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 $\pi^2/6$, 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 terms where $j + k = n$ is indeed $n - 1$. Therefore, maybe I can rewrite the double sum as a single sum over $n$:

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

Wait, so that simplifies the double sum into a single sum. Let me check that again. If I fix $n = j + k$, then for each $n$, the number of pairs $(j, k)$ is $n - 1$, so the inner sum over $j$ and $k$ becomes $(n - 1) \times \frac{1}{n^3}$. Therefore, the entire double sum becomes $\sum_{n=2}^\infty \frac{n - 1}{n^3}$. Let me write that as:

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

Because $\frac{n - 1}{n^3} = \frac{n}{n^3} - \frac{1}{n^3} = \frac{1}{n^2} - \frac{1}{n^3}$. Therefore, splitting the sum:

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

But $p = \sum_{n=1}^\infty \frac{1}{n^2}$, so $\sum_{n=2}^\infty \frac{1}{n^2} = p - 1$. Similarly, $q = \sum_{n=1}^\infty \frac{1}{n^3}$, so $\sum_{n=2}^\infty \frac{1}{n^3} = q - 1$. 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 check the steps again. The original double sum is over all $j, k \geq 1$, so $j + k$ starts at 2. 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 $\sum_{n=2}^\infty \frac{1}{n^2} - \sum_{n=2}^\infty \frac{1}{n^3}$. Then, since $p = \sum_{n=1}^\infty \frac{1}{n^2}$, subtracting the $n=1$ term which is 1, gives $p - 1$. Similarly for $q$, subtracting 1 gives $q - 1$. Therefore, the difference is $p - 1 - (q - 1) = p - q$. So that seems correct.

But wait, let me check with an example. Let me compute partial sums numerically. Let me take the original double sum and compute a few terms, and compare with $p - q$.

But since I don't have exact values for $p$ and $q$, maybe I can check the logic again. Let me think.

Alternatively, maybe there's a mistake in the substitution. Let me think again. The original double sum is over all $j$ and $k$ starting at 1. So, if I let $n = j + k$, then for each $n \geq 2$, the number of pairs $(j, k)$ with $j + k = n$ is indeed $n - 1$. Therefore, the double sum is equal to $\sum_{n=2}^\infty \frac{n - 1}{n^3}$. Then, splitting into $\sum_{n=2}^\infty \frac{1}{n^2} - \sum_{n=2}^\infty \frac{1}{n^3}$. Then, since $p = \sum_{n=1}^\infty \frac{1}{n^2}$, so $\sum_{n=2}^\infty \frac{1}{n^2} = p - 1$, and similarly for $q$. Therefore, the total is $p - 1 - (q - 1) = p - q$. Therefore, the answer is $p - q$?

But wait, let me check with a concrete example. Let me compute the original double sum for small values and see if it aligns with $p - q$.

But since $p$ and $q$ are infinite sums, maybe I can check the partial sums. Let me take the first few terms of the double sum and compare with the partial sums of $p - q$.

But maybe this is too vague. Alternatively, let me think if there's another way to approach this problem, to verify.

Another approach: Let me consider that the original double sum can be written as $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^3}$. Let me make a substitution. Let me set $m = j + k$. But since $j$ and $k$ are positive integers, $m$ ranges from 2 to infinity. However, for each $m$, the number of pairs $(j, k)$ with $j + k = m$ is $m - 1$, as before. Therefore, the same conclusion as before. Therefore, the double sum is $\sum_{m=2}^\infty \frac{m - 1}{m^3} = \sum_{m=2}^\infty \left( \frac{1}{m^2} - \frac{1}{m^3} \right) = (p - 1) - (q - 1) = p - q$. Therefore, the answer is $p - q$.

But wait, let me check if there's an alternative way to think about this. For example, maybe using integral representations or generating functions?

Alternatively, perhaps I can consider that $\frac{1}{(j + k)^3}$ can be related to some integral. For example, since $\frac{1}{n^3} = \int_0^1 \int_0^1 \int_0^1 x^{n - 1} y^{n - 1} z^{n - 1} dx dy dz$ or something like that? Wait, maybe not. Alternatively, using the integral representation $\frac{1}{n^3} = \int_0^1 \int_0^1 \int_0^1 x^{n - 1} y^{n - 1} z^{n - 1} dx dy dz$? Wait, actually, I know that $\frac{1}{n} = \int_0^1 x^{n - 1} dx$, so maybe integrating multiple times?

Alternatively, maybe using the fact that $\frac{1}{(j + k)^3} = \int_0^1 \int_0^1 \int_0^1 x^{j + k - 1} y^{j + k - 1} z^{j + k - 1} dx dy dz$? Wait, not sure. Maybe this is complicating things.

Alternatively, maybe consider switching the order of summation. Let me think. Let me fix $j$ and sum over $k$, but I don't see an immediate simplification. Alternatively, maybe consider that the double sum is symmetric in $j$ and $k$, so maybe we can write it as something else.

Alternatively, think about the original problem again. If the answer is $p - q$, then maybe that's correct. But let me check with specific numbers. Let me compute the first few terms of the double sum and compare with $p - q$.

But since $p$ and $q$ are known in terms of their series, maybe I can compute partial sums. Let me take the partial sum up to $n = 2$ for the transformed sum. For example, the original double sum up to $j, k = 1$:

When $j = 1, k = 1$, term is $1/(1 + 1)^3 = 1/8$.

When $j = 1, k = 2$, term is $1/(1 + 2)^3 = 1/27$.

Similarly, $j = 2, k = 1$ is also 1/27.

Then for $j = 1, k = 3$: 1/64, and $j = 2, k = 2$: 1/64, $j = 3, k = 1$: 1/64. So for n=4, there are 3 terms each of 1/64, etc.

But if I compute the partial sum up to n=2, which corresponds to m=2, which is 1 term: 1/8. Then for m=3, two terms: 2*(1/27). For m=4, three terms: 3*(1/64). So the partial sum up to m=4 is 1/8 + 2/27 + 3/64. Let me compute that numerically:

1/8 = 0.125

2/27 ≈ 0.07407

3/64 ≈ 0.046875

Total ≈ 0.125 + 0.07407 + 0.046875 ≈ 0.245945

Now, let me compute p - q. Well, p is π²/6 ≈ 1.6449, and q is ζ(3) ≈ 1.2020569. Therefore, p - q ≈ 1.6449 - 1.2020569 ≈ 0.4428. But the partial sum up to m=4 is only ≈ 0.2459, which is less than 0.4428. However, the partial sums will approach that value as m increases. Let me check with more terms.

For m=5, there are 4 terms of 1/5³ = 1/125. So 4/125 = 0.032. Adding to previous total: 0.2459 + 0.032 ≈ 0.2779.

For m=6, 5 terms of 1/6³ = 1/216 ≈ 0.00463. So adding 5/216 ≈ 0.0231. Total ≈ 0.2779 + 0.0231 ≈ 0.301.

m=7: 6 terms of 1/343 ≈ 0.0175. So 6/343 ≈ 0.0175. Total ≈ 0.3185.

m=8: 7 terms of 1/512 ≈ 0.0137. Total ≈ 0.3322.

m=9: 8 terms of 1/729 ≈ 0.01097. Total ≈ 0.3432.

m=10: 9 terms of 1/1000 = 0.009. Total ≈ 0.3522.

Continuing this, the partial sums are approaching around 0.4428? Let me check with m=100. But this is tedious. Alternatively, maybe I can compute the sum $\sum_{n=2}^\infty \frac{n - 1}{n^3}$ numerically.

Alternatively, note that $\sum_{n=2}^\infty \frac{n - 1}{n^3} = \sum_{n=2}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right) = \left( \sum_{n=1}^\infty \frac{1}{n^2} - 1 \right) - \left( \sum_{n=1}^\infty \frac{1}{n^3} - 1 \right) = p - 1 - q + 1 = p - q$. Therefore, mathematically, this seems correct. Therefore, the answer is $p - q$.

But wait, let me check again. Suppose I consider the original double sum:

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

If I make substitution $m = j + k$, then for each $m \geq 2$, the number of pairs $(j, k)$ is $m - 1$, so the sum becomes:

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

Which is equal to:

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

Yes, that seems correct. Therefore, the answer is $p - q$. Therefore, the double sum equals $p - q$.

But wait, let me check another way. Let me consider that the original double sum can be written as:

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

Let me consider changing variables. Let me set $n = j + k$, and $m = j$, so that $k = n - m$. However, since $j$ and $k$ are positive integers, $m$ ranges from 1 to $n - 1$, and $n$ ranges from 2 to infinity. Therefore, the sum becomes:

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

Which is the same as before. Therefore, same conclusion. Therefore, the answer is $p - q$.

But let me check if there's a possible mistake in the problem statement. Wait, the problem says "Find a way to write [...] in terms of p and q." So if my answer is $p - q$, then that's the answer. However, I need to make sure that there isn't a different approach that gives a different answer. Let me think again.

Alternatively, maybe I can use some known series or identities. For example, I know that $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s}$ is related to the Riemann zeta function. In fact, for general $s$, this sum is $\zeta(s - 1) \cdot \zeta(s)$. Wait, is that true?

Wait, let me recall. There is an identity for the double sum $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s}$. Let me check for $s = 3$. If this identity holds, then $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s} = \zeta(s - 1) \cdot \zeta(s)$. Wait, but for $s = 2$, this would be $\zeta(1) \cdot \zeta(2)$, but $\zeta(1)$ diverges. However, in our case, $s = 3$, so $\zeta(2) \cdot \zeta(3)$. But according to our previous calculation, the sum is $p - q$, which is $\zeta(2) - \zeta(3)$. But according to this hypothetical identity, it would be $\zeta(2) \cdot \zeta(3)$. These are different. Therefore, either the identity is wrong, or my previous calculation is wrong.

Wait, so which one is correct? Let me check for $s = 2$. Let me compute $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^2}$. According to my previous method, this would be $\sum_{n=2}^\infty \frac{n - 1}{n^2} = \sum_{n=2}^\infty \left( \frac{1}{n} - \frac{1}{n^2} \right) = \sum_{n=2}^\infty \frac{1}{n} - \sum_{n=2}^\infty \frac{1}{n^2} = \zeta(1) - 1 - (p - 1)$. But $\zeta(1)$ diverges, which is consistent with the fact that the original double sum for $s = 2$ diverges. However, if the identity $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s} = \zeta(s - 1) \cdot \zeta(s)$, then for $s = 3$, it would be $\zeta(2) \cdot \zeta(3)$, which is a finite number. However, according to my previous calculation, it's $\zeta(2) - \zeta(3)$. Therefore, there is a contradiction here. Therefore, one of the approaches is wrong.

Wait, so which one is correct? Let me check with $s = 3$. Let me compute the double sum $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^3}$ using another method.

Alternatively, let me consider that $\frac{1}{(j + k)^3}$ can be expressed as an integral. For example, we know that $\frac{1}{n^3} = \int_0^1 \int_0^1 \int_0^1 x^{n - 1} y^{n - 1} z^{n - 1} dx dy dz$? Wait, actually, $\frac{1}{n} = \int_0^1 x^{n - 1} dx$, so integrating three times:

$$
\frac{1}{n^3} = \int_0^1 \int_0^1 \int_0^1 x^{n - 1} y^{n - 1} z^{n - 1} dx dy dz
$$

But maybe that's not helpful. Alternatively, using the integral representation:

$$
\frac{1}{(j + k)^3} = \int_0^1 \int_0^1 \int_0^1 x^{j + k - 1} y^{j + k - 1} z^{j + k - 1} dx dy dz
$$

But then the double sum becomes:

$$
\sum_{j=1}^\infty \sum_{k=1}^\infty \int_0^1 \int_0^1 \int_0^1 x^{j + k - 1} y^{j + k - 1} z^{j + k - 1} dx dy dz
$$

Assuming we can interchange the sum and the integral (which we can, by dominated convergence theorem or monotone convergence theorem), then:

$$
\int_0^1 \int_0^1 \int_0^1 \sum_{j=1}^\infty \sum_{k=1}^\infty (x y z)^{j + k - 1} dx dy dz
$$

Let me compute the inner sum. Let me note that $j$ and $k$ start at 1, so $j + k - 1$ starts at 1 + 1 - 1 = 1. Let me make substitution $m = j + k - 1$. But maybe it's easier to separate the sums:

$$
\sum_{j=1}^\infty \sum_{k=1}^\infty (x y z)^{j + k - 1} = \sum_{j=1}^\infty (x y z)^{j - 1} \sum_{k=1}^\infty (x y z)^{k}
$$

Wait, no. Wait, $j + k - 1 = (j - 1) + k$. Therefore, the sum becomes:

$$
\sum_{j=1}^\infty (x y z)^{j - 1} \sum_{k=1}^\infty (x y z)^k
$$

Wait, no. Wait, $j + k - 1 = (j - 1) + k$. Therefore, the double sum is:

$$
\sum_{j=1}^\infty \sum_{k=1}^\infty (x y z)^{j + k - 1} = \sum_{j=1}^\infty (x y z)^{j - 1} \sum_{k=1}^\infty (x y z)^k
$$

Wait, no. Wait, if I fix $j$, then the inner sum over $k$ is $\sum_{k=1}^\infty (x y z)^{j + k - 1} = (x y z)^{j - 1} \sum_{k=1}^\infty (x y z)^k$. Therefore, the entire double sum is:

$$
\sum_{j=1}^\infty (x y z)^{j - 1} \sum_{k=1}^\infty (x y z)^k = \sum_{j=1}^\infty (x y z)^{j - 1} \cdot \frac{(x y z)}{1 - x y z}
$$

Assuming $|x y z| < 1$, which it is for $x, y, z \in [0, 1]$. Therefore, this becomes:

$$
\frac{x y z}{1 - x y z} \sum_{j=1}^\infty (x y z)^{j - 1} = \frac{x y z}{1 - x y z} \cdot \frac{1}{1 - x y z} = \frac{x y z}{(1 - x y z)^2}
$$

Therefore, the original double sum becomes:

$$
\int_0^1 \int_0^1 \int_0^1 \frac{x y z}{(1 - x y z)^2} dx dy dz
$$

Hmm, this seems complicated. Maybe I can compute this integral? Let me try. Let me make substitution $u = x y z$. But integrating over three variables might be complex. Alternatively, maybe switch to polar coordinates or something else. Alternatively, note that the integral is symmetric in x, y, z, so maybe we can compute it by integrating over one variable at a time.

Alternatively, let me consider integrating with respect to z first. Let me fix x and y, and integrate over z from 0 to 1:

$$
\int_0^1 \frac{x y z}{(1 - x y z)^2} dz
$$

Let me make substitution $t = x y z$, so $dt = x y dz$, so $dz = dt/(x y)$. When z = 0, t = 0; z = 1, t = x y. Therefore, the integral becomes:

$$
\int_0^{x y} \frac{t}{(1 - t)^2} \cdot \frac{dt}{x y} = \frac{1}{x y} \int_0^{x y} \frac{t}{(1 - t)^2} dt
$$

Let me compute the integral $\int \frac{t}{(1 - t)^2} dt$. Let me use substitution. Let me set $u = 1 - t$, then $du = -dt$, and $t = 1 - u$. Therefore:

$$
\int \frac{1 - u}{u^2} (-du) = \int \frac{1 - u}{u^2} du = \int \left( \frac{1}{u^2} - \frac{1}{u} \right) du = -\frac{1}{u} - \ln |u| + C = -\frac{1}{1 - t} - \ln(1 - t) + C
$$

Therefore, the definite integral from 0 to $x y$ is:

$$
\left[ -\frac{1}{1 - t} - \ln(1 - t) \right]_0^{x y} = \left( -\frac{1}{1 - x y} - \ln(1 - x y) \right) - \left( -1 - \ln 1 \right) = -\frac{1}{1 - x y} - \ln(1 - x y) + 1
$$

Therefore, the integral over z is:

$$
\frac{1}{x y} \left( -\frac{1}{1 - x y} - \ln(1 - x y) + 1 \right ) = \frac{1}{x y} \left( 1 - \frac{1}{1 - x y} - \ln(1 - x y) \right ) = \frac{1}{x y} \left( \frac{ - x y }{1 - x y} - \ln(1 - x y) \right ) = - \frac{1}{1 - x y} - \frac{ \ln(1 - x y) }{x y }
$$

Therefore, the original triple integral becomes:

$$
\int_0^1 \int_0^1 \left( - \frac{1}{1 - x y} - \frac{ \ln(1 - x y) }{x y } \right ) dx dy
$$

This seems very complicated. Maybe I made a mistake in the approach. Alternatively, perhaps this integral is not helpful for our purposes. Let me think again.

Alternatively, maybe I can use the fact that the original double sum equals $p - q$, and check if that aligns with known values. For example, if I take $p = \pi^2/6 \approx 1.6449$, $q = \zeta(3) \approx 1.2020569$, then $p - q \approx 0.4428$. Let me check if this value is consistent with the triple integral I was trying to compute. However, without computing the integral, it's hard to tell. Alternatively, maybe I can check with another approach.

Wait, let me think again. If the original double sum is equal to $p - q$, then for $s = 3$, the sum is $\zeta(2) - \zeta(3)$. However, if there is an identity that says $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s} = \zeta(s - 1) \cdot \zeta(s)$, then for $s = 3$, it would be $\zeta(2) \cdot \zeta(3)$. But according to my previous calculation, it's $\zeta(2) - \zeta(3)$. These are different. Therefore, one of the approaches is wrong.

Wait, let me check for $s = 2$. If I use my previous method, the double sum would be $\sum_{n=2}^\infty \frac{n - 1}{n^2} = \sum_{n=2}^\infty \left( \frac{1}{n} - \frac{1}{n^2} \right ) = \sum_{n=2}^\infty \frac{1}{n} - \sum_{n=2}^\infty \frac{1}{n^2} = \zeta(1) - 1 - (p - 1) = \zeta(1) - p$. But $\zeta(1)$ diverges, which is correct because the double sum for $s = 2$ diverges. However, if the identity $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s} = \zeta(s - 1) \cdot \zeta(s)$ is true, then for $s = 2$, it would be $\zeta(1) \cdot \zeta(2)$, which also diverges, so that's consistent. However, for $s = 3$, according to the identity, it would be $\zeta(2) \cdot \zeta(3)$, which is finite, while according to my previous calculation, it's $\zeta(2) - \zeta(3)$. Therefore, there is a contradiction here. Therefore, one of the approaches is wrong.

Wait, let me check with a reference. I recall that there is a formula for the double sum $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s}$. Let me try to derive it.

Let me consider that $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s} = \sum_{n=2}^\infty \frac{1}{n^s} \sum_{j=1}^{n - 1} 1 = \sum_{n=2}^\infty \frac{n - 1}{n^s} = \sum_{n=1}^\infty \frac{n - 1}{n^s} - \frac{0}{1^s} = \sum_{n=1}^\infty \frac{n - 1}{n^s} = \sum_{n=1}^\infty \left( \frac{1}{n^{s - 1}} - \frac{1}{n^s} \right ) = \zeta(s - 1) - \zeta(s)$.

Wait, this is different from my previous conclusion. Wait, according to this derivation, the double sum is $\zeta(s - 1) - \zeta(s)$. Therefore, for $s = 3$, it would be $\zeta(2) - \zeta(3)$, which matches my previous result. Therefore, the identity is $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s} = \zeta(s - 1) - \zeta(s)$. Therefore, the previous idea that it was $\zeta(s - 1) \cdot \zeta(s)$ was incorrect. Therefore, my original calculation is correct, and the answer is $p - q$.

Therefore, the answer is $\boxed{p - q}$.

But let me check again with the example of $s = 3$. If the double sum is $\zeta(2) - \zeta(3)$, then that is $p - q$, which is what I found earlier. Therefore, the answer is indeed $p - q$. Therefore, my initial calculation was correct, and the confusion with the hypothetical identity was due to a wrong assumption.

Therefore, the final answer is $\boxed{p - q}$.

**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 $\pi^2/6$, 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 terms where $j + k = n$ is indeed $n - 1$. Therefore, maybe I can rewrite the double sum as a single sum over $n$:

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

Wait, so that simplifies the double sum into a single sum. Let me check that again. If I fix $n = j + k$, then for each $n$, the number of pairs $(j, k)$ is $n - 1$, so the inner sum over $j$ and $k$ becomes $(n - 1) \times \frac{1}{n^3}$. Therefore, the entire double sum becomes $\sum_{n=2}^\infty \frac{n - 1}{n^3}$. Let me write that as:

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

Because $\frac{n - 1}{n^3} = \frac{n}{n^3} - \frac{1}{n^3} = \frac{1}{n^2} - \frac{1}{n^3}$. Therefore, splitting the sum:

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

But $p = \sum_{n=1}^\infty \frac{1}{n^2}$, so $\sum_{n=2}^\infty \frac{1}{n^2} = p - 1$. Similarly, $q = \sum_{n=1}^\infty \frac{1}{n^3}$, so $\sum_{n=2}^\infty \frac{1}{n^3} = q - 1$. 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 check the steps again. The original double sum is over all $j, k \geq 1$, so $j + k$ starts at 2. 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 $\sum_{n=2}^\infty \frac{1}{n^2} - \sum_{n=2}^\infty \frac{1}{n^3}$. Then, since $p = \sum_{n=1}^\infty \frac{1}{n^2}$, subtracting the $n=1$ term which is 1, gives $p - 1$. Similarly for $q$, subtracting 1 gives $q - 1$. Therefore, the difference is $p - 1 - (q - 1) = p - q$. So that seems correct.

But wait, let me check with an example. Let me compute partial sums numerically. Let me take the original double sum and compute a few terms, and compare with $p - q$.

But since I don't have exact values for $p$ and $q$, maybe I can check the logic again. Let me think.

Alternatively, maybe there's a mistake in the substitution. Let me think again. The original double sum is over all $j$ and $k$ starting at 1. So, if I let $n = j + k$, then for each $n \geq 2$, the number of pairs $(j, k)$ with $j + k = n$ is indeed $n - 1$. Therefore, the double sum is equal to $\sum_{n=2}^\infty \frac{n - 1}{n^3}$. Then, splitting into $\sum_{n=2}^\infty \frac{1}{n^2} - \sum_{n=2}^\infty \frac{1}{n^3}$. Then, since $p = \sum_{n=1}^\infty \frac{1}{n^2}$, so $\sum_{n=2}^\infty \frac{1}{n^2} = p - 1$, and similarly for $q$. Therefore, the total is $p - 1 - (q - 1) = p - q$. Therefore, the answer is $p - q$?

But wait, let me check with a concrete example. Let me compute the original double sum for small values and see if it aligns with $p - q$.

But since $p$ and $q$ are infinite sums, maybe I can check the partial sums. Let me take the first few terms of the double sum and compare with the partial sums of $p - q$.

But maybe this is too vague. Alternatively, let me think if there's another way to approach this problem, to verify.

Another approach: Let me consider that the original double sum can be written as $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^3}$. Let me make a substitution. Let me set $m = j + k$. But since $j$ and $k$ are positive integers, $m$ ranges from 2 to infinity. However, for each $m$, the number of pairs $(j, k)$ with $j + k = m$ is $m - 1$, as before. Therefore, the same conclusion as before. Therefore, the double sum is $\sum_{m=2}^\infty \frac{m - 1}{m^3} = \sum_{m=2}^\infty \left( \frac{1}{m^2} - \frac{1}{m^3} \right) = (p - 1) - (q - 1) = p - q$. Therefore, the answer is $p - q$.

But wait, let me check if there's an alternative way to think about this. For example, maybe using integral representations or generating functions?

Alternatively, perhaps I can consider that $\frac{1}{(j + k)^3}$ can be related to some integral. For example, since $\frac{1}{n^3} = \int_0^1 \int_0^1 \int_0^1 x^{n - 1} y^{n - 1} z^{n - 1} dx dy dz$ or something like that? Wait, maybe not. Alternatively, using the integral representation $\frac{1}{n^3} = \int_0^1 \int_0^1 \int_0^1 x^{n - 1} y^{n - 1} z^{n - 1} dx dy dz$? Wait, actually, I know that $\frac{1}{n} = \int_0^1 x^{n - 1} dx$, so maybe integrating multiple times?

Alternatively, maybe using the fact that $\frac{1}{(j + k)^3} = \int_0^1 \int_0^1 \int_0^1 x^{j + k - 1} y^{j + k - 1} z^{j + k - 1} dx dy dz$? Wait, not sure. Maybe this is complicating things.

Alternatively, maybe consider switching the order of summation. Let me think. Let me fix $j$ and sum over $k$, but I don't see an immediate simplification. Alternatively, maybe consider that the double sum is symmetric in $j$ and $k$, so maybe we can write it as something else.

Alternatively, think about the original problem again. If the answer is $p - q$, then maybe that's correct. But let me check with specific numbers. Let me compute the first few terms of the double sum and compare with $p - q$.

But since $p$ and $q$ are known in terms of their series, maybe I can compute partial sums. Let me take the partial sum up to $n = 2$ for the transformed sum. For example, the original double sum up to $j, k = 1$:

When $j = 1, k = 1$, term is $1/(1 + 1)^3 = 1/8$.

When $j = 1, k = 2$, term is $1/(1 + 2)^3 = 1/27$.

Similarly, $j = 2, k = 1$ is also 1/27.

Then for $j = 1, k = 3$: 1/64, and $j = 2, k = 2$: 1/64, $j = 3, k = 1$: 1/64. So for n=4, there are 3 terms each of 1/64, etc.

But if I compute the partial sum up to n=2, which corresponds to m=2, which is 1 term: 1/8. Then for m=3, two terms: 2*(1/27). For m=4, three terms: 3*(1/64). So the partial sum up to m=4 is 1/8 + 2/27 + 3/64. Let me compute that numerically:

1/8 = 0.125

2/27 ≈ 0.07407

3/64 ≈ 0.046875

Total ≈ 0.125 + 0.07407 + 0.046875 ≈ 0.245945

Now, let me compute p - q. Well, p is π²/6 ≈ 1.6449, and q is ζ(3) ≈ 1.2020569. Therefore, p - q ≈ 1.6449 - 1.2020569 ≈ 0.4428. But the partial sum up to m=4 is only ≈ 0.2459, which is less than 0.4428. However, the partial sums will approach that value as m increases. Let me check with more terms.

For m=5, there are 4 terms of 1/5³ = 1/125. So 4/125 = 0.032. Adding to previous total: 0.2459 + 0.032 ≈ 0.2779.

For m=6, 5 terms of 1/6³ = 1/216 ≈ 0.00463. So adding 5/216 ≈ 0.0231. Total ≈ 0.2779 + 0.0231 ≈ 0.301.

m=7: 6 terms of 1/343 ≈ 0.0175. So 6/343 ≈ 0.0175. Total ≈ 0.3185.

m=8: 7 terms of 1/512 ≈ 0.0137. Total ≈ 0.3322.

m=9: 8 terms of 1/729 ≈ 0.01097. Total ≈ 0.3432.

m=10: 9 terms of 1/1000 = 0.009. Total ≈ 0.3522.

Continuing this, the partial sums are approaching around 0.4428? Let me check with m=100. But this is tedious. Alternatively, maybe I can compute the sum $\sum_{n=2}^\infty \frac{n - 1}{n^3}$ numerically.

Alternatively, note that $\sum_{n=2}^\infty \frac{n - 1}{n^3} = \sum_{n=2}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right) = \left( \sum_{n=1}^\infty \frac{1}{n^2} - 1 \right) - \left( \sum_{n=1}^\infty \frac{1}{n^3} - 1 \right) = p - 1 - q + 1 = p - q$. Therefore, mathematically, this seems correct. Therefore, the answer is $p - q$.

But wait, let me check again. Suppose I consider the original double sum:

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

If I make substitution $m = j + k$, then for each $m \geq 2$, the number of pairs $(j, k)$ is $m - 1$, so the sum becomes:

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

Which is equal to:

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

Yes, that seems correct. Therefore, the answer is $p - q$. Therefore, the double sum equals $p - q$.

But wait, let me check another way. Let me consider that the original double sum can be written as:

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

Let me consider changing variables. Let me set $n = j + k$, and $m = j$, so that $k = n - m$. However, since $j$ and $k$ are positive integers, $m$ ranges from 1 to $n - 1$, and $n$ ranges from 2 to infinity. Therefore, the sum becomes:

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

Which is the same as before. Therefore, same conclusion. Therefore, the answer is $p - q$.

But let me check if there's a possible mistake in the problem statement. Wait, the problem says "Find a way to write [...] in terms of p and q." So if my answer is $p - q$, then that's the answer. However, I need to make sure that there isn't a different approach that gives a different answer. Let me think again.

Alternatively, maybe I can use some known series or identities. For example, I know that $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s}$ is related to the Riemann zeta function. In fact, for general $s$, this sum is $\zeta(s - 1) \cdot \zeta(s)$. Wait, is that true?

Wait, let me recall. There is an identity for the double sum $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s}$. Let me check for $s = 3$. If this identity holds, then $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s} = \zeta(s - 1) \cdot \zeta(s)$. Wait, but for $s = 2$, this would be $\zeta(1) \cdot \zeta(2)$, but $\zeta(1)$ diverges. However, in our case, $s = 3$, so $\zeta(2) \cdot \zeta(3)$. But according to our previous calculation, the sum is $p - q$, which is $\zeta(2) - \zeta(3)$. But according to this hypothetical identity, it would be $\zeta(2) \cdot \zeta(3)$. These are different. Therefore, either the identity is wrong, or my previous calculation is wrong.

Wait, so which one is correct? Let me check for $s = 2$. Let me compute $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^2}$. According to my previous method, this would be $\sum_{n=2}^\infty \frac{n - 1}{n^2} = \sum_{n=2}^\infty \left( \frac{1}{n} - \frac{1}{n^2} \right) = \sum_{n=2}^\infty \frac{1}{n} - \sum_{n=2}^\infty \frac{1}{n^2} = \zeta(1) - 1 - (p - 1)$. But $\zeta(1)$ diverges, which is consistent with the fact that the original double sum for $s = 2$ diverges. However, if the identity $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s} = \zeta(s - 1) \cdot \zeta(s)$, then for $s = 3$, it would be $\zeta(2) \cdot \zeta(3)$, which is a finite number. However, according to my previous calculation, it's $\zeta(2) - \zeta(3)$. Therefore, there is a contradiction here. Therefore, one of the approaches is wrong.

Wait, so which one is correct? Let me check with $s = 3$. Let me compute the double sum $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^3}$ using another method.

Alternatively, let me consider that $\frac{1}{(j + k)^3}$ can be expressed as an integral. For example, we know that $\frac{1}{n^3} = \int_0^1 \int_0^1 \int_0^1 x^{n - 1} y^{n - 1} z^{n - 1} dx dy dz$? Wait, actually, $\frac{1}{n} = \int_0^1 x^{n - 1} dx$, so integrating three times:

$$
\frac{1}{n^3} = \int_0^1 \int_0^1 \int_0^1 x^{n - 1} y^{n - 1} z^{n - 1} dx dy dz
$$

But maybe that's not helpful. Alternatively, using the integral representation:

$$
\frac{1}{(j + k)^3} = \int_0^1 \int_0^1 \int_0^1 x^{j + k - 1} y^{j + k - 1} z^{j + k - 1} dx dy dz
$$

But then the double sum becomes:

$$
\sum_{j=1}^\infty \sum_{k=1}^\infty \int_0^1 \int_0^1 \int_0^1 x^{j + k - 1} y^{j + k - 1} z^{j + k - 1} dx dy dz
$$

Assuming we can interchange the sum and the integral (which we can, by dominated convergence theorem or monotone convergence theorem), then:

$$
\int_0^1 \int_0^1 \int_0^1 \sum_{j=1}^\infty \sum_{k=1}^\infty (x y z)^{j + k - 1} dx dy dz
$$

Let me compute the inner sum. Let me note that $j$ and $k$ start at 1, so $j + k - 1$ starts at 1 + 1 - 1 = 1. Let me make substitution $m = j + k - 1$. But maybe it's easier to separate the sums:

$$
\sum_{j=1}^\infty \sum_{k=1}^\infty (x y z)^{j + k - 1} = \sum_{j=1}^\infty (x y z)^{j - 1} \sum_{k=1}^\infty (x y z)^{k}
$$

Wait, no. Wait, $j + k - 1 = (j - 1) + k$. Therefore, the sum becomes:

$$
\sum_{j=1}^\infty (x y z)^{j - 1} \sum_{k=1}^\infty (x y z)^k
$$

Wait, no. Wait, $j + k - 1 = (j - 1) + k$. Therefore, the double sum is:

$$
\sum_{j=1}^\infty \sum_{k=1}^\infty (x y z)^{j + k - 1} = \sum_{j=1}^\infty (x y z)^{j - 1} \sum_{k=1}^\infty (x y z)^k
$$

Wait, no. Wait, if I fix $j$, then the inner sum over $k$ is $\sum_{k=1}^\infty (x y z)^{j + k - 1} = (x y z)^{j - 1} \sum_{k=1}^\infty (x y z)^k$. Therefore, the entire double sum is:

$$
\sum_{j=1}^\infty (x y z)^{j - 1} \sum_{k=1}^\infty (x y z)^k = \sum_{j=1}^\infty (x y z)^{j - 1} \cdot \frac{(x y z)}{1 - x y z}
$$

Assuming $|x y z| < 1$, which it is for $x, y, z \in [0, 1]$. Therefore, this becomes:

$$
\frac{x y z}{1 - x y z} \sum_{j=1}^\infty (x y z)^{j - 1} = \frac{x y z}{1 - x y z} \cdot \frac{1}{1 - x y z} = \frac{x y z}{(1 - x y z)^2}
$$

Therefore, the original double sum becomes:

$$
\int_0^1 \int_0^1 \int_0^1 \frac{x y z}{(1 - x y z)^2} dx dy dz
$$

Hmm, this seems complicated. Maybe I can compute this integral? Let me try. Let me make substitution $u = x y z$. But integrating over three variables might be complex. Alternatively, maybe switch to polar coordinates or something else. Alternatively, note that the integral is symmetric in x, y, z, so maybe we can compute it by integrating over one variable at a time.

Alternatively, let me consider integrating with respect to z first. Let me fix x and y, and integrate over z from 0 to 1:

$$
\int_0^1 \frac{x y z}{(1 - x y z)^2} dz
$$

Let me make substitution $t = x y z$, so $dt = x y dz$, so $dz = dt/(x y)$. When z = 0, t = 0; z = 1, t = x y. Therefore, the integral becomes:

$$
\int_0^{x y} \frac{t}{(1 - t)^2} \cdot \frac{dt}{x y} = \frac{1}{x y} \int_0^{x y} \frac{t}{(1 - t)^2} dt
$$

Let me compute the integral $\int \frac{t}{(1 - t)^2} dt$. Let me use substitution. Let me set $u = 1 - t$, then $du = -dt$, and $t = 1 - u$. Therefore:

$$
\int \frac{1 - u}{u^2} (-du) = \int \frac{1 - u}{u^2} du = \int \left( \frac{1}{u^2} - \frac{1}{u} \right) du = -\frac{1}{u} - \ln |u| + C = -\frac{1}{1 - t} - \ln(1 - t) + C
$$

Therefore, the definite integral from 0 to $x y$ is:

$$
\left[ -\frac{1}{1 - t} - \ln(1 - t) \right]_0^{x y} = \left( -\frac{1}{1 - x y} - \ln(1 - x y) \right) - \left( -1 - \ln 1 \right) = -\frac{1}{1 - x y} - \ln(1 - x y) + 1
$$

Therefore, the integral over z is:

$$
\frac{1}{x y} \left( -\frac{1}{1 - x y} - \ln(1 - x y) + 1 \right ) = \frac{1}{x y} \left( 1 - \frac{1}{1 - x y} - \ln(1 - x y) \right ) = \frac{1}{x y} \left( \frac{ - x y }{1 - x y} - \ln(1 - x y) \right ) = - \frac{1}{1 - x y} - \frac{ \ln(1 - x y) }{x y }
$$

Therefore, the original triple integral becomes:

$$
\int_0^1 \int_0^1 \left( - \frac{1}{1 - x y} - \frac{ \ln(1 - x y) }{x y } \right ) dx dy
$$

This seems very complicated. Maybe I made a mistake in the approach. Alternatively, perhaps this integral is not helpful for our purposes. Let me think again.

Alternatively, maybe I can use the fact that the original double sum equals $p - q$, and check if that aligns with known values. For example, if I take $p = \pi^2/6 \approx 1.6449$, $q = \zeta(3) \approx 1.2020569$, then $p - q \approx 0.4428$. Let me check if this value is consistent with the triple integral I was trying to compute. However, without computing the integral, it's hard to tell. Alternatively, maybe I can check with another approach.

Wait, let me think again. If the original double sum is equal to $p - q$, then for $s = 3$, the sum is $\zeta(2) - \zeta(3)$. However, if there is an identity that says $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s} = \zeta(s - 1) \cdot \zeta(s)$, then for $s = 3$, it would be $\zeta(2) \cdot \zeta(3)$. But according to my previous calculation, it's $\zeta(2) - \zeta(3)$. These are different. Therefore, one of the approaches is wrong.

Wait, let me check for $s = 2$. If I use my previous method, the double sum would be $\sum_{n=2}^\infty \frac{n - 1}{n^2} = \sum_{n=2}^\infty \left( \frac{1}{n} - \frac{1}{n^2} \right ) = \sum_{n=2}^\infty \frac{1}{n} - \sum_{n=2}^\infty \frac{1}{n^2} = \zeta(1) - 1 - (p - 1) = \zeta(1) - p$. But $\zeta(1)$ diverges, which is correct because the double sum for $s = 2$ diverges. However, if the identity $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s} = \zeta(s - 1) \cdot \zeta(s)$ is true, then for $s = 2$, it would be $\zeta(1) \cdot \zeta(2)$, which also diverges, so that's consistent. However, for $s = 3$, according to the identity, it would be $\zeta(2) \cdot \zeta(3)$, which is finite, while according to my previous calculation, it's $\zeta(2) - \zeta(3)$. Therefore, there is a contradiction here. Therefore, one of the approaches is wrong.

Wait, let me check with a reference. I recall that there is a formula for the double sum $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s}$. Let me try to derive it.

Let me consider that $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s} = \sum_{n=2}^\infty \frac{1}{n^s} \sum_{j=1}^{n - 1} 1 = \sum_{n=2}^\infty \frac{n - 1}{n^s} = \sum_{n=1}^\infty \frac{n - 1}{n^s} - \frac{0}{1^s} = \sum_{n=1}^\infty \frac{n - 1}{n^s} = \sum_{n=1}^\infty \left( \frac{1}{n^{s - 1}} - \frac{1}{n^s} \right ) = \zeta(s - 1) - \zeta(s)$.

Wait, this is different from my previous conclusion. Wait, according to this derivation, the double sum is $\zeta(s - 1) - \zeta(s)$. Therefore, for $s = 3$, it would be $\zeta(2) - \zeta(3)$, which matches my previous result. Therefore, the identity is $\sum_{j=1}^\infty \sum_{k=1}^\infty \frac{1}{(j + k)^s} = \zeta(s - 1) - \zeta(s)$. Therefore, the previous idea that it was $\zeta(s - 1) \cdot \zeta(s)$ was incorrect. Therefore, my original calculation is correct, and the answer is $p - q$.

Therefore, the answer is $\boxed{p - q}$.

But let me check again with the example of $s = 3$. If the double sum is $\zeta(2) - \zeta(3)$, then that is $p - q$, which is what I found earlier. Therefore, the answer is indeed $p - q$. Therefore, my initial calculation was correct, and the confusion with the hypothetical identity was due to a wrong assumption.

Therefore, the final answer is $\boxed{p - q}$.

**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*}