Question

$\displaystyle \mathfrak{ \sum \limits^{ \infty }_{n = 1} \frac{ \zeta(2n + 1)}{ {4}^{n}(2n + 1) } }$​

Answer 1

$\displaystyle \tt \red{{ \sum \limits^{ \infty }_{n = 1} \frac{ \zeta(2n + 1)}{ {4}^{n}(2n + 1) } }}$

Can be written as:-

$\displaystyle \tt \red{{ \sum_{n=1}^{\infty} 4^{- n} \zeta}}$

Pull the constant out of the series:

$\displaystyle \tt \red{{ \color{red}{\sum_{n=1}^{\infty} 4^{- n} \zeta}=\color{red}{\zeta \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^{n}}}}$

$\displaystyle \tt \red{{ \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^{n} } }$ is an infinite geometric series with the first term b=1/4 and the common ratio q=1/4.

By the ratio test, it is convergent.

$\displaystyle\sf \red{Its \: sum \: is \: S=\frac{b}{1-q}=\frac{1}{3}}$

Therefore,

$\displaystyle{\tt \red{ \zeta \color{red}{\left(\sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^{n}\right)}=\zeta \color{red}{\left(\frac{1}{3}\right)}}}$

Hence,

$\displaystyle{\tt \red {\sum_{n=1}^{\infty} 4^{- n} \zeta=\frac{\zeta}{3}}}$

Answer

$\displaystyle \tt \red{{ \sum \limits^{ \infty }_{n = 1} \frac{ \zeta(2n + 1)}{ {4}^{n}(2n + 1)} =\frac{\zeta}{3} }}$