# A Beautiful Convergent Series II

In this post I want to generalize the given formula at my second post as follows

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

To prove this , I first prove following useful identity

$\displaystyle\frac{1}{\sin^2x}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}$

PROOF: (Trigonometric Method) Note that

$\displaystyle{\frac{1}{\sin^2x}=\frac{1}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}=\frac{1}{4}\left(\frac{1}{\sin^2\frac{x}{2}}+\frac{1}{\cos^2\frac{x}{2}}\right)=\frac{1}{4}\left(\frac{1}{\sin^2\frac{x}{2}}+\frac{1}{\sin^2\frac{\pi+x}{2}}\right)}$

$\displaystyle=\frac{1}{4^2}\left(\frac{1}{\sin^2\frac{x}{2^2}}+\frac{1}{\sin^2\frac{2\pi+x}{2^2}}+\frac{1}{\sin^2\frac{\pi+x}{2^2}}+\frac{1}{\sin^2\frac{3\pi+x}{2^2}}\right)$

Repeatedly applying $\displaystyle\frac{1}{\sin^2x}=\frac{1}{4}\left(\frac{1}{\sin^2\frac{x}{2}}+\frac{1}{\sin^2\frac{\pi+x}{2}}\right)$, we arrive at the following formula:

$\displaystyle\frac{1}{\sin^2x}=\frac{1}{4^k}\sum_{n=0}^{2^k-1}\frac{1}{\sin^2\frac{x+n\pi}{2^k}}$

but

$\displaystyle{\frac{1}{4^k}\sum_{n=0}^{2^k-1}\frac{1}{\sin^2\frac{x+n\pi}{2^k}}=\frac{1}{4^k}\sum_{n=-2^{k-1}}^{2^{k-1}-1}\frac{1}{\sin^2\frac{x+n\pi}{2^k}}=\lim_{k\to\infty}\sum_{n=-k}^k\frac{1}{(x+n\pi)^2}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}}$

and now note that

$\displaystyle\sum_{n\in\mathbb{Z}}\frac{1}{x+n}=\pi\cot\pi x$

Because,

$\displaystyle{\cot x=-\int\frac{1}{\sin^2x}dx=-\int\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}dx=\sum_{n\in\mathbb{Z}}\frac{1}{x+n\pi}}$

By using this recent identity we can write

$\displaystyle\sum_{n=1}^\infty\frac{(m-1)^n-1}{m^n}\zeta(n+1)$ $\displaystyle=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(m-1)^n-1}{m^n}\frac1{k^{n+1}}$

$\displaystyle=\sum_{k=1}^\infty\frac1k\sum_{n=1}^\infty\left(\frac{(m-1)^n}{m^nk^n}-\frac1{m^nk^n}\right)$

$\displaystyle=\sum_{k=1}^\infty\frac1k\left(\frac{\frac{m-1}{mk}}{1-\frac{m-1}{mk}}-\frac{\frac1{mk}}{1-\frac1{mk}}\right)$

$\displaystyle=\sum_{k=1}^\infty\left(\frac1{\frac1m-k}+\frac1{\frac1m+k-1}\right)$

$\displaystyle=\sum_{k\in\mathbb{Z}}\frac1{\frac1m+k}$ $\displaystyle=\pi\cot\frac{\pi}{m}.$