TITLE: How to prove that $\frac{\zeta(2) }{2}+\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}+\frac{\zeta (8)}{2^7}+\cdots=1$?
QUESTION [87 upvotes]: How can one prove this identity? 

$$\frac{\zeta(2) }{2}+\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}+\frac{\zeta (8)}{2^7}+\cdots=1$$


There is a formula for $\zeta$ values at even integers, but it involves Bernoulli numbers; simply plugging it in does not appear to be an efficient approach.

REPLY [13 votes]: $\newcommand{\angles}[1]{\left\langle{#1}\right\rangle}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}{\zeta\pars{2n} \over 2^{2n - 1}}} =
\sum_{n = 2}^{\infty}{\zeta\pars{n} \over 2^{n - 1}} - \sum_{n = 1}^{\infty}{\zeta\pars{2n + 1} \over 2^{2n}}
\\[3mm] = &\
-\sum_{n = 2}^{\infty}\pars{-1}^{n}\zeta\pars{n}\pars{-\,\half}^{n - 1} -
\sum_{n = 1}^{\infty}\bracks{\zeta\pars{2n + 1} - 1}\pars{\half}^{2n}\ -\
\underbrace{\sum_{n = 1}^{\infty}\pars{\half}^{2n}}_{\ds{1 \over 3}}
\\ = &\
-\bracks{\Psi\pars{1 + z} + \gamma}_{\ z\ =\ -1/2}
\\[3mm] & - \bracks{%
{1 \over 2z} - \half\,\pi\cot\pars{\pi z} - {1 \over 1 - z^{2}} + 1 - \gamma - \Psi\pars{1 + z}}_{\ z\ =\ 1/2} - {1 \over 3}
\\[8mm] = &\
-\Psi\pars{\half} - {2 \over 3}\ +\ 
\underbrace{\Psi\pars{3 \over 2}}_{\ds{\Psi\pars{1/2} + 1/\pars{1/2}}} -
{1 \over 3} = \color{#f00}{1}
\end{align}

$\Psi$ and $\gamma$ are the Digamma function and the Euler-Mascheroni constant, respectively. We used the Digamma Recurrence Formula $\ds{\Psi\pars{z + 1} = \Psi\pars{z} + 1/z}$ and the identities $\mathbf{6.3.14}$ and $\mathbf{6.3.15}$ in this link.