What is the proof for this sum of sum generalized harmonic number?

Question

I believe this sum: $$\sum_{m=2}^k\sum_{n=1}^{m-1}(nm)^{-s}$$ to be equal to $$\frac 12((H_k^{s})^2-H_k^{(2s)})$$ where $H_k^{s}$ is the generalized harmonic number. I only discovered this by experimenting on Wolfram.com. Can someone please show a proof of this?

Answer 1

This can be proved straightforwardly by working backwards. Note that the desired sum can be rewritten as $$ S \equiv \sum_{m=2}^k \sum_{n=1}^{m-1} (nm)^{-s} = \sum_{1\le n<m\le k} (nm)^{-s} $$ Starting with \begin{eqnarray} (H_k^s)^2 &=& \left(\sum_{n=1}^k n^{-s}\right)^2 = \sum_{n,m=1}^k (nm)^{-s} \\ &=& \sum_{1\le n<m\le k} (nm)^{-s} + \sum_{1\le n=m \le k} (nm)^{-s} + \sum_{1\le m<n\le k} (nm)^{-s} \\ &=& 2\sum_{1\le n<m\le k} (nm)^{-s} + \sum_{n=1}^k n^{-2s} \\ &=& 2S + H_k^{2s} \end{eqnarray} where we decomposed the sum over the region $(n,m)\in\{1,\ldots,k\}^2$ into three sums, and used the symmetry of $n\leftrightarrow m$ in the last sum. It's then clear that $$S=\frac12 \left((H_k^s)^2-H_k^{2s}\right)$$