$$\frac{\zeta (2)}{2}+\frac{\zeta (3)}{2^2}+\frac{\zeta (4)}{2^3}+\frac{\zeta (5)}{2^4}+...=\log(4)$$
I tried to prove it, but the problem with the odd zeta terms so that I don't have a function which deals with odd terms.I used the WolframAlpha but it couldn't recognize the calculated value is $\log(4)$.
Note that $$\sum_{k\geqslant2}\frac{\zeta(k)}{2^{k-1}}=\sum_{k\geqslant2}\sum_{n\geqslant1}\frac{1}{n^k2^{k-1}}=\sum_{n\geqslant1}\frac1n\sum_{k\geqslant2}\frac{1}{(2n)^{k-1}}=\sum_{n\geqslant1}\frac1n\frac1{2n}\frac1{1-\frac1{2n}},$$ hence $$\sum_{k\geqslant2}\frac{\zeta(k)}{2^{k-1}}=\sum_{n\geqslant1}\frac2{2n(2n-1)}=2\sum_{n\geqslant1}\frac1{2n-1}-\frac1{2n}=2\sum_{i\geqslant1}\frac{(-1)^{i-1}}i=2\log2.$$ Which is exactly one of the approaches explained to you à propos your former question (do you study the answers you receive, to understand them?).
Edit: More generally, for every integer $\ell\geqslant2$, $$\sum_{k\geqslant2}\frac{\zeta(k)}{\ell^{k-1}}=\sum_z(1-z)\log(1-z),$$ where the sum in the RHS is over every $\ell$th root of unity $z\ne1$. For example, $$\sum_{k\geqslant2}\frac{\zeta(k)}{3^{k-1}}=\frac32\log3-\sqrt3\frac\pi6\approx0.741,$$ and $$\sum_{k\geqslant2}\frac{\zeta(k)}{4^{k-1}}=3\log2-\frac\pi2\approx0.509.$$
