TITLE: Infinite Series $‎\sum\limits_{n=2}^{\infty}\frac{\zeta(n)}{k^n}$
QUESTION [41 upvotes]: ‎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 $\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}$?

REPLY [6 votes]: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\color{#c00000}{\sum_{n = 2}^{\infty}{\zeta\pars{n} \over k^{n}}}
=\sum_{n = 2}^{\infty}{1 \over k^{n}}
\sum_{\ell = 0}^{\infty}{1 \over \pars{\ell + 1}^{n}}
=\sum_{\ell = 0}^{\infty}\sum_{n = 2}^{\infty}{1 \over \bracks{k\pars{\ell + 1}}^{n}}
=\sum_{\ell = 0}^{\infty}
{\bracks{k\pars{\ell + 1}}^{-2} \over 1 - \bracks{k\pars{\ell + 1}}^{-1}}
\\[3mm]&={1 \over k^{2}}\sum_{\ell = 0}^{\infty}
{1 \over \pars{\ell + 1}\pars{\ell + 1 - 1/k}}
={1 \over k^{2}}\,{\Psi\pars{1} - \Psi\pars{1 - 1/k} \over 1 - \pars{1 - 1/k}}
\\[3mm]&=\color{#c00000}{%
{1 \over k}\bracks{\Psi\pars{1} - \Psi\pars{1 - {1 \over k}}}}
\end{align}
where $\ds{\Psi\pars{z}}$ is the Digamma Function.

$$
\color{#00f}{\large\sum_{n = 2}^{\infty}{\zeta\pars{n} \over k^{n}}
=-\,{1 \over k}\,\bracks{\gamma + \Psi\pars{1 - {1 \over k}}}}\,,\qquad\verts{k} > 1
$$
  since $\ds{\Psi\pars{1} = -\gamma.\quad}$ $\gamma$ is the
  Euler-Mascheroni Constant. 

$\large\tt\mbox{Just one of your questions !!!}$:
\begin{align}
\sum_{n = 1}^{\infty}{1 \over 3n\pars{3n - 1}}&
={1 \over 9}\sum_{n = 1}^{\infty}{1 \over n\pars{n - 1/3}}
={1 \over 9}\sum_{n = 0}^{\infty}{1 \over \pars{n + 1}\pars{n + 2/3}}
={1 \over 9}\,{\Psi\pars{1} - \Psi\pars{2/3} \over 1 - 2/3}
\\[3mm]&={1 \over 3}\,\bracks{\Psi\pars{1} - \Psi\pars{{2 \over 3}}}
\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\pars{1}
\end{align}
Also,

$$
\Psi\pars{{2 \over 3}} = -\gamma + {\root{3} \over 6}\,\pi - {3 \over 2}\,\ln\pars{3}
$$ 
  By replacing in $\pars{1}$, we'll find:

$$
\color{#00f}{\large\sum_{n = 1}^{\infty}{1 \over 3n\pars{3n - 1}}
={1 \over 18}\bracks{9\ln\pars{3} - \root{3}\pi}} \approx 0.2470
$$