How can I prove $\sum\limits_{n=2}^{\infty}\frac{\zeta(n)}{2^n}=\ln(2)$? Playing with values of $\zeta(n)$ in excel I find those two interesting results
$$\sum\limits_{n=2}^{\infty}\frac{\zeta(n)}{2^n}=\ln(2)$$
$$\sum\limits_{n=2}^{\infty}\frac{\zeta(n)}{(-2)^n}=1-\ln(2)$$
How is it obvious? How can I prove it?
 A: Using absolute convergence to justify changing the order of summation together with the sum of a geometric series:
\begin{align}
\sum_{n = 2}^{\infty} \frac{\zeta(n)}{2^n} &= \sum_{n = 2}^{\infty} \frac 1 {2^n} \sum_{k = 1}^{\infty} \frac 1 {k^n} \\
&= \sum_{k = 1}^{\infty} \sum_{n = 2}^{\infty} \frac 1 {(2k)^n} \\
&= \sum_{k = 1}^{\infty} \frac{1}{4k^2} \frac{1}{1 - \frac 1 {2k}} \\
&= \sum_{k = 1}^{\infty} \frac{1}{2k(2k - 1)} \\
&= \log 2
\end{align}
A similar argument works for the alternating sum.
A: $$\sum_{n=2}^{\infty}\frac{\zeta(n)}{2^n}=\sum_{n=2}^{\infty}\frac{1}{2^n}\sum_{k=1}^{\infty}\frac{1}{k^n}=\sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(2k)^n}=\sum_{k=1}^{\infty}\sum_{n=2}^{\infty}\frac{1}{(2k)^n}\\=\sum_{k=1}^{\infty}\frac{1}{4k^2}\sum_{n=0}^{\infty}\frac{1}{(2k)^n}=\sum_{k=1}^{\infty}\frac{1}{4k^2}\frac{1}{1-\frac{1}{2k}}=\sum_{k=1}^{\infty}\frac{1}{2k(2k-1)}\\
=\sum_{k=1}^{\infty}\Big(\frac{1}{2k-1}-\frac{1}{2k}\Big)=1-\frac{1}{2}+\frac{1}{3}-...=\ln(1+1)=\ln 2$$
Analogue for the other sum. Note that interchange of the summations is justified by absolute convergence of the series involved.  
A: You may also exploit the integral representation for $\zeta(s)$:
$$\begin{eqnarray*}\sum_{n\geq 2}\frac{\zeta(n)}{2^n}&=&\int_{0}^{+\infty}\sum_{n\geq 2}\frac{x^{n-1}}{(n-1)!2^n(e^x-1)}\,dx\\&=&\int_{0}^{+\infty}\frac{e^{x/2}-1}{2(e^x-1)}\,dx=\frac{1}{2}\int_{0}^{+\infty}\frac{dx}{e^{x/2}+1}\\&=&\int_{0}^{+\infty}\frac{dz}{e^z+1}=\int_{1}^{+\infty}\frac{du}{u(u+1)}=\int_{0}^{1}\frac{dv}{v+1}=\color{blue}{\log 2}.\end{eqnarray*}$$
