I proved the following equation
$$\sum_{n\geq 1} \frac{H^{(k)}_n}{n^k}\, = \frac{\zeta{(2k)}+\zeta^{2}(k)}{2}$$
We define
$$H^{(k)}_n=\sum_{m= 1}^n \frac{1}{m^k}$$
I am looking forward to seeing what approaches would you use .
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You case is a special case of the more general case B(p,q). – Mhenni Benghorbal Aug 09 '13 at 19:42
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Yes, but the general case has no closed form ! The evaluation of the multiple integrals might be impossible – Zaid Alyafeai Aug 09 '13 at 19:45
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Just to be clear, you are asking for others to prove this identity, right, to see if they approach it the same way or some novel and surprising way? – Thomas Andrews Aug 09 '13 at 19:46
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@ThomasAndrews exactly, please tell me if this is against the rules here ! – Zaid Alyafeai Aug 09 '13 at 19:48
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2No, that sort of question is okay, but it is always good to include an actual "question" in your question, to make it clear what you are asking. "How would you prove this?" "I am looking forward,..." sounds like a side-musing. – Thomas Andrews Aug 09 '13 at 19:53
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Could you work your proof a little more? – draks ... Aug 09 '13 at 20:02
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This is a special case of a more general formula .Proving it would take several papers ! – Zaid Alyafeai Aug 09 '13 at 20:16
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Isn't this just $$\begin{align} \zeta^2(k) &= \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{1}{n^km^k} \\&= 2\sum_{1\leq m\leq n} \frac{1}{(nm)^k} - \sum_{m=1}^\infty\frac{1}{(n^k)^2}\\ &=2\sum_{n=1}^\infty \frac{1}{n^k}\sum_{m=1}^n \frac{1}{m^k} - \zeta(2k)\\ &=2\sum_{n=1}^\infty \frac{H_n^{(k)}}{n^k} - \zeta(2k) \end{align} $$?
You could also figure that:
$$\zeta^2(k) = 2\sum_{n=1}^\infty \frac{H_{n-1}^{(k)}}{n^k} +\zeta(2k)$$ Every rearrangement here can be done due to absolute convergence.
Thomas Andrews
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Hey, if you are interested , can you solve the following http://math.stackexchange.com/questions/459973/generalized-alternating-harmonic-sum-sum-n-geq-1-frac-left1-frac12-f?rq=1 for p is odd . – Zaid Alyafeai Aug 09 '13 at 20:19