 + \omega^k)}{(1 + \omega^k)(1 + \omega^{-k})} = \frac{2 + \omega^k + \omega^{-k}}{2 + \omega^k + \omega^{-k}} = 1.
$$

So, each such pair sums to 1. There are $998$ such pairs (since $1996$ terms from $k = 1$ to $1996$), and the last term is $\frac{1}{2}$.

Thus, the total sum is:

$$
S = 998 \cdot 1 + \frac{1}{2} = 998 + \frac{1}{2} = \frac{1997}{2}.
$$

---

### Step 3: Verification via Complex Analysis

Alternatively, we can use a more general approach. Let $P(x) = x^{1997} - 1$, which has roots $\omega^k$ for $k = 0$ to $1996$. Then, the sum over all roots of $1/(1 + x)$ is:

$$
\sum_{k=0}^{1996} \frac{1}{1 + \omega^k}.
$$

Using the identity:

$$
\sum_{x^n = 1} \frac{1}{1 + x} = \frac{n}{2},
$$

which holds for any odd $n$, we get:

$$
\sum_{k=0}^{1996} \frac{1}{1 + \omega^k} = \frac{1997}{2}.
$$

Subtracting the term for $k = 0$, which is $\frac{1}{1 + 1} = \frac{1}{2}$, we get:

$$
\sum_{k=1}^{1996} \frac{1}{1 + \omega^k} = \frac{1997}{2} - \frac{1}{2} = \frac{1996}{2} = 998.
$$

Adding the last term $\frac{1}{2}$, we get:

$$
S = 998 + \frac{1}{2} = \frac{1997}{2}.
$$

---

### Final Answer

$$
\boxed{\dfrac{1997}{2}}
$$

Solution:
Note that
\begin{align*}
\frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{1997 - k}} &= \frac{1}{1 + \omega^k} + \frac{\omega^k}{\omega^k + \omega^{1997}} \\
&= \frac{1}{1 + \omega^k} + \frac{\omega^k}{\omega^k + 1} \\
&= \frac{1 + \omega^k}{1 + \omega^k} = 1.
\end{align*}Thus, we can pair the terms
\[\frac{1}{1 + \omega}, \ \frac{1}{1 + \omega^2}, \ \dots, \ \frac{1}{1 + \omega^{1995}}, \ \frac{1}{1 + \omega^{1996}}\]into $1996/2 = 998$ pairs, so that the sum of the numbers in each pair is 1.  Also, $\frac{1}{1 + \omega^{1997}} = \frac{1}{2},$ so the sum works out to $998 + \frac{1}{2} = \boxed{\frac{1997}{2}}.$


Explanation:
Okay, so I need to figure out the real values of x that are not in the domain of the function f(x) = 1 divided by (|x² + 3x - 4| + |x² + 9x + 20|). Hmm, the domain of a function is all the real numbers for which the function is defined. Since this function is a fraction, the denominator can't be zero. So, the values of x that make the denominator zero are not in the domain. Therefore, I need to find all real numbers x where |x² + 3x - 4| + |x² + 9x + 20| equals zero. But wait, the sum of two absolute values can't be zero unless both absolute values are zero individually. Because absolute values are always non-negative, right? So if their sum is zero, each must be zero. Therefore, I need to solve the system:

|x² + 3x - 4| = 0 and |x² + 9x + 20| = 0.

Which implies:

x² + 3x - 4 = 0 and x² + 9x + 20 = 0.

So, solving these two quadratic equations separately and finding their common solutions. Let me start with the first equation: x² + 3x - 4 = 0.

Using the quadratic formula, x = [-b ± √(b² - 4ac)] / 2a. Here, a = 1, b = 3, c = -4. So discriminant D = 3² - 4*1*(-4) = 9 + 16 = 25. So sqrt(25) = 5. Therefore, x = [-3 ± 5]/2. That gives two solutions: (-3 + 5)/2 = 2/2 = 1, and (-3 -5)/2 = -8/2 = -4. So the solutions for the first equation are x = 1 and x = -4.

Now, moving on to the second equation: x² + 9x + 20 = 0. Again, quadratic formula. Here, a = 1, b = 9, c = 20. Discriminant D = 9² - 4*1*20 = 81 - 80 = 1. So sqrt(1) = 1. Therefore, x = [-9 ± 1]/2. That gives two solutions: (-9 + 1)/2 = -8/2 = -4, and (-9 -1)/2 = -10/2 = -5. So the solutions for the second equation are x = -4 and x = -5.

So, the first equation has solutions x = 1 and x = -4, the second equation has solutions x = -4 and x = -5. Therefore, the only common solution between the two equations is x = -4. Therefore, the denominator is zero only when x = -4. Therefore, the domain of f(x) is all real numbers except x = -4. Therefore, the real value not in the domain is x = -4. Wait, but let me check if that's correct. Let me verify.

Wait, but maybe I made a mistake here. Because the denominator is |x² + 3x - 4| + |x² + 9x + 20|. So, if I want the denominator to be zero, both absolute values must be zero. Therefore, solving both equations. So, the only x that satisfies both equations is x = -4. Therefore, only x = -4 makes the denominator zero. Therefore, the domain is all real numbers except x = -4. Therefore, the answer is x = -4. But let me check if there are other possibilities. For example, maybe there are some x where one absolute value is positive and the other is negative? But no, absolute values are always non-negative. Therefore, their sum can only be zero if both are zero. Therefore, only x = -4. Therefore, the answer is x = -4. But let me check if x = -4 actually makes both expressions zero.

Let me plug x = -4 into the first quadratic: (-4)^2 + 3*(-4) - 4 = 16 - 12 - 4 = 0. Correct. Then the second quadratic: (-4)^2 + 9*(-4) + 20 = 16 - 36 + 20 = 0. Correct. So yes, x = -4 is the only value where denominator is zero. Therefore, the domain is all real numbers except x = -4. Therefore, the real value not in the domain is x = -4. Therefore, the answer is \boxed{-4}.

Wait, but let me think again. Maybe there are other points where the denominator is zero? For example, suppose that one of the absolute values is negative? But no, absolute values are always non-negative. Therefore, the sum can't be zero unless both are zero. Therefore, only x = -4. Therefore, the answer is x = -4. But let me check if there are other x where maybe one of the quadratics is negative? Wait, no. The absolute value makes them non-negative. Therefore, the sum is zero only when both are zero. Therefore, only x = -4. Therefore, the answer is x = -4. Therefore, the real value not in the domain is -4. Therefore, the answer is \boxed{-4}.

But wait, let me check if maybe there are other solutions where the quadratics inside the absolute values are zero. For example, maybe x = 1? Let me check. If x = 1, then the first quadratic is zero, but the second quadratic is 1² + 9*1 + 20 = 1 + 9 + 20 = 30, which is not zero. Therefore, the denominator would be |0| + |30| = 30, which is not zero. Similarly, x = -5: first quadratic is (-5)^2 + 3*(-5) - 4 = 25 - 15 - 4 = 6, which is not zero. The second quadratic is zero. Therefore, denominator is |6| + |0| = 6, which is not zero. Therefore, only x = -4 makes both zero. Therefore, only x = -4 is excluded. Therefore, the answer is \boxed{-4}.

But wait, let me think again. Maybe there's a mistake in assuming that the sum of absolute values can only be zero if both are zero. Let me think. Suppose that |A| + |B| = 0. Since |A| ≥ 0 and |B| ≥ 0, their sum is zero only if |A| = 0 and |B| = 0. Therefore, A = 0 and B = 0. Therefore, yes, only x = -4. Therefore, the answer is x = -4. Therefore, the real value not in the domain is -4. Therefore, the answer is \boxed{-4}.

But let me check if there are other points where maybe the expressions inside the absolute values are negative? Wait, but even if they are negative, the absolute value makes them positive. Therefore, the sum is still positive. Therefore, the denominator is always positive except when both expressions are zero. Therefore, only x = -4. Therefore, the answer is \boxed{-4}.

But let me check if maybe there are other x where the expressions inside the absolute values are zero. For example, the first quadratic x² + 3x - 4 factors as (x + 4)(x - 1). So roots at x = -4 and x = 1. The second quadratic x² + 9x + 20 factors as (x + 4)(x + 5). So roots at x = -4 and x = -5. Therefore, only x = -4 is common. Therefore, only x = -4. Therefore, the answer is \boxed{-4}.

Therefore, I think that's the answer. But let me check once more. Suppose someone might think that maybe the denominator can be zero if one of the absolute values is negative? But no, absolute value is always non-negative. Therefore, the sum can't be zero unless both are zero. Therefore, only x = -4. Therefore, the answer is \boxed{-4}.

**Final Answer**
The real value not in the domain of $ f(x) $ is \boxed{-4}.

To determine the real values of $ x $ that are **not in the domain** of the function

$$
f(x) = \frac{1}{|x^2 + 3x - 4| + |x^2 + 9x + 20|},
$$

we must identify the values of $ x $ for which the **denominator becomes zero**, since division by zero is undefined.

---

### Step 1: Understand the Denominator

The denominator is:

$$
|x^2 + 3x - 4| + |x^2 + 9x + 20|.
$$

Since **absolute values are always non-negative**, the sum of two absolute values is **zero only if both absolute values are zero**. That is:

$$
|x^2 + 3x - 4| = 0 \quad \text{and} \quad |x^2 + 9x + 20| = 0.
$$

This implies:

$$
x^2 + 3x - 4 = 0 \quad \text{and} \quad x^2 + 9x + 20 = 0.
$$

---

### Step 2: Solve the Quadratic Equations

#### First Equation: $ x^2 + 3x - 4 = 0 $

Factor:

$$
x^2 + 3x - 4 = (x + 4)(x - 1) = 0 \Rightarrow x = -4 \text{ or } x = 1.
$$

#### Second Equation: $ x^2 + 9x + 20 = 0 $

Factor:

$$
x^2 + 9x + 20 = (x + 4)(x + 5) = 0 \Rightarrow x = -4 \text{ or } x = -5.
$$

---

### Step 3: Find Common Solutions

We are looking for values of $ x $ that satisfy **both** equations simultaneously. The only common solution is:

$$
x = -4.
$$

---

### Step 4: Verify the Denominator is Zero at $ x = -4 $

Check both expressions:

- $ x^2 + 3x - 4 = (-4)^2 + 3(-4) - 4 = 16 - 12 - 4 = 0 $
- $ x^2 + 9x + 20 = (-4)^2 + 9(-4) + 20 = 16 - 36 + 20 = 0 $

So, the denominator becomes:

$$
|0| + |0| = 0,
$$

which makes the function undefined at $ x = -4 $.

---

### Step 5: Conclusion

The only real value of $ x $ that makes the denominator zero is $ x = -4 $. Therefore, this value is **not in the domain** of the function.

---

### Final Answer

$$
\boxed{-4}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the real values of x that are not in the domain of the function f(x) = 1 divided by (|x² + 3x - 4| + |x² + 9x + 20|). Hmm, the domain of a function is all the real numbers for which the function is defined. Since this function is a fraction, the denominator can't be zero. So, the values of x that make the denominator zero are not in the domain. Therefore, I need to find all real numbers x where |x² + 3x - 4| + |x² + 9x + 20| equals zero. But wait, the sum of two absolute values can't be zero unless both absolute values are zero individually. Because absolute values are always non-negative, right? So if their sum is zero, each must be zero. Therefore, I need to solve the system:

|x² + 3x - 4| = 0 and |x² + 9x + 20| = 0.

Which implies:

x² + 3x - 4 = 0 and x² + 9x + 20 = 0.

So, solving these two quadratic equations separately and finding their common solutions. Let me start with the first equation: x² + 3x - 4 = 0.

Using the quadratic formula, x = [-b ± √(b² - 4ac)] / 2a. Here, a = 1, b = 3, c = -4. So discriminant D = 3² - 4*1*(-4) = 9 + 16 = 25. So sqrt(25) = 5. Therefore, x = [-3 ± 5]/2. That gives two solutions: (-3 + 5)/2 = 2/2 = 1, and (-3 -5)/2 = -8/2 = -4. So the solutions for the first equation are x = 1 and x = -4.

Now, moving on to the second equation: x² + 9x + 20 = 0. Again, quadratic formula. Here, a = 1, b = 9, c = 20. Discriminant D = 9² - 4*1*20 = 81 - 80 = 1. So sqrt(1) = 1. Therefore, x = [-9 ± 1]/2. That gives two solutions: (-9 + 1)/2 = -8/2 = -4, and (-9 -1)/2 = -10/2 = -5. So the solutions for the second equation are x = -4 and x = -5.

So, the first equation has solutions x = 1 and x = -4, the second equation has solutions x = -4 and x = -5. Therefore, the only common solution between the two equations is x = -4. Therefore, the denominator is zero only when x = -4. Therefore, the domain of f(x) is all real numbers except x = -4. Therefore, the real value not in the domain is x = -4. Wait, but let me check if that's correct. Let me verify.

Wait, but maybe I made a mistake here. Because the denominator is |x² + 3x - 4| + |x² + 9x + 20|. So, if I want the denominator to be zero, both absolute values must be zero. Therefore, solving both equations. So, the only x that satisfies both equations is x = -4. Therefore, only x = -4 makes the denominator zero. Therefore, the domain is all real numbers except x = -4. Therefore, the answer is x = -4. But let me check if there are other possibilities. For example, maybe there are some x where one absolute value is positive and the other is negative? But no, absolute values are always non-negative. Therefore, their sum can only be zero if both are zero. Therefore, only x = -4. Therefore, the answer is x = -4. But let me check if x = -4 actually makes both expressions zero.

Let me plug x = -4 into the first quadratic: (-4)^2 + 3*(-4) - 4 = 16 - 12 - 4 = 0. Correct. Then the second quadratic: (-4)^2 + 9*(-4) + 20 = 16 - 36 + 20 = 0. Correct. So yes, x = -4 is the only value where denominator is zero. Therefore, the domain is all real numbers except x = -4. Therefore, the real value not in the domain is x = -4. Therefore, the answer is \boxed{-4}.

Wait, but let me think again. Maybe there are other points where the denominator is zero? For example, suppose that one of the absolute values is negative? But no, absolute values are always non-negative. Therefore, the sum can't be zero unless both are zero. Therefore, only x = -4. Therefore, the answer is x = -4. But let me check if there are other x where maybe one of the quadratics is negative? Wait, no. The absolute value makes them non-negative. Therefore, the sum is zero only when both are zero. Therefore, only x = -4. Therefore, the answer is x = -4. Therefore, the real value not in the domain is -4. Therefore, the answer is \boxed{-4}.

But wait, let me check if maybe there are other solutions where the quadratics inside the absolute values are zero. For example, maybe x = 1? Let me check. If x = 1, then the first quadratic is zero, but the second quadratic is 1² + 9*1 + 20 = 1 + 9 + 20 = 30, which is not zero. Therefore, the denominator would be |0| + |30| = 30, which is not zero. Similarly, x = -5: first quadratic is (-5)^2 + 3*(-5) - 4 = 25 - 15 - 4 = 6, which is not zero. The second quadratic is zero. Therefore, denominator is |6| + |0| = 6, which is not zero. Therefore, only x = -4 makes both zero. Therefore, only x = -4 is excluded. Therefore, the answer is \boxed{-4}.

But wait, let me think again. Maybe there's a mistake in assuming that the sum of absolute values can only be zero if both are zero. Let me think. Suppose that |A| + |B| = 0. Since |A| ≥ 0 and |B| ≥ 0, their sum is zero only if |A| = 0 and |B| = 0. Therefore, A = 0 and B = 0. Therefore, yes, only x = -4. Therefore, the answer is x = -4. Therefore, the real value not in the domain is -4. Therefore, the answer is \boxed{-4}.

But let me check if there are other points where maybe the expressions inside the absolute values are negative? Wait, but even if they are negative, the absolute value makes them positive. Therefore, the sum is still positive. Therefore, the denominator is always positive except when both expressions are zero. Therefore, only x = -4. Therefore, the answer is \boxed{-4}.

But let me check if maybe there are other x where the expressions inside the absolute values are zero. For example, the first quadratic x² + 3x - 4 factors as (x + 4)(x - 1). So roots at x = -4 and x = 1. The second quadratic x² + 9x + 20 factors as (x + 4)(x + 5). So roots at x = -4 and x = -5. Therefore, only x = -4 is common. Therefore, only x = -4. Therefore, the answer is \boxed{-4}.

Therefore, I think that's the answer. But let me check once more. Suppose someone might think that maybe the denominator can be zero if one of the absolute values is negative? But no, absolute value is always non-negative. Therefore, the sum can't be zero unless both are zero. Therefore, only x = -4. Therefore, the answer is \boxed{-4}.

**Final Answer**
The real value not in the domain of $ f(x) $ is \boxed{-4}.

To determine the real values of $ x $ that are **not in the domain** of the function

$$
f(x) = \frac{1}{|x^2 + 3x - 4| + |x^2 + 9x + 20|},
$$

we must identify the values of $ x $ for which the **denominator becomes zero**, since division by zero is undefined.

---

### Step 1: Understand the Denominator

The denominator is:

$$
|x^2 + 3x - 4| + |x^2 + 9x + 20|.
$$

Since **absolute values are always non-negative**, the sum of two absolute values is **zero only if both absolute values are zero**. That is:

$$
|x^2 + 3x - 4| = 0 \quad \text{and} \quad |x^2 + 9x + 20| = 0.
$$

This implies:

$$
x^2 + 3x - 4 = 0 \quad \text{and} \quad x^2 + 9x + 20 = 0.
$$

---

### Step 2: Solve the Quadratic Equations

#### First Equation: $ x^2 + 3x - 4 = 0 $

Factor:

$$
x^2 + 3x - 4 = (x + 4)(x - 1) = 0 \Rightarrow x = -4 \text{ or } x = 1.
$$

#### Second Equation: $ x^2 + 9x + 20 = 0 $

Factor:

$$
x^2 + 9x + 20 = (x + 4)(x + 5) = 0 \Rightarrow x = -4 \text{ or } x = -5.
$$

---

### Step 3: Find Common Solutions

We are looking for values of $ x $ that satisfy **both** equations simultaneously. The only common solution is:

$$
x = -4.
$$

---

### Step 4: Verify the Denominator is Zero at $ x = -4 $

Check both expressions:

- $ x^2 + 3x - 4 = (-4)^2 + 3(-4) - 4 = 16 - 12 - 4 = 0 $
- $ x^2 + 9x + 20 = (-4)^2 + 9(-4) + 20 = 16 - 36 + 20 = 0 $

So, the denominator becomes:

$$
|0| + |0| = 0,
$$

which makes the function undefined at $ x = -4 $.

---

### Step 5: Conclusion

The only real value of $ x $ that makes the denominator zero is $ x = -4 $. Therefore, this value is **not in the domain** of the function.

---

### Final Answer

$$
\boxed{-4}
$$

Solution:
$x$ is not in the domain of $f$ if the denominator is zero.  Since both absolute values are nonnegative, both must be zero for the denominator to be zero.  So

\begin{align*}
0=x^2+3x-4=(x+4)(x-1)&\Rightarrow x=-4\text{ or }x=1\\
0=x^2+9x+20=(x+4)(x+5)&\Rightarrow x=-4\text{ or }x=-5
\end{align*}

The only value of $x$ which makes both absolute values zero is $x=\boxed{-4}$.