TITLE: Infinite Series $\sum\limits_{n=1}^{\infty}\frac{(m-1)^n-1}{m^n}\zeta(n+1)$
QUESTION [10 upvotes]: How to prove the following identity?

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

REPLY [4 votes]: Using the identity $(7)$ proven in this answer, we get
$$
\begin{align}
\sum_{n=1}^\infty\frac{(m-1)^n-1}{m^n}\zeta(n+1)
&=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(m-1)^n-1}{m^n}\frac1{k^{n+1}}\\
&=\sum_{k=1}^\infty\frac1k\sum_{n=1}^\infty\left(\frac{(m-1)^n}{m^nk^n}-\frac1{m^nk^n}\right)\\
&=\sum_{k=1}^\infty\frac1k\left(\frac{\frac{m-1}{mk}}{1-\frac{m-1}{mk}}-\frac{\frac1{mk}}{1-\frac1{mk}}\right)\\
&=\sum_{k=1}^\infty\left(\frac1{\frac1m-k}+\frac1{\frac1m+k-1}\right)\\
&=\sum_{k\in\mathbb{Z}}\frac1{\frac1m+k}\\
&=\pi\cot\left(\frac\pi m\right)
\end{align}
$$