<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] We have:
$$\zeta(2n)\left(1-\frac{1}{4^n}\right)=\sum_{k=0}^{+\infty}\frac{1}{(2k+1)^{2n}}\tag{1}$$
hence:
$$\sum_{n=1}^{+\infty}\frac{\zeta(2n)}{4^{n-1}}\left(1-\frac{1}{4^n}\right)=4\sum_{n=1}^{+\infty}\sum_{k=0}^{+\infty}\frac{1}{(4k+2)^{2n}}=4\sum_{k=0}^{+\infty}\frac{1}{(4k+2)^2-1}\tag{2}$$
and the result follows from:
$$\sum_{k=0}^{+\infty}\frac{1}{(4k+2)^2-1}=\frac{1}{2}\sum_{k=0}^{+\infty}\left(\frac{1}{4k+1}-\frac{1}{4k+3}\right)=\frac{1}{2}\arctan 1 = \color{red}{\frac{\pi}{8}}.\tag{3}$$
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