Question

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

$\color{blue}\boxed{\boxed{ \boxed{\boxed{\begin{matrix}\displaystyle \tt \red{{ \sum \limits^{ \infty }_{n = 1} \frac{ \zeta(2n + 1)}{ {4}^{n}(2n + 1) } }} \\ \red{\sf{Can \: be \: written \: as:-}} \\ \displaystyle \tt \red{{ \sum_{n=1}^{\infty} 4^{- n} \zeta}} \\ \red{\small \sf{Pull \: the \: constant \: out \: of \: the \: series:}} \\ \displaystyle \tt \red{{ \color{red}{\sum_{n=1}^{\infty} 4^{- n} \zeta}=\color{red}{\zeta \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^{n}}}} \\ \displaystyle \tt \red{{ \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^{n} } } \sf \red{ is \: an \: infinite \: geometric \: series} \\ \sf \red{ with \: the \: first \: term \: b= \frac{1}{4} \: and \: the} \\ \sf \red{common \: ratio \: q= \frac{1}{4} } \\ \sf \red{By \: the \: ratio \: test, \: it \: is \: convergent} \\ \displaystyle\sf \red{Its \: sum \: is \: S=\frac{b}{1-q}=\frac{1}{3}} \\ \sf \red{Therefore,} \\ \displaystyle{\tt \red{ \zeta \color{red}{\left(\sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^{n}\right)}=\zeta \color{red}{\left(\frac{1}{3}\right)}}} \\ \sf \red{Hence, }\\ \displaystyle{\tt \red {\sum_{n=1}^{\infty} 4^{- n} \zeta=\frac{\zeta}{3}}} \end{matrix}}}}}$