NICE INFINITE SUM INVOLVING THE RIEMANN ZETA FUNCTION

This is a quick post sharing the proof of a nice infinite sum involving the Riemann Zeta function that I came across yesterday, namely:

$S=\sum_{n=2}^{\infty} \frac{\zeta(n)}{k^n}$ (1)

Rewrite (1) as

$\sum_{n=2}^{\infty} \frac{\zeta(n)}{k^n}=\sum_{n=2}^{\infty} \frac{1}{k^n}\sum_{m=1}^{\infty}\frac{1}{m^n}$

$=\sum_{n=2}^{\infty} \sum_{m=1}^{\infty}\frac{1}{(km)^n}$

$=\sum_{m=1}^{\infty}\sum_{n=2}^{\infty} \frac{1}{(km)^n}$

let n=r+2 , then

$=\sum_{m=1}^{\infty}\frac{1}{(km)^2}\sum_{r=0}^{\infty} \frac{1}{(km)^r}$

$=\sum_{m=1}^{\infty}\frac{1}{(km)^2} \frac{1}{\left(1-\frac{1}{km}\right)}$

$=\sum_{m=1}^{\infty} \frac{1}{\left((km)^2-km\right)}$

$=\sum_{m=1}^{\infty}\frac{1}{km} \frac{1}{\left(km-1\right)}$ (2)

$=\frac{1}{k^2}\sum_{m=1}^{\infty}\frac{1}{m} \frac{1}{\left(m-\frac{1}{k}\right)}$

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

$\boxed{\sum_{n=2}^{\infty} \frac{\zeta(n)}{k^n}=-\frac{1}{k}\left(\gamma+\psi\left(1-\frac{1}{k} \right) \right)}$ (3)

Now for particular values we can calculate for k=2 , and from (1), (2) and (3) we have

$\sum_{n=2}^{\infty} \frac{\zeta(n)}{2^n}=\sum_{n=1}^{\infty}\frac{1}{2n} \frac{1}{\left(2n-1\right)}$

$\sum_{n=2}^{\infty} \frac{\zeta(n)}{2^n}=-\frac{1}{2}\left(\gamma+\psi\left(\frac{1}{2} \right) \right)$

The value of $\psi\left(\frac{1}{2} \right)$ can be found here, therefore

$\sum_{n=2}^{\infty} \frac{\zeta(n)}{2^n}=-\frac{1}{2}\left(\gamma-\gamma -2 \ln 2\right) \right)$

$\boxed{\sum_{n=2}^{\infty} \frac{\zeta(n)}{2^n}= \ln 2}$ (4)

$\boxed{\sum_{n=1}^{\infty}\frac{1}{2n} \frac{1}{\left(2n-1\right)}= \ln 2}$ (5)

For k=3

$\sum_{n=2}^{\infty} \frac{\zeta(n)}{3^n}=\sum_{n=1}^{\infty}\frac{1}{3n} \frac{1}{\left(3n-1\right)}$

$\sum_{n=2}^{\infty} \frac{\zeta(n)}{3^n}=-\frac{1}{3}\left(\gamma+\psi\left(\frac{2}{3} \right) \right)$

$\sum_{n=2}^{\infty} \frac{\zeta(n)}{3^n}=-\frac{1}{3}\left(\gamma-\gamma+\frac{\pi}{2 \sqrt{3}} -\frac{3 \ln 3}{2}\right)$

$\boxed{\sum_{n=2}^{\infty} \frac{\zeta(n)}{3^n}=\frac{ \ln 3}{2}-\frac{\pi}{6 \sqrt{3}}}$ (6)

$\boxed{\sum_{n=1}^{\infty}\frac{1}{3n} \frac{1}{\left(3n-1\right)}=\frac{ \ln 3}{2}-\frac{\pi}{6 \sqrt{3}}}$ (7)

For k=4

$\sum_{n=2}^{\infty} \frac{\zeta(n)}{4^n}=\sum_{n=1}^{\infty}\frac{1}{4n} \frac{1}{\left(4n-1\right)}$

$\sum_{n=2}^{\infty} \frac{\zeta(n)}{4^n}=-\frac{1}{4}\left(\gamma+\psi\left(\frac{3}{4} \right) \right)$

$\sum_{n=2}^{\infty} \frac{\zeta(n)}{4^n}=-\frac{1}{4}\left(\gamma-\gamma-3 \ln 2 +\frac{\pi}{2} \right)$

$\boxed{\sum_{n=2}^{\infty} \frac{\zeta(n)}{4^n}=\frac{3 \ln 2 }{4}-\frac{\pi}{8} \right)}$ (8)

$\boxed{\sum_{n=1}^{\infty}\frac{1}{4n} \frac{1}{\left(4n-1\right)}=\frac{3 \ln 2 }{4}-\frac{\pi}{8} \right)}$ (9)

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