<s>[INST] How can prove that $\sum_{n=1}^{\infty }\frac{\zeta (2n)}{4^{n-1}}(1-\frac{1}{4^n})=\frac{\pi }{2}$ 
$$\zeta (2)(1-\frac{1}{4})+\frac{\zeta (4)}{4}(1-\frac{1}{4^2})+\frac{\zeta (6)}{4^2}(1-\frac{1}{4^3})+...=\frac{\pi }{2}$$

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.
 [/INST] Since, 
$\begin{align}\sum_{k=1}^\infty\zeta(2k)\,x^{2k} &= \sum_{k=1}^\infty\sum_{n=1}^\infty\frac{x^{2k}}{n^{2k}}\\
&=\sum_{n=1}^\infty\frac{\frac{x^2}{n^2}}{1-\frac{x^2}{n^2}}\\
&=\sum_{n=1}^\infty\frac{x^2}{n^2-x^2}\\
&=-\frac{x}{2}\sum_{n=1}^\infty\left(\frac{1}{x+n} + \frac{1}{x-n}\right)\\
&=-\frac{x}{2}\left(\pi\cot(\pi x)-\frac1x\right)\\
&=\frac{1}{2}\left(1-\pi x\cot(\pi x)\right)
\end{align}$
Plug in the particular cases $x = \dfrac{1}{2}$ and $x = \dfrac{1}{4}$ in the series.
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