Problem
Evaluate
\[S=\sum_{n=1}^\infty\frac{1}{n^2}\sum_{k=1}^n\frac{1}{k^2}.\]Solution
The formula for summation by parts tells us that if we define
\[B_n=\sum_{k=0}^n b_k\]so that for every $n>0$, $b_n=B_n-B_{n-1}$ we have
\[\sum_{n=0}^N a_nb_n=a_Nb_N-\sum_{n=0}^{N-1}B_n(a_{n+1}-a_n).\]In order to evaluate $S$, consider the partial sum
\[S_N=\sum_{n=1}^N\frac{1}{n^2}\sum_{k=1}^n\frac{1}{k^2}\]and let $b_n=1/n^2$ and $a_n=\sum_{k=1}^n 1/k^2.$ Applying summation by parts, we acquire an alternate formula for $S_N$,
\[\begin{align*}S_N &=\left(\sum_{k=1}^N\frac{1}{k^2} \right )^2-\sum_{n=1}^{N-1}\frac{1}{(n+1)^2}\sum_{k=1}^n\frac{1}{k^2} \\ &=\left(\sum_{k=1}^N\frac{1}{k^2} \right )^2-\sum_{n=1}^{N-1}\frac{1}{(n+1)^2}\left( -\frac{1}{(n+1)^2}+\sum_{k=1}^{n+1}\frac{1}{k^2}\right ) \\ &= \left(\sum_{k=1}^N\frac{1}{k^2} \right )^2-\sum_{n=1}^{N-1}\frac{1}{(n+1)^2}\sum_{k=1}^{n+1}\frac{1}{k^2}+\sum_{n=1}^{N-1}\frac{1}{(n+1)^4}.\end{align*}\]In the limit as $N\to\infty$ this becomes
\[S=\zeta^2(2)-(S-1)+(\zeta(4)-1)\]so $S=7\pi^4/360.$ In general one can show that
\[\sum_{n=1}^\infty\frac{1}{n^m}\sum_{k=1}^n\frac{1}{k^m}=\frac{1}{2}\zeta^2(m)+\frac{1}{2}\zeta(2m).\]