Cfrgtkky

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.

Since
$$zeta(2n) = frac{1}{(2n-1)!}int_{0}^{+infty}frac{x^{2n-1}}{e^x-1},dx $$
we have:
$$sum_{n=1}^{+infty}frac{zeta(2n)}{2^{2n-1}} = int_{0}^{+infty}frac{sinh(x/2)}{e^x-1},dx =frac{1}{2}int_{0}^{+infty}e^{-x/2},dx = color{red}{1}.$$

$$
begin{align}
sum_{n=1}^inftyfrac{zeta(2n)}{2^{2n-1}}
&=sum_{n=1}^inftysum_{k=1}^inftyfrac2{k^{2n}2^{2n}}tag{1}\
&=2sum_{k=1}^inftysum_{n=1}^inftyfrac1{(4k^2)^n}tag{2}\
&=2sum_{k=1}^inftyfrac1{4k^2-1}tag{3}\
&=sum_{k=1}^inftyleft(frac1{2k-1}-frac1{2k+1}right)tag{4}\[6pt]
&=1tag{5}
end{align}
$$
Explanation:

$(1)$: expand $zeta(2n)=sumlimits_{k=1}^inftyfrac1{k^{2n}}$

$(2)$: change the order of summation

$(3)$: sum of a geometric series

$(4)$: partial fractions

$(5)$: telescoping sum

The Laurent expansion of $cot (z)$ at the origin in terms of the Riemann zeta function is $$ cot (z) = frac{1}{z} - 2 sum_{k=1}^{infty}zeta(2k) frac{z^{2k-1}}{pi^{2k}} , 0 < |z| < pi. $$

Letting $ displaystyle z= frac{pi}{2}$, $$cot left(frac{pi}{2} right) = frac{2}{pi} - frac{2}{pi} sum_{k=1}^{infty} frac{zeta(2k)}{2^{2k-1}}.$$

But $cot left(frac{pi}{2} right)=0$.

Therefore,

$$ sum_{k=1}^{infty} frac{zeta(2k)}{2^{2k-1}} = 1.$$

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begin{align}
&bbox[10px,#ffd]{sum_{n = 1}^{infty}{zetapars{2n} over 2^{2n - 1}}} =
sum_{n = 2}^{infty}{zetapars{n} over 2^{n - 1}} - sum_{n = 1}^{infty}{zetapars{2n + 1} over 2^{2n}}
\[3mm] = &
-sum_{n = 2}^{infty}pars{-1}^{n}zetapars{n}pars{-,half}^{n - 1} -
sum_{n = 1}^{infty}bracks{zetapars{2n + 1} - 1}pars{half}^{2n} -
underbrace{sum_{n = 1}^{infty}pars{half}^{2n}}_{ds{1 over 3}}
\ = &
-bracks{Psipars{1 + z} + gamma}_{ z = -1/2}
\[3mm] & - bracks{%
{1 over 2z} - half,picotpars{pi z} - {1 over 1 - z^{2}} + 1 - gamma - Psipars{1 + z}}_{ z = 1/2} - {1 over 3}
\[8mm] = &
-Psipars{half} - {2 over 3} +
underbrace{Psipars{3 over 2}}_{ds{Psipars{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 $ds{Psipars{z + 1} = Psipars{z} + 1/z}$ and the identities $mathbf{6.3.14}$ and $mathbf{6.3.15}$ in this link.