# 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.

• Compare with the answers here, and here. – Dietrich Burde Jul 4 '16 at 12:54
• @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 $\frac{A}{3k}$? – Transcendental Jul 4 '16 at 12:58
• Look at the second link how to do it with three terms. See here for almost the same question. – Dietrich Burde Jul 4 '16 at 12:59
• @DietrichBurde Oh yes... I got so catched on that harmonic series that i thought i couldn't solve this...Thank you. – Transcendental Jul 4 '16 at 13:02

$$S =\sum_{k=1}^{\infty} { \frac{1}{3k(3k-2)(3k-1)} }=\sum_{k=1}^{\infty}\frac{1}{2} \left(\frac{1}{3k}-\frac{2}{3k-1}+\frac{1}{3k-2}\right)$$

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

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

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

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.)

• Yes, this is almost the same answer as here. – Dietrich Burde Jul 4 '16 at 13:02
• Good to hear there's precedent. I must have started writing before I saw your post above - thanks in any case. – πr8 Jul 4 '16 at 13:05

Euler's beta function gives an approach with a straightforward generalization:

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


$\ds{\gamma}$ 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}