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

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

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