Calculating $\sum_{k=1}^{\infty} { \frac{(3k-3)!}{(3k)!} }$ I was studying series ( accent on power series ) and came across this one:
$$S =\sum_{k=1}^{\infty} { \frac{(3k-3)!}{(3k)!} }$$
To be precise, the problem originally states:

Inspect the series convergence for $\frac{1}{2*3} + \frac{1}{4*5*6} + \frac{1}{7*8*9}+ \dots$ and calculate series sum.

Of course the sum above can be written as:
$$S =\sum_{k=1}^{\infty} { \frac{1}{3k(3k-2)(3k-1)} }$$
Which seems like a neat thing to separate into partial fractions..But wait. $\sum_{k=0}^\infty\frac{A}{3k}$ diverges for any $A\in \mathbb{R}\backslash \{0\}. $
And of course, this is the slippery slope that causes sum of divergent series to be convergent. But how do I calculate the sum of this series. I hoped that taking a peek at the final result will give me some ideas but when i saw 
$$S = \frac{1}{12}(\sqrt{3}\pi - 3\ln{3})$$
I decided to give up.  Can anyone give me a hint on where to start with this.
 A: $$S =\sum_{k=1}^{\infty} { \frac{1}{3k(3k-2)(3k-1)} }=\sum_{k=1}^{\infty}\frac{1}{2} \left(\frac{1}{3k}-\frac{2}{3k-1}+\frac{1}{3k-2}\right)$$
$$=\sum_{k\ge1}\frac{1}{2}\int_0^1(x^{3k-1}-2x^{3k-2}+x^{3k-3})\,dx=\sum_{k\ge1}\frac{1}{2}\int_0^1x^{3k-3}(x^2-2x+1)\,dx$$
$$=\frac{1}{2}\int_0^1\sum_{k\ge1}(1-x)^2x^{3k-3}\,dx=\frac{1}{2}\int_0^1\frac{(1-x)^2}{1-x^3}\,dx$$
$$=\frac{1}{2}\int_0^1\frac{1-x}{1+x+x^2}\,dx$$
This can then be finished off through a sequence of standard techniques for dealing with such integrals ($x=\frac{-1}{2}+\frac{\sqrt{3}}{2}\tan\theta$, etc.)
A: Euler's beta function gives an approach with a straightforward generalization:
$$\begin{eqnarray*}\sum_{k\geq 1}\frac{(3k-3)!}{(3k)!}=\sum_{k\geq 1}\frac{\Gamma(3k-2)}{\Gamma(3k+1)}&=&\frac{1}{\Gamma(3)}\sum_{k\geq 1}B(3k-2,3)\\&=&\frac{1}{\Gamma(3)}\int_{0}^{1}\sum_{k\geq 1}(1-t)^2 t^{3k-3}\,dt\\&=&\frac{1}{2}\int_{0}^{1}\frac{(1-t)^2}{1-t^3}\,dt\\[0.2cm]&=&\color{red}{\frac{\pi\sqrt{3}-3\log(3)}{12}}.\end{eqnarray*}$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{S} & =
\sum_{k = 1}^{\infty}{\pars{3k - 3}! \over \pars{3k}!} =
\sum_{k = 0}^{\infty}{1 \over \pars{3k + 3}\pars{3k + 2}\pars{3k + 1}}
\\[3mm] & =
\sum_{k = 0}\bracks{%
{1 \over 6\pars{k + 1}} + {1 \over 2\pars{3k + 1}} - {1 \over 3k + 2}}
\\[3mm] & =
{1 \over 6}\sum_{k = 0}^{\infty}\pars{{1 \over k + 1} - {1 \over k + 1/3}} +
{1 \over 3}\sum_{k = 0}^{\infty}\pars{{1 \over k + 1/3} - {1 \over k + 2/3}}
\\[3mm] & =
{1 \over 6}\bracks{\Psi\pars{1 \over 3} - \Psi\pars{1}} +
{1 \over 3}\bracks{\Psi\pars{2 \over 3} - \Psi\pars{1 \over 3}}
\\[3mm] & =
\fbox{$\ds{\
\color{#f00}{-\,{1 \over 6}\,\Psi\pars{1 \over 3} + {1 \over 6}\,\gamma +
{1 \over 3}\,\Psi\pars{2 \over 3}}\ }$}
\end{align}

$\ds{\gamma}$ and $\ds{\Psi}$ are the Euler-Mascheroni constant and the Digamma Function, respectively. Note that $\ds{\Psi\pars{1} = -\gamma}$.


Also $\ds{\pars{~\mbox{see}\ \mathbf{8.366}.6.\ \mbox{and}\ \mathbf{8.366}.7.\ \mbox{in  Gradshteyn & Rizhik, page}\ 905,\ 7^{\mathrm{th}}\ \mbox{ed.}~}}$,
\begin{align}
&\left\lbrace\begin{array}{rcl}
\ds{\Psi\pars{1 \over 3}} & \ds{=} &
\ds{-\gamma - {\root{3} \over 6}\,\pi - {3 \over 2}\,\ln\pars{3}}
\\[2mm]
\ds{\Psi\pars{2 \over 3}} & \ds{=} &
\ds{-\gamma + {\root{3} \over 6}\,\pi - {3 \over 2}\,\ln\pars{3}}
\end{array}\right.
\\[5mm]
\mbox{which leads to}\quad &\
\color{#f00}{S} =
\sum_{k = 1}^{\infty}{\pars{3k - 3}! \over \pars{3k}!} =
\color{#f00}{{1 \over 12}\bracks{\root{3}\pi - 3\ln\pars{3}}} \approx 0.1788
\end{align}
