Define
\[p = \sum_{k = 1}^\infty \frac{1}{k^2} \quad \text{and} \quad q = \sum_{k = 1}^\infty \frac{1}{k^3}.\]Find a way to write
\[\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3}\]in terms of $p$ and $q.$
I notice that the double sum involves adding the indices $j$ and $k$, so I wonder if there is a way to change the order of summation and make it easier to work with. To do that, I need to find the range of possible values for $j + k$, and how many times each value appears in the sum. The smallest possible value for $j + k$ is $2$, when $j = k = 1$. For a fixed value of $j + k$, say $n$, how many pairs of positive integers $(j, k)$ satisfy $j + k = n$? Well, one possibility is $j = 1$ and $k = n - 1$. Another is $j = 2$ and $k = n - 2$. And so on, until $j = n - 1$ and $k = 1$. So there are $n - 1$ such pairs. This means I can rewrite the double sum as a single sum over $n$, where I multiply each term by the number of times it appears, which is $n - 1$:
\[\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3} = \sum_{n = 2}^\infty (n - 1) \frac{1}{n^3}.\] Now I try to simplify this expression by splitting it into two sums and using some properties of series. I have
\[\sum_{n = 2}^\infty (n - 1) \frac{1}{n^3} = \sum_{n = 2}^\infty \frac{1}{n^2} - \sum_{n = 2}^\infty \frac{1}{n^3}.\] The first sum looks very similar to the definition of $p$, except it is missing the term $1/1^2$. I can add and subtract that term to make it match:
\[\sum_{n = 2}^\infty \frac{1}{n^2} = \sum_{n = 1}^\infty \frac{1}{n^2} - \frac{1}{1^2} = p - 1.\] The second sum also looks similar to the definition of $q$, except it is missing the term $1/1^3$. I can do the same trick as before and get:
\[\sum_{n = 2}^\infty \frac{1}{n^3} = \sum_{n = 1}^\infty \frac{1}{n^3} - \frac{1}{1^3} = q - 1.\] Putting these together, I get
\[\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3} = p - 1 - (q - 1) = p - q.\] 