Explanation:
Okay, so I need to find a way to express the double sum $\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3}$ in terms of $p$ and $q$, where $p = \sum_{k = 1}^\infty \frac{1}{k^2}$ and $q = \sum_{k = 1}^\infty \frac{1}{k^3}$. Hmm, let me think about how to approach this.

First, I know that $p$ is the famous Basel problem, which converges to $\frac{\pi^2}{6}$, and $q$ is the Riemann zeta function at 3, which doesn't have a simple closed-form expression like $p$, but maybe I don't need to know the exact values since the problem just wants it in terms of $p$ and $q$.

The double sum given is over all positive integers $j$ and $k$, and each term is $\frac{1}{(j + k)^3}$. I need to relate this to $p$ and $q$. Let me try to manipulate the double sum.

One idea is to change variables. Let me consider substituting $n = j + k$. For each fixed $n$, the number of pairs $(j, k)$ such that $j + k = n$ is $n - 1$, since $j$ can range from 1 to $n - 1$ and $k$ would be $n - j$. However, in our case, both $j$ and $k$ start from 1, so for each $n \geq 2$, the number of pairs is $n - 1$. Therefore, maybe I can rewrite the double sum as a single sum over $n$:

$$
\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^3} = \sum_{n=2}^\infty \frac{1}{n^3} \sum_{\substack{j=1 \\ j + k = n}}^\infty 1
$$

Wait, but the inner sum is over all $j$ and $k$ such that $j + k = n$. Since $j$ and $k$ are positive integers, for each $n \geq 2$, the number of such pairs is $n - 1$. Therefore, the double sum becomes:

$$
\sum_{n=2}^\infty \frac{n - 1}{n^3} = \sum_{n=2}^\infty \frac{n - 1}{n^3} = \sum_{n=2}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right)
$$

Wait, let me check that. If I split $\frac{n - 1}{n^3}$ into $\frac{n}{n^3} - \frac{1}{n^3} = \frac{1}{n^2} - \frac{1}{n^3}$. Yes, that's correct. Therefore, the entire sum becomes:

$$
\sum_{n=2}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right) = \left( \sum_{n=2}^\infty \frac{1}{n^2} \right) - \left( \sum_{n=2}^\infty \frac{1}{n^3} \right)
$$

But we know that $p = \sum_{n=1}^\infty \frac{1}{n^2}$, so $\sum_{n=2}^\infty \frac{1}{n^2} = p - 1$. Similarly, $\sum_{n=2}^\infty \frac{1}{n^3} = q - 1$. Therefore, substituting back:

$$
(p - 1) - (q - 1) = p - 1 - q + 1 = p - q
$$

Wait, so does that mean the double sum is equal to $p - q$? But let me check if I did everything correctly. Let me verify with small values.

Wait, let me think again. The original double sum is over all $j$ and $k$ starting at 1. When I change variables to $n = j + k$, then for each $n \geq 2$, the number of pairs is $n - 1$. Therefore, the sum becomes $\sum_{n=2}^\infty \frac{n - 1}{n^3}$. Then splitting into two sums:

$$
\sum_{n=2}^\infty \frac{1}{n^2} - \sum_{n=2}^\infty \frac{1}{n^3} = (p - 1) - (q - 1) = p - q
$$

So according to this, the answer is $p - q$. But wait, let me check with an example. Let me compute partial sums numerically to see if this makes sense.

Take the original double sum. Let me compute the first few terms. For j and k from 1 to 2:

When j=1, k=1: 1/(1+1)^3 = 1/8

j=1, k=2: 1/(1+2)^3 = 1/27

j=2, k=1: 1/(2+1)^3 = 1/27

j=2, k=2: 1/(2+2)^3 = 1/64

So total for n=2: 1/8 + 1/27 + 1/27 + 1/64? Wait no, wait. Wait, when n=2, j + k = 2, so only (1,1). Then n=3: (1,2) and (2,1). n=4: (1,3), (2,2), (3,1). So for n=2, the number of terms is 1, n=3: 2, n=4: 3, etc. So the sum over n=2 to infinity of (n-1)/n^3. So for n=2, term is 1/8, n=3: 2/27, n=4: 3/64, etc. So the total sum is 1/8 + 2/27 + 3/64 + ...

Now, if I compute p - q, where p is pi^2/6 ≈ 1.6449, and q is zeta(3) ≈ 1.2020569. So p - q ≈ 0.4428. Let me compute the partial sum up to n=4:

1/8 ≈ 0.125

+ 2/27 ≈ 0.07407 → total ≈ 0.19907

+ 3/64 ≈ 0.046875 → total ≈ 0.2459

+ 4/125 = 0.032 → total ≈ 0.2779

+ 5/216 ≈ 0.0231 → total ≈ 0.301

+ 6/343 ≈ 0.0175 → total ≈ 0.3185

Continuing this, it seems like the sum is approaching something around 0.44... which is p - q ≈ 0.4428. So that seems plausible. Therefore, maybe my initial conclusion is correct?

But wait, let me check again. Wait, the original problem is to write the double sum in terms of p and q. According to my calculation, it's p - q. But let me think again if there's a mistake in the reasoning.

Wait, the original double sum is over all j and k ≥ 1. When changing variables to n = j + k, then for each n ≥ 2, the number of pairs (j, k) with j + k = n is n - 1. Therefore, the double sum is equal to sum_{n=2}^\infty (n - 1)/n^3. Then splitting into sum_{n=2}^\infty 1/n^2 - sum_{n=2}^\infty 1/n^3. Which is (p - 1) - (q - 1) = p - q. So that seems correct.

But wait, let me check another way. Let me consider that the original double sum can be written as sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. Let me try to switch the order of summation or find another representation.

Alternatively, maybe I can consider that for each j and k, j + k is the same as m, but I think that's what I did before.

Alternatively, maybe I can use integral representations? For example, since 1/n^3 can be represented as an integral. But maybe that complicates things. Alternatively, think about generating functions?

Alternatively, think about symmetry. Let me note that the double sum is symmetric in j and k, so maybe I can write it as 2 times the sum over j ≥ 1, k ≥ 1, j < k, plus the diagonal terms where j = k. Wait, but maybe that complicates.

Alternatively, think of the double sum as sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. Let me make substitution m = j + k. But that's similar to before.

Alternatively, think of the sum as sum_{m=2}^\infty (number of ways to write m as j + k) * 1/m^3. Which is exactly what I did before. So that gives the same result.

Alternatively, maybe I can check with known values. For example, if I consider the sum over j and k of 1/(j + k)^s, which is known? Maybe there's a formula for that?

Alternatively, think of the double sum as sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. Let me fix j and sum over k. For fixed j, sum_{k=1}^\infty 1/(j + k)^3 = sum_{n=j + 1}^\infty 1/n^3 = q - sum_{n=1}^j 1/n^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{n=1}^j 1/n^3] = sum_{j=1}^\infty q - sum_{j=1}^\infty sum_{n=1}^j 1/n^3.

But the first term is q * sum_{j=1}^\infty 1, which diverges. Wait, that can't be. Wait, but that suggests that my approach here is wrong? Wait, but the original double sum converges, since for each term, 1/(j + k)^3 is O(1/(j + k)^3), and the double sum over j and k is similar to sum_{n=2}^\infty (n - 1)/n^3, which converges. But if I try to split it as above, I get an infinite sum of q, which diverges. Therefore, there must be a mistake here.

Wait, so my mistake is that when I fix j and sum over k, the inner sum is sum_{k=1}^\infty 1/(j + k)^3 = sum_{m=j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty sum_{m=1}^j 1/m^3. The first term is q * sum_{j=1}^\infty 1, which diverges, but the original double sum converges. Therefore, this suggests that there is a mistake in my reasoning here. Wait, but how?

Wait, no, actually, the original double sum is convergent. Therefore, there must be some cancellation here. Wait, but how? Let me check again. If I write the double sum as sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3], then this is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty sum_{m=1}^j 1/m^3. However, the first term is divergent, but the original double sum is convergent. Therefore, this suggests that my approach is wrong. Wait, but why?

Wait, no, actually, the inner sum is not equal to q - sum_{m=1}^j 1/m^3. Wait, let me check. The inner sum over k is sum_{k=1}^\infty 1/(j + k)^3. Let me make substitution m = j + k, so when k = 1, m = j + 1, and as k approaches infinity, m approaches infinity. Therefore, the inner sum is sum_{m = j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, that is correct. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this seems to be divergent? But the original double sum is convergent. Therefore, there must be a mistake here. Wait, but how?

Wait, maybe I made a mistake in the substitution. Let me check with specific numbers. Let me take j=1. Then the inner sum is sum_{k=1}^\infty 1/(1 + k)^3 = sum_{m=2}^\infty 1/m^3 = q - 1. Similarly, for j=2, it's sum_{m=3}^\infty 1/m^3 = q - 1 - 1/2^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. Therefore, the first term when j=1 is q - 1, j=2 is q - 1 - 1/8, j=3 is q - 1 - 1/8 - 1/27, etc. Therefore, the entire sum is sum_{j=1}^\infty [q - H_j^{(3)}], where H_j^{(3)} is the j-th harmonic number of order 3. Therefore, the double sum is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty H_j^{(3)}. But the first term is q * sum_{j=1}^\infty 1, which diverges. However, the original double sum is convergent. Therefore, there must be a mistake here. Wait, but how?

Wait, no! Wait, the original double sum is convergent, but when I split it into two sums, one of them diverges? That can't be. Therefore, my mistake must be in the splitting. Wait, but how?

Wait, actually, the original double sum is equal to sum_{j=1}^\infty [q - H_j^{(3)}], but this is not correct. Wait, no. Wait, the inner sum is sum_{k=1}^\infty 1/(j + k)^3 = sum_{m=j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty sum_{m=1}^j 1/m^3. But the first term is divergent. However, the original double sum is convergent. Therefore, this suggests that my approach is wrong. But why?

Wait, but the original double sum is convergent. Therefore, there must be some cancellation. Wait, but how? Let me think again. Wait, maybe I need to consider that the original double sum is equal to sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. Let me consider that this is equal to sum_{n=2}^\infty (n - 1)/n^3, which is convergent. Therefore, there must be an error in the other approach. Wait, but why?

Wait, perhaps the problem is that when I write sum_{j=1}^\infty [q - H_j^{(3)}], the term q is actually not a constant? Wait, no. Wait, q is defined as sum_{m=1}^\infty 1/m^3. Therefore, q is a constant. Therefore, sum_{j=1}^\infty q is divergent. However, the original double sum is convergent. Therefore, there must be a mistake in my reasoning here. Therefore, my previous approach where I changed variables to n = j + k must be correct, and this other approach is wrong. Therefore, why?

Wait, maybe I made a mistake in the substitution. Let me check again. Let me take the original double sum:

sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3.

Let me fix j and sum over k. For each fixed j, sum_{k=1}^\infty 1/(j + k)^3 = sum_{m=j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty sum_{m=1}^j 1/m^3. The first term is q * sum_{j=1}^\infty 1, which diverges. But the original double sum is convergent. Therefore, this suggests that there is a contradiction. Therefore, my mistake must be in assuming that the inner sum is equal to q - sum_{m=1}^j 1/m^3. Wait, but that seems correct. Wait, unless there is a misunderstanding in the definition of q. Wait, q is defined as sum_{k=1}^\infty 1/k^3. Therefore, sum_{m=j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, that is correct. Therefore, why is there a contradiction?

Wait, unless the original double sum is not equal to that? Wait, but how? Let me check with j=1. For j=1, the inner sum is sum_{k=1}^\infty 1/(1 + k)^3 = sum_{m=2}^\infty 1/m^3 = q - 1. Similarly, for j=2, it's sum_{m=3}^\infty 1/m^3 = q - 1 - 1/8. Therefore, the double sum is (q - 1) + (q - 1 - 1/8) + (q - 1 - 1/8 - 1/27) + ... So this is equal to sum_{j=1}^\infty (q - H_j^{(3)}), where H_j^{(3)} is the j-th harmonic number of order 3. Therefore, the double sum is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty H_j^{(3)}. However, the first term is divergent. But the original double sum is convergent. Therefore, this suggests that there is a mistake in my reasoning. But how?

Wait, maybe I need to consider that the original double sum is equal to sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3 = sum_{n=2}^\infty (n - 1)/n^3, which is convergent. Therefore, there must be a mistake in the other approach. Wait, but how?

Wait, perhaps the problem is that when I write sum_{j=1}^\infty [q - H_j^{(3)}], the term q is actually not a constant? Wait, no. Wait, q is a constant. Therefore, sum_{j=1}^\infty q is divergent. Therefore, this suggests that my two approaches are giving conflicting results. Therefore, one of them must be wrong. But which one?

Wait, let me check with the first approach. If I use the first approach, changing variables to n = j + k, then the double sum is sum_{n=2}^\infty (n - 1)/n^3 = sum_{n=2}^\infty (1/n^2 - 1/n^3) = (p - 1) - (q - 1) = p - q. Therefore, according to this, the answer is p - q. However, according to the second approach, I get an expression that seems to diverge. Therefore, there must be a mistake in the second approach. But why?

Wait, maybe I need to consider that when I write the double sum as sum_{j=1}^\infty [q - H_j^{(3)}], the term q is actually not the same as the q defined in the problem? Wait, no. The problem defines q as sum_{k=1}^\infty 1/k^3. Therefore, that is correct. Therefore, why is there a discrepancy?

Wait, maybe I need to check the convergence. Let me compute the partial sums. Let me take partial sums up to some N. Let me compute S_N = sum_{j=1}^N [q - H_j^{(3)}]. Then S_N = N*q - sum_{j=1}^N H_j^{(3)}. However, the original double sum up to j and k up to N would be sum_{j=1}^N sum_{k=1}^N 1/(j + k)^3. But according to the first approach, this should be approximately p - q when N approaches infinity. However, according to the second approach, S_N = N*q - sum_{j=1}^N H_j^{(3)}. Let me check if these two expressions are equal?

Wait, let me think. The original double sum up to N is sum_{j=1}^N sum_{k=1}^N 1/(j + k)^3. Let me change variables to n = j + k. For each n from 2 to 2N, the number of pairs (j, k) with j + k = n and j, k ≤ N is:

If n ≤ N + 1, then the number of pairs is n - 1.

If n > N + 1, then the number of pairs is 2N + 1 - n.

Therefore, the partial sum up to N is sum_{n=2}^{N + 1} (n - 1)/n^3 + sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3.

But as N approaches infinity, the first sum becomes sum_{n=2}^\infty (n - 1)/n^3 - sum_{n=N + 2}^\infty (n - 1)/n^3, and the second sum becomes sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3. However, as N approaches infinity, the second sum tends to zero because each term is O(1/n^3) and the number of terms is O(N). Therefore, the partial sum tends to sum_{n=2}^\infty (n - 1)/n^3, which is p - q. Therefore, according to this, the original double sum is p - q.

But according to the second approach, the partial sum S_N = N*q - sum_{j=1}^N H_j^{(3)}. Let me check if this is equal to the partial sum from the first approach. Let me compute S_N = N*q - sum_{j=1}^N H_j^{(3)}. Let me note that sum_{j=1}^N H_j^{(3)} = sum_{j=1}^N sum_{m=1}^j 1/m^3 = sum_{m=1}^N sum_{j=m}^N 1/m^3 = sum_{m=1}^N (N - m + 1)/m^3. Therefore, S_N = N*q - sum_{m=1}^N (N - m + 1)/m^3 = N*q - sum_{m=1}^N (N + 1 - m)/m^3.

But the original partial sum up to N is sum_{j=1}^N sum_{k=1}^N 1/(j + k)^3. Let me compare this with S_N. Let me note that:

sum_{j=1}^N sum_{k=1}^N 1/(j + k)^3 = sum_{n=2}^{2N} c_n /n^3, where c_n is the number of pairs (j, k) with j + k = n and j, k ≤ N. As I mentioned before, for n from 2 to N + 1, c_n = n - 1, and for n from N + 2 to 2N, c_n = 2N + 1 - n.

Therefore, the partial sum is sum_{n=2}^{N + 1} (n - 1)/n^3 + sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3.

On the other hand, S_N = N*q - sum_{m=1}^N (N + 1 - m)/m^3. Let me compute this:

sum_{m=1}^N (N + 1 - m)/m^3 = sum_{m=1}^N (N + 1)/m^3 - sum_{m=1}^N 1/m^3 = (N + 1) * sum_{m=1}^N 1/m^3 - sum_{m=1}^N 1/m^3 = N * sum_{m=1}^N 1/m^3 + sum_{m=1}^N 1/m^3 - sum_{m=1}^N 1/m^3 = N * sum_{m=1}^N 1/m^3.

Therefore, S_N = N*q - N * sum_{m=1}^N 1/m^3 = N [ q - sum_{m=1}^N 1/m^3 ].

But q is sum_{m=1}^\infty 1/m^3, so q - sum_{m=1}^N 1/m^3 = sum_{m=N + 1}^\infty 1/m^3. Therefore, S_N = N * sum_{m=N + 1}^\infty 1/m^3.

But how does this relate to the original partial sum?

Wait, the original partial sum is sum_{n=2}^{N + 1} (n - 1)/n^3 + sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3. Let me compare this with S_N = N * sum_{m=N + 1}^\infty 1/m^3.

Wait, let me compute S_N:

S_N = N * sum_{m=N + 1}^\infty 1/m^3 = N * [ sum_{m=N + 1}^{2N} 1/m^3 + sum_{m=2N + 1}^\infty 1/m^3 ].

But the original partial sum is sum_{n=2}^{N + 1} (n - 1)/n^3 + sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3. Let me note that sum_{n=2}^{N + 1} (n - 1)/n^3 = sum_{n=2}^{N + 1} [1/n^2 - 1/n^3] = [sum_{n=2}^{N + 1} 1/n^2] - [sum_{n=2}^{N + 1} 1/n^3]. Similarly, the second sum is sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3. Let me make substitution m = n - N - 1. Wait, maybe not. Alternatively, note that for n from N + 2 to 2N, 2N + 1 - n = (N + 1) - (n - N - 1). Wait, maybe not helpful.

Alternatively, note that sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3 = sum_{k=1}^{N - 1} (k)/ ( (N + 1 + k)^3 ). Wait, let me set k = n - (N + 1). Then when n = N + 2, k = 1; when n = 2N, k = N - 1. Therefore, the sum becomes sum_{k=1}^{N - 1} k / (N + 1 + k)^3. Not sure if helpful.

Alternatively, compare with S_N. Let me think about the original partial sum and S_N. If I can show that they are equal, then there is no contradiction. However, according to the first approach, the original partial sum tends to p - q, and according to the second approach, S_N = N * sum_{m=N + 1}^\infty 1/m^3. Let me compute the limit as N approaches infinity of S_N. Since sum_{m=N + 1}^\infty 1/m^3 ~ ∫_{N}^\infty 1/x^3 dx = 1/(2N^2). Therefore, S_N ~ N * 1/(2N^2) = 1/(2N) → 0 as N → ∞. But according to the first approach, the original partial sum tends to p - q ≈ 0.4428. Therefore, there is a contradiction here. Therefore, my second approach must be wrong. But why?

Wait, this suggests that there is a mistake in the second approach. But where?

Wait, earlier I thought that the original double sum is equal to sum_{j=1}^\infty [q - H_j^{(3)}], but according to the partial sums, this leads to a contradiction. Therefore, my mistake must be in that step. Let me check again.

Original double sum: sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. For fixed j, sum_{k=1}^\infty 1/(j + k)^3 = sum_{m=j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty sum_{m=1}^j 1/m^3. But the first term is divergent. However, the original double sum is convergent. Therefore, this suggests that my assumption that the inner sum is equal to q - sum_{m=1}^j 1/m^3 is wrong? But how?

Wait, no. Wait, q is defined as sum_{m=1}^\infty 1/m^3. Therefore, sum_{m=j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, that is correct. Therefore, the problem must be that the original double sum is not equal to that? But how?

Wait, unless there is a misunderstanding in the definition of the double sum. Wait, the original double sum is over all j and k ≥ 1. However, when I fix j and sum over k, I get sum_{k=1}^\infty 1/(j + k)^3. But if I then sum over j, I get sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this seems to be divergent. But the original double sum is convergent. Therefore, there must be a mistake in the logic here. Wait, but how?

Wait, maybe the problem is that when I write sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3], I need to consider that q is not a constant? Wait, no. q is a constant. Therefore, this suggests that there is a contradiction. Therefore, my first approach must be correct, and the second approach is wrong. But why?

Wait, maybe the second approach is correct, but the partial sums are not matching? Wait, earlier I thought that the partial sum S_N = N * sum_{m=N + 1}^\infty 1/m^3 tends to zero, but according to the first approach, the original partial sum tends to p - q. Therefore, there must be a mistake in the second approach. Therefore, I need to figure out where.

Wait, let me think again. The original double sum is sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. Let me consider that this is equal to sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. Let me make substitution m = j + k, and n = j - k or something? Wait, no. Alternatively, think of it as sum_{m=2}^\infty (number of pairs) * 1/m^3, which is sum_{m=2}^\infty (m - 1)/m^3, which is convergent. Therefore, this must be correct. Therefore, the second approach must have an error. Therefore, my mistake in the second approach is that I incorrectly assumed that the double sum is equal to sum_{j=1}^\infty [q - H_j^{(3)}], but actually, this is not the case. Wait, but why?

Wait, no. If I fix j and sum over k, then the inner sum is sum_{k=1}^\infty 1/(j + k)^3 = q - sum_{m=1}^j 1/m^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this seems to be divergent. But the original double sum is convergent. Therefore, there must be a mistake in the assumption that the inner sum is equal to q - sum_{m=1}^j 1/m^3. Wait, but how?

Wait, let me check with j=1. The inner sum is sum_{k=1}^\infty 1/(1 + k)^3 = sum_{m=2}^\infty 1/m^3 = q - 1. Therefore, that is correct. Similarly, for j=2, it's sum_{m=3}^\infty 1/m^3 = q - 1 - 1/8. Therefore, that is correct. Therefore, the double sum is sum_{j=1}^\infty [q - H_j^{(3)}]. But this seems to be divergent. However, the original double sum is convergent. Therefore, this suggests that there is a contradiction. Therefore, I must have made a mistake in the first approach?

Wait, let me check the first approach again. The first approach says that the double sum is equal to sum_{n=2}^\infty (n - 1)/n^3. Let me compute this sum:

sum_{n=2}^\infty (n - 1)/n^3 = sum_{n=2}^\infty (1/n^2 - 1/n^3) = (sum_{n=1}^\infty 1/n^2 - 1) - (sum_{n=1}^\infty 1/n^3 - 1) = (p - 1) - (q - 1) = p - q. Therefore, this is correct. Therefore, the double sum is p - q. Therefore, the second approach must be wrong. But why?

Wait, maybe the second approach is correct, but the partial sums are not matching because of some misunderstanding? Let me think. If the double sum is p - q, then according to the second approach, the partial sum S_N = N * sum_{m=N + 1}^\infty 1/m^3 should approach p - q as N approaches infinity. But earlier I thought that S_N tends to zero, but that was incorrect. Let me re-examine that.

Wait, S_N = N * sum_{m=N + 1}^\infty 1/m^3. Let me approximate this sum. For large N, sum_{m=N + 1}^\infty 1/m^3 ≈ ∫_{N}^\infty 1/x^3 dx = 1/(2N^2). Therefore, S_N ≈ N * 1/(2N^2) = 1/(2N) → 0 as N → ∞. But according to the first approach, the double sum is p - q ≈ 0.4428. Therefore, there is a contradiction. Therefore, my second approach must be wrong. But why?

Wait, I think the mistake is that in the second approach, I considered the partial sum up to N as S_N = N * sum_{m=N + 1}^\infty 1/m^3, but this is not correct. Wait, earlier I derived that S_N = N * [ q - sum_{m=1}^N 1/m^3 ] = N * sum_{m=N + 1}^\infty 1/m^3. But according to the first approach, the partial sum up to N is sum_{n=2}^{N + 1} (n - 1)/n^3 + sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3. Therefore, if I compute the limit as N approaches infinity of this partial sum, it should be p - q. However, according to the second approach, the partial sum S_N tends to zero. Therefore, there must be a mistake in the derivation of S_N.

Wait, let me re-examine the derivation of S_N. Earlier, I said that sum_{j=1}^N sum_{k=1}^N 1/(j + k)^3 = sum_{j=1}^N [q - H_j^{(3)}]. But this is not correct. Wait, no. Wait, for each j, sum_{k=1}^N 1/(j + k)^3 is not equal to q - H_j^{(3)}, because q is the infinite sum. Instead, it is equal to sum_{m=j + 1}^{j + N} 1/m^3. Therefore, it is equal to sum_{m=j + 1}^\infty 1/m^3 - sum_{m=j + N + 1}^\infty 1/m^3 = [q - H_j^{(3)}] - [q - H_{j + N}^{(3)}] = H_{j + N}^{(3)} - H_j^{(3)}. Therefore, the double sum up to N is sum_{j=1}^N [H_{j + N}^{(3)} - H_j^{(3)}]. Therefore, this is a telescoping sum!

Ah! This is the key mistake. Earlier, I incorrectly assumed that the inner sum up to N is q - H_j^{(3)}, but actually, it is H_{j + N}^{(3)} - H_j^{(3)}. Therefore, the double sum up to N is sum_{j=1}^N [H_{j + N}^{(3)} - H_j^{(3)}]. This is a telescoping series. Let me write out the terms:

For j=1: H_{N + 1}^{(3)} - H_1^{(3)}

For j=2: H_{N + 2}^{(3)} - H_2^{(3)}

...

For j=N: H_{2N}^{(3)} - H_N^{(3)}

Therefore, summing over j=1 to N:

[H_{N + 1}^{(3)} - H_1^{(3)}] + [H_{N + 2}^{(3)} - H_2^{(3)}] + ... + [H_{2N}^{(3)} - H_N^{(3)}]

This can be rearranged as:

sum_{m=N + 1}^{2N} H_m^{(3)} - sum_{m=1}^N H_m^{(3)}.

But this is equal to [sum_{m=1}^{2N} H_m^{(3)} - sum_{m=1}^N H_m^{(3)}] - sum_{m=1}^N H_m^{(3)}? Wait, no. Wait, the first sum is from m=N + 1 to 2N of H_m^{(3)}, and the second sum is from m=1 to N of H_m^{(3)}. Therefore, the total is sum_{m=N + 1}^{2N} H_m^{(3)} - sum_{m=1}^N H_m^{(3)}.

But this is not a telescoping sum. However, if I consider the original double sum up to N, it is equal to this expression. However, this seems complicated. However, if I take the limit as N approaches infinity, then sum_{m=N + 1}^{2N} H_m^{(3)} - sum_{m=1}^N H_m^{(3)}.

But as N approaches infinity, sum_{m=N + 1}^{2N} H_m^{(3)} ≈ sum_{m=N + 1}^{2N} (ζ(3) + o(1)) = ζ(3) * N * (1/2) + ... = (ζ(3)/2) * N. Similarly, sum_{m=1}^N H_m^{(3)} ≈ ζ(3) * N + ... Therefore, the difference would be approximately (ζ(3)/2) * N - ζ(3) * N = -ζ(3)/2 * N, which tends to -infty. But this contradicts the first approach where the double sum converges to p - q. Therefore, there must be a mistake here.

Wait, this is getting too convoluted. Let me go back. The first approach seems straightforward and gives a finite result, and the numerical check with partial sums suggests that it is correct. Therefore, I think the mistake in the second approach is due to incorrect manipulation of the sums, possibly due to improper handling of the infinite series. Therefore, I will trust the first approach, which gives the result p - q.

Therefore, the answer is p - q. Therefore, the double sum is equal to p - q.

But let me check with another method. Let me recall that in some cases, sums over j and k can be related to known series. For example, consider that:

sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3 = sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3.

Let me consider that this is equal to sum_{n=2}^\infty (n - 1)/n^3, as before. Which is equal to sum_{n=2}^\infty (1/n^2 - 1/n^3) = (p - 1) - (q - 1) = p - q. Therefore, this seems correct.

Therefore, I think the correct answer is p - q. Therefore, the double sum is equal to p - q.

But to be thorough, let me check with another perspective. Let me recall that in general, for sums of the form sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^s, there might be known formulas. For example, for s = 2, this sum is known to be 2ζ(2) - 1, but I need to check.

Wait, for s = 2, the double sum would be sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^2. Using the same method, this would be sum_{n=2}^\infty (n - 1)/n^2 = sum_{n=2}^\infty (1/n - 1/n^2) = (sum_{n=2}^\infty 1/n) - (sum_{n=2}^\infty 1/n^2). But sum_{n=2}^\infty 1/n diverges, which contradicts the fact that the original double sum for s=2 should converge? Wait, no. Wait, for s=2, the double sum is sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^2. This is similar to the previous case, but for s=2. However, this sum is actually divergent, because for each fixed j, sum_{k=1}^\infty 1/(j + k)^2 ~ ∫_{j}^\infty 1/x^2 dx = 1/j, so the double sum is sum_{j=1}^\infty 1/j, which diverges. Therefore, for s=2, the double sum diverges, which matches with our previous result. However, for s=3, the double sum converges, as we saw. Therefore, in our problem, s=3, and the double sum converges to p - q.

Therefore, this seems consistent. Therefore, I think my initial conclusion is correct. Therefore, the answer is p - q.

**Final Answer**
\boxed{p - q}

To evaluate the double sum

$$
\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3},
$$

we begin by observing that the expression involves a sum over all positive integers $ j $ and $ k $, and the term depends on the sum $ j + k $. This suggests a change of variables to simplify the structure of the sum.

---

### Step 1: Change of Variables

Let us define a new variable $ n = j + k $. For each fixed $ n \geq 2 $, the number of integer pairs $ (j, k) $ such that $ j + k = n $ and $ j, k \geq 1 $ is $ n - 1 $. This is because $ j $ can range from 1 to $ n - 1 $, and $ k = n - j $.

Thus, the double sum can be rewritten as:

$$
\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3} = \sum_{n = 2}^\infty \frac{n - 1}{n^3}.
$$

---

### Step 2: Simplify the Sum

We now simplify the expression:

$$
\sum_{n = 2}^\infty \frac{n - 1}{n^3} = \sum_{n = 2}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right).
$$

This is a telescoping-like sum, and we can split it into two separate sums:

$$
\sum_{n = 2}^\infty \frac{1}{n^2} - \sum_{n = 2}^\infty \frac{1}{n^3}.
$$

Now, recall the definitions:

- $ p = \sum_{n = 1}^\infty \frac{1}{n^2} = \zeta(2) $,
- $ q = \sum_{n = 1}^\infty \frac{1}{n^3} = \zeta(3) $.

So, we can write:

$$
\sum_{n = 2}^\infty \frac{1}{n^2} = p - 1, \quad \sum_{n = 2}^\infty \frac{1}{n^3} = q - 1.
$$

Substituting back, we get:

$$
\sum_{n = 2}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right) = (p - 1) - (q - 1) = p - q.
$$

---

### Final Result

Thus, the original double sum simplifies to:

$$
\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3} = p - q.
$$

---

### Final Answer

$$
\boxed{p - q}
$$

[Assistant]
Explanation:
Okay, so I need to find a way to express the double sum $\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3}$ in terms of $p$ and $q$, where $p = \sum_{k = 1}^\infty \frac{1}{k^2}$ and $q = \sum_{k = 1}^\infty \frac{1}{k^3}$. Hmm, let me think about how to approach this.

First, I know that $p$ is the famous Basel problem, which converges to $\frac{\pi^2}{6}$, and $q$ is the Riemann zeta function at 3, which doesn't have a simple closed-form expression like $p$, but maybe I don't need to know the exact values since the problem just wants it in terms of $p$ and $q$.

The double sum given is over all positive integers $j$ and $k$, and each term is $\frac{1}{(j + k)^3}$. I need to relate this to $p$ and $q$. Let me try to manipulate the double sum.

One idea is to change variables. Let me consider substituting $n = j + k$. For each fixed $n$, the number of pairs $(j, k)$ such that $j + k = n$ is $n - 1$, since $j$ can range from 1 to $n - 1$ and $k$ would be $n - j$. However, in our case, both $j$ and $k$ start from 1, so for each $n \geq 2$, the number of pairs is $n - 1$. Therefore, maybe I can rewrite the double sum as a single sum over $n$:

$$
\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^3} = \sum_{n=2}^\infty \frac{1}{n^3} \sum_{\substack{j=1 \\ j + k = n}}^\infty 1
$$

Wait, but the inner sum is over all $j$ and $k$ such that $j + k = n$. Since $j$ and $k$ are positive integers, for each $n \geq 2$, the number of such pairs is $n - 1$. Therefore, the double sum becomes:

$$
\sum_{n=2}^\infty \frac{n - 1}{n^3} = \sum_{n=2}^\infty \frac{n - 1}{n^3} = \sum_{n=2}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right)
$$

Wait, let me check that. If I split $\frac{n - 1}{n^3}$ into $\frac{n}{n^3} - \frac{1}{n^3} = \frac{1}{n^2} - \frac{1}{n^3}$. Yes, that's correct. Therefore, the entire sum becomes:

$$
\sum_{n=2}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right) = \left( \sum_{n=2}^\infty \frac{1}{n^2} \right) - \left( \sum_{n=2}^\infty \frac{1}{n^3} \right)
$$

But we know that $p = \sum_{n=1}^\infty \frac{1}{n^2}$, so $\sum_{n=2}^\infty \frac{1}{n^2} = p - 1$. Similarly, $\sum_{n=2}^\infty \frac{1}{n^3} = q - 1$. Therefore, substituting back:

$$
(p - 1) - (q - 1) = p - 1 - q + 1 = p - q
$$

Wait, so does that mean the double sum is equal to $p - q$? But let me check if I did everything correctly. Let me verify with small values.

Wait, let me think again. The original double sum is over all $j$ and $k$ starting at 1. When I change variables to $n = j + k$, then for each $n \geq 2$, the number of pairs is $n - 1$. Therefore, the sum becomes $\sum_{n=2}^\infty \frac{n - 1}{n^3}$. Then splitting into two sums:

$$
\sum_{n=2}^\infty \frac{1}{n^2} - \sum_{n=2}^\infty \frac{1}{n^3} = (p - 1) - (q - 1) = p - q
$$

So according to this, the answer is $p - q$. But wait, let me check with an example. Let me compute partial sums numerically to see if this makes sense.

Take the original double sum. Let me compute the first few terms. For j and k from 1 to 2:

When j=1, k=1: 1/(1+1)^3 = 1/8

j=1, k=2: 1/(1+2)^3 = 1/27

j=2, k=1: 1/(2+1)^3 = 1/27

j=2, k=2: 1/(2+2)^3 = 1/64

So total for n=2: 1/8 + 1/27 + 1/27 + 1/64? Wait no, wait. Wait, when n=2, j + k = 2, so only (1,1). Then n=3: (1,2) and (2,1). n=4: (1,3), (2,2), (3,1). So for n=2, the number of terms is 1, n=3: 2, n=4: 3, etc. So the sum over n=2 to infinity of (n-1)/n^3. So for n=2, term is 1/8, n=3: 2/27, n=4: 3/64, etc. So the total sum is 1/8 + 2/27 + 3/64 + ...

Now, if I compute p - q, where p is pi^2/6 ≈ 1.6449, and q is zeta(3) ≈ 1.2020569. So p - q ≈ 0.4428. Let me compute the partial sum up to n=4:

1/8 ≈ 0.125

+ 2/27 ≈ 0.07407 → total ≈ 0.19907

+ 3/64 ≈ 0.046875 → total ≈ 0.2459

+ 4/125 = 0.032 → total ≈ 0.2779

+ 5/216 ≈ 0.0231 → total ≈ 0.301

+ 6/343 ≈ 0.0175 → total ≈ 0.3185

Continuing this, it seems like the sum is approaching something around 0.44... which is p - q ≈ 0.4428. So that seems plausible. Therefore, maybe my initial conclusion is correct?

But wait, let me check again. Wait, the original problem is to write the double sum in terms of p and q. According to my calculation, it's p - q. But let me think again if there's a mistake in the reasoning.

Wait, the original double sum is over all j and k ≥ 1. When changing variables to n = j + k, then for each n ≥ 2, the number of pairs (j, k) with j + k = n is n - 1. Therefore, the double sum is equal to sum_{n=2}^\infty (n - 1)/n^3. Then splitting into sum_{n=2}^\infty 1/n^2 - sum_{n=2}^\infty 1/n^3. Which is (p - 1) - (q - 1) = p - q. So that seems correct.

But wait, let me check another way. Let me consider that the original double sum can be written as sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. Let me try to switch the order of summation or find another representation.

Alternatively, maybe I can consider that for each j and k, j + k is the same as m, but I think that's what I did before.

Alternatively, maybe I can use integral representations? For example, since 1/n^3 can be represented as an integral. But maybe that complicates things. Alternatively, think about generating functions?

Alternatively, think about symmetry. Let me note that the double sum is symmetric in j and k, so maybe I can write it as 2 times the sum over j ≥ 1, k ≥ 1, j < k, plus the diagonal terms where j = k. Wait, but maybe that complicates.

Alternatively, think of the double sum as sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. Let me make substitution m = j + k. But that's similar to before.

Alternatively, think of the sum as sum_{m=2}^\infty (number of ways to write m as j + k) * 1/m^3. Which is exactly what I did before. So that gives the same result.

Alternatively, maybe I can check with known values. For example, if I consider the sum over j and k of 1/(j + k)^s, which is known? Maybe there's a formula for that?

Alternatively, think of the double sum as sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. Let me fix j and sum over k. For fixed j, sum_{k=1}^\infty 1/(j + k)^3 = sum_{n=j + 1}^\infty 1/n^3 = q - sum_{n=1}^j 1/n^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{n=1}^j 1/n^3] = sum_{j=1}^\infty q - sum_{j=1}^\infty sum_{n=1}^j 1/n^3.

But the first term is q * sum_{j=1}^\infty 1, which diverges. Wait, that can't be. Wait, but that suggests that my approach here is wrong? Wait, but the original double sum converges, since for each term, 1/(j + k)^3 is O(1/(j + k)^3), and the double sum over j and k is similar to sum_{n=2}^\infty (n - 1)/n^3, which converges. But if I try to split it as above, I get an infinite sum of q, which diverges. Therefore, there must be a mistake here.

Wait, so my mistake is that when I fix j and sum over k, the inner sum is sum_{k=1}^\infty 1/(j + k)^3 = sum_{m=j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty sum_{m=1}^j 1/m^3. The first term is q * sum_{j=1}^\infty 1, which diverges, but the original double sum converges. Therefore, this suggests that there is a mistake in my reasoning here. Wait, but how?

Wait, no, actually, the original double sum is convergent. Therefore, there must be some cancellation here. Wait, but how? Let me check again. If I write the double sum as sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3], then this is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty sum_{m=1}^j 1/m^3. However, the first term is divergent, but the original double sum is convergent. Therefore, this suggests that my approach is wrong. Wait, but why?

Wait, no, actually, the inner sum is not equal to q - sum_{m=1}^j 1/m^3. Wait, let me check. The inner sum over k is sum_{k=1}^\infty 1/(j + k)^3. Let me make substitution m = j + k, so when k = 1, m = j + 1, and as k approaches infinity, m approaches infinity. Therefore, the inner sum is sum_{m = j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, that is correct. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this seems to be divergent? But the original double sum is convergent. Therefore, there must be a mistake here. Wait, but how?

Wait, maybe I made a mistake in the substitution. Let me check with specific numbers. Let me take j=1. Then the inner sum is sum_{k=1}^\infty 1/(1 + k)^3 = sum_{m=2}^\infty 1/m^3 = q - 1. Similarly, for j=2, it's sum_{m=3}^\infty 1/m^3 = q - 1 - 1/2^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. Therefore, the first term when j=1 is q - 1, j=2 is q - 1 - 1/8, j=3 is q - 1 - 1/8 - 1/27, etc. Therefore, the entire sum is sum_{j=1}^\infty [q - H_j^{(3)}], where H_j^{(3)} is the j-th harmonic number of order 3. Therefore, the double sum is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty H_j^{(3)}. But the first term is q * sum_{j=1}^\infty 1, which diverges. However, the original double sum is convergent. Therefore, there must be a mistake here. Wait, but how?

Wait, no! Wait, the original double sum is convergent, but when I split it into two sums, one of them diverges? That can't be. Therefore, my mistake must be in the splitting. Wait, but how?

Wait, actually, the original double sum is equal to sum_{j=1}^\infty [q - H_j^{(3)}], but this is not correct. Wait, no. Wait, the inner sum is sum_{k=1}^\infty 1/(j + k)^3 = sum_{m=j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty sum_{m=1}^j 1/m^3. But the first term is divergent. However, the original double sum is convergent. Therefore, this suggests that my approach is wrong. But why?

Wait, but the original double sum is convergent. Therefore, there must be some cancellation. Wait, but how? Let me think again. Wait, maybe I need to consider that the original double sum is equal to sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. Let me consider that this is equal to sum_{n=2}^\infty (n - 1)/n^3, which is convergent. Therefore, there must be an error in the other approach. Wait, but why?

Wait, perhaps the problem is that when I write sum_{j=1}^\infty [q - H_j^{(3)}], the term q is actually not a constant? Wait, no. Wait, q is defined as sum_{m=1}^\infty 1/m^3. Therefore, q is a constant. Therefore, sum_{j=1}^\infty q is divergent. However, the original double sum is convergent. Therefore, there must be a mistake in my reasoning here. Therefore, my previous approach where I changed variables to n = j + k must be correct, and this other approach is wrong. Therefore, why?

Wait, maybe I made a mistake in the substitution. Let me check again. Let me take the original double sum:

sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3.

Let me fix j and sum over k. For each fixed j, sum_{k=1}^\infty 1/(j + k)^3 = sum_{m=j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty sum_{m=1}^j 1/m^3. The first term is q * sum_{j=1}^\infty 1, which diverges. But the original double sum is convergent. Therefore, this suggests that there is a contradiction. Therefore, my mistake must be in assuming that the inner sum is equal to q - sum_{m=1}^j 1/m^3. Wait, but that seems correct. Wait, unless there is a misunderstanding in the definition of q. Wait, q is defined as sum_{k=1}^\infty 1/k^3. Therefore, sum_{m=j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, that is correct. Therefore, why is there a contradiction?

Wait, unless the original double sum is not equal to that? Wait, but how? Let me check with j=1. For j=1, the inner sum is sum_{k=1}^\infty 1/(1 + k)^3 = sum_{m=2}^\infty 1/m^3 = q - 1. Similarly, for j=2, it's sum_{m=3}^\infty 1/m^3 = q - 1 - 1/8. Therefore, the double sum is (q - 1) + (q - 1 - 1/8) + (q - 1 - 1/8 - 1/27) + ... So this is equal to sum_{j=1}^\infty (q - H_j^{(3)}), where H_j^{(3)} is the j-th harmonic number of order 3. Therefore, the double sum is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty H_j^{(3)}. However, the first term is divergent. But the original double sum is convergent. Therefore, this suggests that there is a mistake in my reasoning. But how?

Wait, maybe I need to consider that the original double sum is equal to sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3 = sum_{n=2}^\infty (n - 1)/n^3, which is convergent. Therefore, there must be a mistake in the other approach. Wait, but how?

Wait, perhaps the problem is that when I write sum_{j=1}^\infty [q - H_j^{(3)}], the term q is actually not a constant? Wait, no. Wait, q is a constant. Therefore, sum_{j=1}^\infty q is divergent. Therefore, this suggests that my two approaches are giving conflicting results. Therefore, one of them must be wrong. But which one?

Wait, let me check with the first approach. If I use the first approach, changing variables to n = j + k, then the double sum is sum_{n=2}^\infty (n - 1)/n^3 = sum_{n=2}^\infty (1/n^2 - 1/n^3) = (p - 1) - (q - 1) = p - q. Therefore, according to this, the answer is p - q. However, according to the second approach, I get an expression that seems to diverge. Therefore, there must be a mistake in the second approach. But why?

Wait, maybe I need to consider that when I write the double sum as sum_{j=1}^\infty [q - H_j^{(3)}], the term q is actually not the same as the q defined in the problem? Wait, no. The problem defines q as sum_{k=1}^\infty 1/k^3. Therefore, that is correct. Therefore, why is there a discrepancy?

Wait, maybe I need to check the convergence. Let me compute the partial sums. Let me take partial sums up to some N. Let me compute S_N = sum_{j=1}^N [q - H_j^{(3)}]. Then S_N = N*q - sum_{j=1}^N H_j^{(3)}. However, the original double sum up to j and k up to N would be sum_{j=1}^N sum_{k=1}^N 1/(j + k)^3. But according to the first approach, this should be approximately p - q when N approaches infinity. However, according to the second approach, S_N = N*q - sum_{j=1}^N H_j^{(3)}. Let me check if these two expressions are equal?

Wait, let me think. The original double sum up to N is sum_{j=1}^N sum_{k=1}^N 1/(j + k)^3. Let me change variables to n = j + k. For each n from 2 to 2N, the number of pairs (j, k) with j + k = n and j, k ≤ N is:

If n ≤ N + 1, then the number of pairs is n - 1.

If n > N + 1, then the number of pairs is 2N + 1 - n.

Therefore, the partial sum up to N is sum_{n=2}^{N + 1} (n - 1)/n^3 + sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3.

But as N approaches infinity, the first sum becomes sum_{n=2}^\infty (n - 1)/n^3 - sum_{n=N + 2}^\infty (n - 1)/n^3, and the second sum becomes sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3. However, as N approaches infinity, the second sum tends to zero because each term is O(1/n^3) and the number of terms is O(N). Therefore, the partial sum tends to sum_{n=2}^\infty (n - 1)/n^3, which is p - q. Therefore, according to this, the original double sum is p - q.

But according to the second approach, the partial sum S_N = N*q - sum_{j=1}^N H_j^{(3)}. Let me check if this is equal to the partial sum from the first approach. Let me compute S_N = N*q - sum_{j=1}^N H_j^{(3)}. Let me note that sum_{j=1}^N H_j^{(3)} = sum_{j=1}^N sum_{m=1}^j 1/m^3 = sum_{m=1}^N sum_{j=m}^N 1/m^3 = sum_{m=1}^N (N - m + 1)/m^3. Therefore, S_N = N*q - sum_{m=1}^N (N - m + 1)/m^3 = N*q - sum_{m=1}^N (N + 1 - m)/m^3.

But the original partial sum up to N is sum_{j=1}^N sum_{k=1}^N 1/(j + k)^3. Let me compare this with S_N. Let me note that:

sum_{j=1}^N sum_{k=1}^N 1/(j + k)^3 = sum_{n=2}^{2N} c_n /n^3, where c_n is the number of pairs (j, k) with j + k = n and j, k ≤ N. As I mentioned before, for n from 2 to N + 1, c_n = n - 1, and for n from N + 2 to 2N, c_n = 2N + 1 - n.

Therefore, the partial sum is sum_{n=2}^{N + 1} (n - 1)/n^3 + sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3.

On the other hand, S_N = N*q - sum_{m=1}^N (N + 1 - m)/m^3. Let me compute this:

sum_{m=1}^N (N + 1 - m)/m^3 = sum_{m=1}^N (N + 1)/m^3 - sum_{m=1}^N 1/m^3 = (N + 1) * sum_{m=1}^N 1/m^3 - sum_{m=1}^N 1/m^3 = N * sum_{m=1}^N 1/m^3 + sum_{m=1}^N 1/m^3 - sum_{m=1}^N 1/m^3 = N * sum_{m=1}^N 1/m^3.

Therefore, S_N = N*q - N * sum_{m=1}^N 1/m^3 = N [ q - sum_{m=1}^N 1/m^3 ].

But q is sum_{m=1}^\infty 1/m^3, so q - sum_{m=1}^N 1/m^3 = sum_{m=N + 1}^\infty 1/m^3. Therefore, S_N = N * sum_{m=N + 1}^\infty 1/m^3.

But how does this relate to the original partial sum?

Wait, the original partial sum is sum_{n=2}^{N + 1} (n - 1)/n^3 + sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3. Let me compare this with S_N = N * sum_{m=N + 1}^\infty 1/m^3.

Wait, let me compute S_N:

S_N = N * sum_{m=N + 1}^\infty 1/m^3 = N * [ sum_{m=N + 1}^{2N} 1/m^3 + sum_{m=2N + 1}^\infty 1/m^3 ].

But the original partial sum is sum_{n=2}^{N + 1} (n - 1)/n^3 + sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3. Let me note that sum_{n=2}^{N + 1} (n - 1)/n^3 = sum_{n=2}^{N + 1} [1/n^2 - 1/n^3] = [sum_{n=2}^{N + 1} 1/n^2] - [sum_{n=2}^{N + 1} 1/n^3]. Similarly, the second sum is sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3. Let me make substitution m = n - N - 1. Wait, maybe not. Alternatively, note that for n from N + 2 to 2N, 2N + 1 - n = (N + 1) - (n - N - 1). Wait, maybe not helpful.

Alternatively, note that sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3 = sum_{k=1}^{N - 1} (k)/ ( (N + 1 + k)^3 ). Wait, let me set k = n - (N + 1). Then when n = N + 2, k = 1; when n = 2N, k = N - 1. Therefore, the sum becomes sum_{k=1}^{N - 1} k / (N + 1 + k)^3. Not sure if helpful.

Alternatively, compare with S_N. Let me think about the original partial sum and S_N. If I can show that they are equal, then there is no contradiction. However, according to the first approach, the original partial sum tends to p - q, and according to the second approach, S_N = N * sum_{m=N + 1}^\infty 1/m^3. Let me compute the limit as N approaches infinity of S_N. Since sum_{m=N + 1}^\infty 1/m^3 ~ ∫_{N}^\infty 1/x^3 dx = 1/(2N^2). Therefore, S_N ~ N * 1/(2N^2) = 1/(2N) → 0 as N → ∞. But according to the first approach, the original partial sum tends to p - q ≈ 0.4428. Therefore, there is a contradiction here. Therefore, my second approach must be wrong. But why?

Wait, this suggests that there is a mistake in the second approach. But where?

Wait, earlier I thought that the original double sum is equal to sum_{j=1}^\infty [q - H_j^{(3)}], but according to the partial sums, this leads to a contradiction. Therefore, my mistake must be in that step. Let me check again.

Original double sum: sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. For fixed j, sum_{k=1}^\infty 1/(j + k)^3 = sum_{m=j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this is equal to sum_{j=1}^\infty q - sum_{j=1}^\infty sum_{m=1}^j 1/m^3. But the first term is divergent. However, the original double sum is convergent. Therefore, this suggests that my assumption that the inner sum is equal to q - sum_{m=1}^j 1/m^3 is wrong? But how?

Wait, no. Wait, q is defined as sum_{m=1}^\infty 1/m^3. Therefore, sum_{m=j + 1}^\infty 1/m^3 = q - sum_{m=1}^j 1/m^3. Therefore, that is correct. Therefore, the problem must be that the original double sum is not equal to that? But how?

Wait, unless there is a misunderstanding in the definition of the double sum. Wait, the original double sum is over all j and k ≥ 1. However, when I fix j and sum over k, I get sum_{k=1}^\infty 1/(j + k)^3. But if I then sum over j, I get sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this seems to be divergent. But the original double sum is convergent. Therefore, there must be a mistake in the logic here. Wait, but how?

Wait, maybe the problem is that when I write sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3], I need to consider that q is not a constant? Wait, no. q is a constant. Therefore, this suggests that there is a contradiction. Therefore, my first approach must be correct, and the second approach is wrong. But why?

Wait, maybe the second approach is correct, but the partial sums are not matching? Wait, earlier I thought that the partial sum S_N = N * sum_{m=N + 1}^\infty 1/m^3 tends to zero, but according to the first approach, the original partial sum tends to p - q. Therefore, there must be a mistake in the second approach. Therefore, I need to figure out where.

Wait, let me think again. The original double sum is sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. Let me consider that this is equal to sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3. Let me make substitution m = j + k, and n = j - k or something? Wait, no. Alternatively, think of it as sum_{m=2}^\infty (number of pairs) * 1/m^3, which is sum_{m=2}^\infty (m - 1)/m^3, which is convergent. Therefore, this must be correct. Therefore, the second approach must have an error. Therefore, my mistake in the second approach is that I incorrectly assumed that the double sum is equal to sum_{j=1}^\infty [q - H_j^{(3)}], but actually, this is not the case. Wait, but why?

Wait, no. If I fix j and sum over k, then the inner sum is sum_{k=1}^\infty 1/(j + k)^3 = q - sum_{m=1}^j 1/m^3. Therefore, the double sum is sum_{j=1}^\infty [q - sum_{m=1}^j 1/m^3]. However, this seems to be divergent. But the original double sum is convergent. Therefore, there must be a mistake in the assumption that the inner sum is equal to q - sum_{m=1}^j 1/m^3. Wait, but how?

Wait, let me check with j=1. The inner sum is sum_{k=1}^\infty 1/(1 + k)^3 = sum_{m=2}^\infty 1/m^3 = q - 1. Therefore, that is correct. Similarly, for j=2, it's sum_{m=3}^\infty 1/m^3 = q - 1 - 1/8. Therefore, that is correct. Therefore, the double sum is sum_{j=1}^\infty [q - H_j^{(3)}]. But this seems to be divergent. However, the original double sum is convergent. Therefore, this suggests that there is a contradiction. Therefore, I must have made a mistake in the first approach?

Wait, let me check the first approach again. The first approach says that the double sum is equal to sum_{n=2}^\infty (n - 1)/n^3. Let me compute this sum:

sum_{n=2}^\infty (n - 1)/n^3 = sum_{n=2}^\infty (1/n^2 - 1/n^3) = (sum_{n=1}^\infty 1/n^2 - 1) - (sum_{n=1}^\infty 1/n^3 - 1) = (p - 1) - (q - 1) = p - q. Therefore, this is correct. Therefore, the double sum is p - q. Therefore, the second approach must be wrong. But why?

Wait, maybe the second approach is correct, but the partial sums are not matching because of some misunderstanding? Let me think. If the double sum is p - q, then according to the second approach, the partial sum S_N = N * sum_{m=N + 1}^\infty 1/m^3 should approach p - q as N approaches infinity. But earlier I thought that S_N tends to zero, but that was incorrect. Let me re-examine that.

Wait, S_N = N * sum_{m=N + 1}^\infty 1/m^3. Let me approximate this sum. For large N, sum_{m=N + 1}^\infty 1/m^3 ≈ ∫_{N}^\infty 1/x^3 dx = 1/(2N^2). Therefore, S_N ≈ N * 1/(2N^2) = 1/(2N) → 0 as N → ∞. But according to the first approach, the double sum is p - q ≈ 0.4428. Therefore, there is a contradiction. Therefore, my second approach must be wrong. But why?

Wait, I think the mistake is that in the second approach, I considered the partial sum up to N as S_N = N * sum_{m=N + 1}^\infty 1/m^3, but this is not correct. Wait, earlier I derived that S_N = N * [ q - sum_{m=1}^N 1/m^3 ] = N * sum_{m=N + 1}^\infty 1/m^3. But according to the first approach, the partial sum up to N is sum_{n=2}^{N + 1} (n - 1)/n^3 + sum_{n=N + 2}^{2N} (2N + 1 - n)/n^3. Therefore, if I compute the limit as N approaches infinity of this partial sum, it should be p - q. However, according to the second approach, the partial sum S_N tends to zero. Therefore, there must be a mistake in the derivation of S_N.

Wait, let me re-examine the derivation of S_N. Earlier, I said that sum_{j=1}^N sum_{k=1}^N 1/(j + k)^3 = sum_{j=1}^N [q - H_j^{(3)}]. But this is not correct. Wait, no. Wait, for each j, sum_{k=1}^N 1/(j + k)^3 is not equal to q - H_j^{(3)}, because q is the infinite sum. Instead, it is equal to sum_{m=j + 1}^{j + N} 1/m^3. Therefore, it is equal to sum_{m=j + 1}^\infty 1/m^3 - sum_{m=j + N + 1}^\infty 1/m^3 = [q - H_j^{(3)}] - [q - H_{j + N}^{(3)}] = H_{j + N}^{(3)} - H_j^{(3)}. Therefore, the double sum up to N is sum_{j=1}^N [H_{j + N}^{(3)} - H_j^{(3)}]. Therefore, this is a telescoping sum!

Ah! This is the key mistake. Earlier, I incorrectly assumed that the inner sum up to N is q - H_j^{(3)}, but actually, it is H_{j + N}^{(3)} - H_j^{(3)}. Therefore, the double sum up to N is sum_{j=1}^N [H_{j + N}^{(3)} - H_j^{(3)}]. This is a telescoping series. Let me write out the terms:

For j=1: H_{N + 1}^{(3)} - H_1^{(3)}

For j=2: H_{N + 2}^{(3)} - H_2^{(3)}

...

For j=N: H_{2N}^{(3)} - H_N^{(3)}

Therefore, summing over j=1 to N:

[H_{N + 1}^{(3)} - H_1^{(3)}] + [H_{N + 2}^{(3)} - H_2^{(3)}] + ... + [H_{2N}^{(3)} - H_N^{(3)}]

This can be rearranged as:

sum_{m=N + 1}^{2N} H_m^{(3)} - sum_{m=1}^N H_m^{(3)}.

But this is equal to [sum_{m=1}^{2N} H_m^{(3)} - sum_{m=1}^N H_m^{(3)}] - sum_{m=1}^N H_m^{(3)}? Wait, no. Wait, the first sum is from m=N + 1 to 2N of H_m^{(3)}, and the second sum is from m=1 to N of H_m^{(3)}. Therefore, the total is sum_{m=N + 1}^{2N} H_m^{(3)} - sum_{m=1}^N H_m^{(3)}.

But this is not a telescoping sum. However, if I consider the original double sum up to N, it is equal to this expression. However, this seems complicated. However, if I take the limit as N approaches infinity, then sum_{m=N + 1}^{2N} H_m^{(3)} - sum_{m=1}^N H_m^{(3)}.

But as N approaches infinity, sum_{m=N + 1}^{2N} H_m^{(3)} ≈ sum_{m=N + 1}^{2N} (ζ(3) + o(1)) = ζ(3) * N * (1/2) + ... = (ζ(3)/2) * N. Similarly, sum_{m=1}^N H_m^{(3)} ≈ ζ(3) * N + ... Therefore, the difference would be approximately (ζ(3)/2) * N - ζ(3) * N = -ζ(3)/2 * N, which tends to -infty. But this contradicts the first approach where the double sum converges to p - q. Therefore, there must be a mistake here.

Wait, this is getting too convoluted. Let me go back. The first approach seems straightforward and gives a finite result, and the numerical check with partial sums suggests that it is correct. Therefore, I think the mistake in the second approach is due to incorrect manipulation of the sums, possibly due to improper handling of the infinite series. Therefore, I will trust the first approach, which gives the result p - q.

Therefore, the answer is p - q. Therefore, the double sum is equal to p - q.

But let me check with another method. Let me recall that in some cases, sums over j and k can be related to known series. For example, consider that:

sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3 = sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^3.

Let me consider that this is equal to sum_{n=2}^\infty (n - 1)/n^3, as before. Which is equal to sum_{n=2}^\infty (1/n^2 - 1/n^3) = (p - 1) - (q - 1) = p - q. Therefore, this seems correct.

Therefore, I think the correct answer is p - q. Therefore, the double sum is equal to p - q.

But to be thorough, let me check with another perspective. Let me recall that in general, for sums of the form sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^s, there might be known formulas. For example, for s = 2, this sum is known to be 2ζ(2) - 1, but I need to check.

Wait, for s = 2, the double sum would be sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^2. Using the same method, this would be sum_{n=2}^\infty (n - 1)/n^2 = sum_{n=2}^\infty (1/n - 1/n^2) = (sum_{n=2}^\infty 1/n) - (sum_{n=2}^\infty 1/n^2). But sum_{n=2}^\infty 1/n diverges, which contradicts the fact that the original double sum for s=2 should converge? Wait, no. Wait, for s=2, the double sum is sum_{j=1}^\infty sum_{k=1}^\infty 1/(j + k)^2. This is similar to the previous case, but for s=2. However, this sum is actually divergent, because for each fixed j, sum_{k=1}^\infty 1/(j + k)^2 ~ ∫_{j}^\infty 1/x^2 dx = 1/j, so the double sum is sum_{j=1}^\infty 1/j, which diverges. Therefore, for s=2, the double sum diverges, which matches with our previous result. However, for s=3, the double sum converges, as we saw. Therefore, in our problem, s=3, and the double sum converges to p - q.

Therefore, this seems consistent. Therefore, I think my initial conclusion is correct. Therefore, the answer is p - q.

**Final Answer**
\boxed{p - q}

To evaluate the double sum

$$
\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3},
$$

we begin by observing that the expression involves a sum over all positive integers $ j $ and $ k $, and the term depends on the sum $ j + k $. This suggests a change of variables to simplify the structure of the sum.

---

### Step 1: Change of Variables

Let us define a new variable $ n = j + k $. For each fixed $ n \geq 2 $, the number of integer pairs $ (j, k) $ such that $ j + k = n $ and $ j, k \geq 1 $ is $ n - 1 $. This is because $ j $ can range from 1 to $ n - 1 $, and $ k = n - j $.

Thus, the double sum can be rewritten as:

$$
\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3} = \sum_{n = 2}^\infty \frac{n - 1}{n^3}.
$$

---

### Step 2: Simplify the Sum

We now simplify the expression:

$$
\sum_{n = 2}^\infty \frac{n - 1}{n^3} = \sum_{n = 2}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right).
$$

This is a telescoping-like sum, and we can split it into two separate sums:

$$
\sum_{n = 2}^\infty \frac{1}{n^2} - \sum_{n = 2}^\infty \frac{1}{n^3}.
$$

Now, recall the definitions:

- $ p = \sum_{n = 1}^\infty \frac{1}{n^2} = \zeta(2) $,
- $ q = \sum_{n = 1}^\infty \frac{1}{n^3} = \zeta(3) $.

So, we can write:

$$
\sum_{n = 2}^\infty \frac{1}{n^2} = p - 1, \quad \sum_{n = 2}^\infty \frac{1}{n^3} = q - 1.
$$

Substituting back, we get:

$$
\sum_{n = 2}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right) = (p - 1) - (q - 1) = p - q.
$$

---

### Final Result

Thus, the original double sum simplifies to:

$$
\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3} = p - q.
$$

---

### Final Answer

$$
\boxed{p - q}
$$

Solution:
We count the number of times $\frac{1}{n^3}$ appears in the sum
\[\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3},\]where $n$ is a fixed positive integer.  (In other words, we are conditioning the sum on $j + k$.)  We get a term of $\frac{1}{n^3}$ each time $j + k = n.$  The pairs $(j,k)$ that work are $(1,n - 1),$ $(2,n - 2),$ $\dots,$ $(n - 1,1),$ for a total of $n - 1$ pairs.  Therefore,
\begin{align*}
\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3} &= \sum_{n = 1}^\infty \frac{n - 1}{n^3} \\
&= \sum_{n = 1}^\infty \left( \frac{n}{n^3} - \frac{1}{n^3} \right) \\
&= \sum_{n = 1}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right) \\
&= \sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{n^3} \\
&= \boxed{p - q}.
\end{align*}