TITLE: How to calculate the value of $\sum\limits_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}$?
QUESTION [7 upvotes]: How do I calculate the value of the series $$\sum_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}= \frac{1}{1\cdot2\cdot3}+\frac{1}{4\cdot5\cdot6}+\frac{1}{7\cdot8\cdot9}+\cdots?$$

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\ds{%
\sum_{k = 0}^{\infty}{1 \over \pars{3k + 1}\pars{3k + 2}\pars{3k + 3}}}} =
{1 \over 27}\sum_{k = 0}^{\infty}
{1 \over \pars{k + 1/3}\pars{k + 2/3}\pars{k + 1}}
\\[5mm] = &
{1 \over 6}\
\underbrace{\sum_{k = 0}^{\infty}\pars{{1 \over k + 1/3} - {1 \over k + 2/3}}}
_{\ds{\underbrace{H_{-1/3} - H_{-2/3}}
_{Euler\ Reflection\ Formula:\\ \ds{=\ \pi\cot\pars{\pi/3} =\root{3}\pi/3}}}}\ +\
{1 \over 6}
\underbrace{\sum_{k = 0}^{\infty}\pars{{1 \over k + 1} - {1 \over k + 2/3}}}
_{\ds{H_{-1/3} - H_{0}}}\quad
\pars{~H_{z}:\ Harmonic\ Number~}
\\[5mm] = &\
{\root{3} \over 18}\,\pi + {1 \over 6}\,H_{-1/3}
\end{align}

Moreover, $\ds{H_{-1/3} = H_{2/3} - 3/2\ \pars{~recurrence~}}$.
  
  $\ds{H_{-1/3} = \overbrace{\braces{3\bracks{1 - \ln\pars{3}}/2 + \root{3}\pi/6}}^{\ds{H_{2/3}}}\ -\ 3/2 =
-3\ln\pars{3}/2 + \root{3}\pi/6}$. The $\ds{H_{2/3}}$ value is
  given in a table. Otherwise, it can be evaluated by means of the
  Gauss Digamma Theorem.

Finally,
\begin{align}
&\bbox[10px,#ffd]{\ds{%
\sum_{k = 0}^{\infty}{1 \over \pars{3k + 1}\pars{3k + 2}\pars{3k + 3}}}} =
{\root{3} \over 18}\,\pi +
{1 \over 6}\bracks{- {3\ln\pars{3} \over 2} + {\root{3} \over 6}\,\pi}
\\[5mm] = &\
\bbx{\root{3}\pi - 3\ln\pars{3} \over 12} \approx 0.1788
\end{align}