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How to prove the following identity?

$$\sum_{n=1}^{\infty}\frac{(m-1)^n-1}{m^n}\zeta(n+1)=\pi\cot\left(\frac{\pi}{m}\right)$$

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Using the identity $(7)$ proven in this answer, we get $$ \begin{align} \sum_{n=1}^\infty\frac{(m-1)^n-1}{m^n}\zeta(n+1) &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(m-1)^n-1}{m^n}\frac1{k^{n+1}}\\ &=\sum_{k=1}^\infty\frac1k\sum_{n=1}^\infty\left(\frac{(m-1)^n}{m^nk^n}-\frac1{m^nk^n}\right)\\ &=\sum_{k=1}^\infty\frac1k\left(\frac{\frac{m-1}{mk}}{1-\frac{m-1}{mk}}-\frac{\frac1{mk}}{1-\frac1{mk}}\right)\\ &=\sum_{k=1}^\infty\left(\frac1{\frac1m-k}+\frac1{\frac1m+k-1}\right)\\ &=\sum_{k\in\mathbb{Z}}\frac1{\frac1m+k}\\ &=\pi\cot\left(\frac\pi m\right) \end{align} $$

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  • $\begingroup$ @Artin: It is one of my favorites :-) $\endgroup$ – robjohn Aug 25 '13 at 14:40
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For $m>1$ we have : \begin{align} \sum_{n=1}^{\infty}\frac{(m-1)^n-1}{m^n}\zeta(n+1)&=\sum_{n=1}^{\infty}\sum_{k=1}^\infty\frac{(m-1)^n-1}{m^n}\frac 1{k^{n+1}}\\ &=\sum_{k=1}^\infty\frac 1k \sum_{n=1}^{\infty}\left(\frac{m-1}{m\,k}\right)^n-\left(\frac{1}{m\,k}\right)^n\\ &=\sum_{k=1}^\infty\frac 1k \left[\frac{\left(\frac{m-1}{m\,k}\right)}{1-\left(\frac{m-1}{m\,k}\right)}-\frac{\left(\frac{1}{m\,k}\right)}{1-\left(\frac 1{m\,k}\right)}\right]\\ &=\sum_{k=1}^\infty\frac 1k \left[\frac{m-1}{mk-m+1}-\frac{1}{m\,k-1}\right]\\ &=\sum_{k=1}^\infty\frac 1k \left[\frac{1-\frac 1m}{k-\left(1-\frac 1m\right)}-\frac{\frac 1m}{k-\frac 1m}\right]\\ &=\psi\left(1-\frac 1m\right)-\psi\left(\frac 1m\right)\\ &=\pi\,\cot\left(\frac{\pi}{m}\right)\\ \end{align}

using the series for digamma and the reflection formula to conclude.

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  • $\begingroup$ You are welcome @Artin ! $\endgroup$ – Raymond Manzoni Aug 25 '13 at 10:28

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