Infinite Series $‎\sum\limits_{n=2}^{\infty}\frac{\zeta(n)}{k^n}$ ‎If $f\left(z \right)=\sum_{n=2}^{\infty}a_{n}z^n$ and $\sum_{n=2}^{\infty}|a_n|$ converges then‎,
$$\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=2}^{\infty}a_n\zeta\left(n\right)‎.$$
‎Since if we set $C:=\sum_{m=2}^{\infty}|a_m|<\infty$‎, ‎then‎
$$\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}|a_m\frac{1}{n^m}|\leq\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}|a_m|\frac{1}{n^2}\leq C\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty‎$$
and ‎by Cauchy's double series theorem‎, ‎we can switch the order of summation‎:
$$\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}a_m\frac{1}{n^m}=\sum_{m=2}^{\infty}a_m\sum_{n=1}^{\infty}\frac{1}{n^m}=\sum_{n=2}^{\infty}a_n\zeta\left(n\right)‎.$$
This shows that $‎\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}=\sum_{n=1}^{\infty}\frac{1}{kn(kn-1)}$.
My Questions:
1) It's obvious that $\sum_{n=1}^{\infty}\frac{1}{2n(2n-1)}=\log(2)$, but how can I evaluate $\sum_{n=1}^{\infty}\frac{1}{3n(3n-1)}$?
2) Is there another method to evaluate $\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}$? 

Answer to Second Question
We remark that
$$\Gamma(z)\zeta(z)=\int_0^\infty \frac{u^{z-1}}{e^u-1}du\tag{1}$$
$$\sum_{n=2}^\infty \frac{t^n}{(n-1)!}=(e^t-1)t \tag{2}$$
$$\psi_0(s+1)=-\gamma+\int_0^1 \frac{1-x^s}{1-x}dx \tag{3}$$ 
where $\psi_0(s)$ is Digamma Function and $\gamma$ is the Euler's Constant.
Then
$$\begin{align*}
\sum_{n=2}^\infty \frac{\zeta(n)}{k^n} &= \sum_{n=2}^\infty \frac{1}{k^n \Gamma(n)}\int_0^\infty \frac{u^{n-1}}{e^u-1}du\\
&=\int_0^\infty\frac{1}{u(e^u-1)}\left( \sum_{n=2}^\infty \frac{1}{(n-1)!}\left(\frac{u}{k}\right)^n\right)du \\
&= \frac{1}{k}\int_0^\infty \frac{e^{\frac{u}{k}}-1}{e^u-1}du \tag{4}
\end{align*}$$
Substituting $t=e^{-u}$, we get
$$
\begin{align*}
\sum_{n=2}^\infty \frac{\zeta(n)}{k^n}&=-\frac{1}{k}\int_0^1 \frac{1-t^{-1/k}}{1-t}dt \\
&= \frac{-1}{k}\left\{ \gamma+\psi_0 \left(1-\frac{1}{k} \right)\right\} \tag{5}
\end{align*}
$$

If $k(\geq 2)$ is an integer, equation $(5)$ can be further simplified using Gauss' Digamma Theorem.
$$\sum_{n=2}^\infty \frac{\zeta(n)}{k^n}=-\frac{\pi}{2k}\cot \left( \frac{\pi}{k}\right)+\frac{\log k}{k}-\frac{1}{k}\sum_{m=1}^{k-1}\cos \left(\frac{2\pi m}{k} \right) \log \left( 2\sin \frac{\pi m}{2}\right)$$
