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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?

tyobrien
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1 Answers1

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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 $$S=\frac12 \left((H_k^s)^2-H_k^{2s}\right)$$

Leo
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  • Thanks @Leo! I assume this would work for any series of numbers, right? Not just the natural numbers? Say if i used the primes instead of the naturals and instead of the harmonic series, I would get the prime zeta function. Would all of this still check out? – tyobrien Jul 02 '15 at 01:32
  • Yup, since this identity is true as long as your original sum can be written as $\sum_{n<m: (n,m)\in D^2} (nm)^{-s}$, irrespective of what the domain $D$ actually is (be it natural numbers or primes) – Leo Jul 02 '15 at 01:42
  • How would I go about finding formulae in terms of the harmonic number like this for similar expressions with more sums? For example, how would I express $$\sum_{q=3}^k \sum_{m=2}^{q-1} \sum_{n=1}^{m-1} (nmq)^{-s}$$ or $$\sum_{t=4}^k \sum_{q=3}^{t-1} \sum_{m=2}^{q-1} \sum_{n=1}^{m-1} (nmqt)^{-s}$$ in terms of harmonic numbers like we did with two sums? – tyobrien Jul 03 '15 at 02:46
  • @joriki maybe you could give me some advice about my comment above since you already probably know where I'm going with this. – tyobrien Jul 03 '15 at 06:09
  • BTW, I made another post specifically for this question. http://math.stackexchange.com/questions/1348409/how-do-i-calculate-these-sum-of-sum-expressions-in-terms-of-the-generalized-harm – tyobrien Jul 03 '15 at 18:31