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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.

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    $\begingroup$ Compare with the answers here, and here. $\endgroup$ – Dietrich Burde Jul 4 '16 at 12:54
  • $\begingroup$ @DietrichBurde Interesting. The accepted answer does split the expression into partial fractions and then ( i assume ) uses the power series representation of $\frac{1}{3n+1}$ but what do i do with the term $\frac{A}{3k}$? $\endgroup$ – Transcendental Jul 4 '16 at 12:58
  • $\begingroup$ Look at the second link how to do it with three terms. See here for almost the same question. $\endgroup$ – Dietrich Burde Jul 4 '16 at 12:59
  • $\begingroup$ @DietrichBurde Oh yes... I got so catched on that harmonic series that i thought i couldn't solve this...Thank you. $\endgroup$ – Transcendental Jul 4 '16 at 13:02
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$$S =\sum_{k=1}^{\infty} { \frac{1}{3k(3k-2)(3k-1)} }=\sum_{k=1}^{\infty}\frac{1}{2} \left(\frac{1}{3k}-\frac{2}{3k-1}+\frac{1}{3k-2}\right)$$

$$=\sum_{k\ge1}\frac{1}{2}\int_0^1(x^{3k-1}-2x^{3k-2}+x^{3k-3})\,dx=\sum_{k\ge1}\frac{1}{2}\int_0^1x^{3k-3}(x^2-2x+1)\,dx$$

$$=\frac{1}{2}\int_0^1\sum_{k\ge1}(1-x)^2x^{3k-3}\,dx=\frac{1}{2}\int_0^1\frac{(1-x)^2}{1-x^3}\,dx$$

$$=\frac{1}{2}\int_0^1\frac{1-x}{1+x+x^2}\,dx$$

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.)

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    $\begingroup$ Yes, this is almost the same answer as here. $\endgroup$ – Dietrich Burde Jul 4 '16 at 13:02
  • $\begingroup$ Good to hear there's precedent. I must have started writing before I saw your post above - thanks in any case. $\endgroup$ – πr8 Jul 4 '16 at 13:05
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Euler's beta function gives an approach with a straightforward generalization:

$$\begin{eqnarray*}\sum_{k\geq 1}\frac{(3k-3)!}{(3k)!}=\sum_{k\geq 1}\frac{\Gamma(3k-2)}{\Gamma(3k+1)}&=&\frac{1}{\Gamma(3)}\sum_{k\geq 1}B(3k-2,3)\\&=&\frac{1}{\Gamma(3)}\int_{0}^{1}\sum_{k\geq 1}(1-t)^2 t^{3k-3}\,dt\\&=&\frac{1}{2}\int_{0}^{1}\frac{(1-t)^2}{1-t^3}\,dt\\[0.2cm]&=&\color{red}{\frac{\pi\sqrt{3}-3\log(3)}{12}}.\end{eqnarray*}$$

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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \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}

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