$\sum_{s=2}^{\infty}\frac{\zeta(s)}{4^s}=\frac{\log 64 - \pi}{8}$

Question

In my files found: $$\sum_{s=2}^{\infty}\frac{\zeta(s)}{4^s}=\frac{\log 64 - \pi}{8}$$

And I assumed that this is something obvious. Now trying to prove this but can not. Could you please suggest any simple way of proving this?

Answer 1

$$\sum\limits_{s=2}^{+\infty}\frac 1{4^s}\sum\limits_{n=1}^{+\infty}\frac 1{n^s}=\sum\limits_{n=1}^{+\infty}\sum\limits_{s=2}^{+\infty}\frac 1{(4n)^s}=\frac 14\sum\limits_{n=1}^{+\infty}\frac 1{n(4n-1)}=\frac 14\log 8-\frac {\pi}8$$