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.

**Answer**

$$

\begin{align}

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

&=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac2{k^{2n}2^{2n}}\tag{1}\\

&=2\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{(4k^2)^n}\tag{2}\\

&=2\sum_{k=1}^\infty\frac1{4k^2-1}\tag{3}\\

&=\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k+1}\right)\tag{4}\\[6pt]

&=1\tag{5}

\end{align}

$$

Explanation:

$(1)$: expand $\zeta(2n)=\sum\limits_{k=1}^\infty\frac1{k^{2n}}$

$(2)$: change the order of summation

$(3)$: sum of a geometric series

$(4)$: partial fractions

$(5)$: telescoping sum

**Attribution***Source : Link , Question Author : E.H.E , Answer Author : robjohn*