TITLE: How can prove that $\sum_{n=1}^{\infty }\frac{\zeta (2n)}{4^{n-1}}(1-\frac{1}{4^n})=\frac{\pi }{2}$
QUESTION [11 upvotes]: $$\zeta (2)(1-\frac{1}{4})+\frac{\zeta (4)}{4}(1-\frac{1}{4^2})+\frac{\zeta (6)}{4^2}(1-\frac{1}{4^3})+...=\frac{\pi }{2}$$

The WolframAlph couldn't recognize the closed-form which is $\pi/2$ when I gave the series,

so I used the WolframAlph again to compute many terms of infinite series.

I think that the WolfarmAlph cannot say the value is $\pi/2$,So we need to prove it.

REPLY [7 votes]: $$\zeta(2n) = \frac{1}{\Gamma(2n)}\int_{0}^{+\infty}\frac{x^{2n-1}}{e^x-1}\,dx $$
so we have:
$$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{4^{n-1}} =\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2^{2n-2}}= 2\int_{0}^{\infty}\frac{\sinh(x/2)}{e^x-1}\,dx =\int_{0}^{\infty}e^{-x/2}\,dx = 2$$
also we have :
$$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{4^{2n-1}} = \int_{0}^{\infty}\frac{\sinh(x/4)}{e^x-1}\,dx =\frac12\int_{0}^{\infty}\frac{1}{e^{\frac{3x}{4}}+e^\frac{x}{4}}\,dx =2-\frac{\pi}{2}$$
last integral we can slove using $u=e^\frac{x}{4}$
putting all together :
$$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{4^{n-1}} -\sum_{n=1}^{\infty}\frac{\zeta(2n)}{4^{2n-1}} =\sum_{n=1}^{\infty}\frac{\zeta(2n)}{4^{n-1}}(1-\frac{1}{4^{n}})=\frac{\pi}{2}$$