How to calculate the value of $\sum\limits_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}$? How do I calculate the value of the series $$\sum_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}= \frac{1}{1\cdot2\cdot3}+\frac{1}{4\cdot5\cdot6}+\frac{1}{7\cdot8\cdot9}+\cdots?$$
 A: The answer is $\frac{\pi\sqrt3}{12}-\frac{\ln3}{4}.$
See the similar problem (problem 2) here:
http://www.imc-math.org.uk/imc2010/imc2010-day1-solutions.pdf
A: You can calculate the partial sum using
$$
\frac{1}{(3k+1)(3k+2)(3k+3)}=\frac{1}{2}\frac{1}{3k+1}-\frac{1}{3k+2}+\frac{1}{2}\frac{1}{3k+3}
$$
A: $\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}\,}\,}
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\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}
A: By making use of the integral
$$\int_{0}^{1} \frac{(1-x)^2}{1-x^3} \, dx = \frac{1}{2} \, \left(\frac{\pi}{\sqrt{3}} - \ln 3 \right)$$
one can take the following path.
\begin{align}
S &= \sum_{k=0}^{\infty} \frac{1}{(3k+1)(3k+2)(3k+3)} \\
&= \sum_{k=0}^{\infty} \frac{\Gamma(3k+1)}{\Gamma(3k+4)} = \frac{1}{2} \, \sum_{k=0}^{\infty} B(3, 3k+1),
\end{align}
where $B(n,m)$ is the Beta function, which leads to
\begin{align}
S &= \frac{1}{2} \, \sum_{k=0}^{\infty} \, \int_{0}^{1} t^{2} \, (1-t)^{3k} \, dt \\
&= \frac{1}{2} \, \int_{0}^{1} \frac{t^{2} \, dt}{1- (1-t)^{3}} \\
&= \frac{1}{2} \, \int_{0}^{1} \frac{(1-x)^{2} \, dx}{1- x^3} \hspace{15mm} x = 1 - t \\
&= \frac{1}{4} \, \left(\frac{\pi}{\sqrt{3}} - \ln 3 \right). 
\end{align}
