jmc

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, $$\begin{array}{rl}\sum _{j=1}^{\infty }\sum _{k=1}^{\infty }\frac{1}{(j+k{)}^3}& =\sum _{n=1}^{\infty }\frac{n-1}{n^3}\\ & =\sum _{n=1}^{\infty }\left(\frac{n}{n^3}-\frac{1}{n^3}\right)\\ & =\sum _{n=1}^{\infty }\left(\frac{1}{n^2}-\frac{1}{n^3}\right)\\ & =\sum _{n=1}^{\infty }\frac{1}{n^2}-\sum _{n=1}^{\infty }\frac{1}{n^3}\\ & =\boxed{p-q}.\end{array}$$