Define \[p = \sum_{k = 1}^\infty \frac{1}{k^2} \quad \text{a...
Prompt
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 $q.$
Answer guidance
Correct answer: 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*}
