# A question related to compute infinite summation

How to solve this: $$1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot5\cdot8}{6\cdot12\cdot18}+\cdots$$ So, the $n^{th}$ term of the sum can be written as $$a_n=\frac{2\cdot5\cdot8\cdots (2+3(n-1))}{6\cdot12\cdot18\cdots (6n)} = \frac{2\cdot5\cdot8 \cdots (2+3(n-1))}{6^n (n!)}$$ So the above sum can be written as $$1+\sum_{n=1}^\infty a_n .$$ Now how should I proceed to solve this. Please help me.

• $a_n$ can be written in terms of a lot of factorials, given some work. – Simply Beautiful Art Sep 3 '16 at 22:36
• That I have included in my question after your comment. That I have already tried but didn't help anymore. – Sachchidanand Prasad Sep 3 '16 at 22:44
• No, I meant the numerator can be worked on as well. Its like the double factorial, but a triple factorial. – Simply Beautiful Art Sep 3 '16 at 22:45
• Okay let me give a try. – Sachchidanand Prasad Sep 3 '16 at 22:46

more than a hint...

Consider the binomial expansion of $$(1-\frac 12)^{-\frac 23}$$

• Aw you sneaky binomial expansions in disguise. Note "generalized binomial expansion" to be more specific. – Simply Beautiful Art Sep 3 '16 at 22:49
• Taking the first $50$ terms of the sum and using the Inverse Symbolic Calculator points at essentially the same result – Henry Sep 3 '16 at 22:54
• @David But can you please tell me the idea behind your approach. – Sachchidanand Prasad Sep 3 '16 at 23:00
• @SachchidanandPrasad. The number of terms on the top and bottom increasing by one each time, and forming an arithmetic progression, is an indication that the Binomial Theorem for fractional powers is in play. Look at the fourth term: the sequence $2,5,8$ can only be generated if the successive factors are $-\frac 23$, $-\frac 53$ and $-\frac 83$, so that means the power of the bracket must be $-\frac 23$. Then it's just a matter of factorising these terms out to see what's left. I hope this helps. – David Quinn Sep 4 '16 at 8:54
• Okay Thank you so much sir. – Sachchidanand Prasad Sep 5 '16 at 12:53

Consider $$1+\sum^\infty_{n=1}\prod^n_{i=1}\frac{2+3(i-1)}{6i}=1+\sum^\infty_{n=1}\prod^n_{i=1}\left(\frac{1}{2}-\frac{1}{6i}\right)=1+\sum^\infty_{n=1}2^{-n}\prod^n_{i=1}\left(1-\frac{1}{3i}\right)$$ Product above can be expressed in terms of Gamma-functions: $$\prod^n_{i=1}\left(1-\frac{1}{3i}\right)=\frac{\Gamma(n+\frac{2}{3}) }{\Gamma(\frac{2}{3})\Gamma(n+1)}$$ Thus, we have $$1+\sum^\infty_{n=1}2^{-n}\frac{\Gamma(n+\frac{2}{3}) }{\Gamma(\frac{2}{3})\Gamma(n+1)}=\sum^\infty_{n=0}2^{-n}\frac{\Gamma(n+\frac{2}{3}) }{\Gamma(\frac{2}{3})\Gamma(n+1)}$$ By definition $$\left(\! \begin{array}{c} x \\ y \end{array} \!\right) = \frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}$$ In our case $x=n-\frac{1}{3},\quad y=-\frac{1}{3}$. Which leads us to: $$\sum^\infty_{n=0}\left(\! \begin{array}{c} n-\frac{1}{3} \\ -\frac{1}{3} \end{array} \!\right)2^{-n}$$ Recall generating function: $$\frac{1}{(1-z)^{m+1}}=\sum_{n\geq 0}\left(\! \begin{array}{c} m+n \\ m \end{array} \!\right)z^n$$ Setting $z=\frac{1}{2}$ and $m=-\frac{1}{3}$ gives us: $$\sum^\infty_{n=0}\left(\! \begin{array}{c} n-\frac{1}{3} \\ -\frac{1}{3} \end{array} \!\right)2^{-n}=\left(1-\frac{1}{2}\right)^{-\frac{2}{3}}=2^{\frac{2}{3}}$$


$\ds{\pars{a}_{m} = {\Gamma\pars{a + m} \over \Gamma\pars{a}}}$ is a Pochhammer Symbol.

\begin{align} &\color{#f00}{1 + \sum_{n = 1}^{\infty} {\prod_{k = 1}^{n}\pars{3k - 1} \over 6^{n}\,\, n!}} = 1 + \sum_{n = 1}^{\infty} {3^{n}\prod_{k = 1}^{n}\pars{k - 1/3} \over 6^{n}\,\, n!} = 1 + \sum_{n = 1}^{\infty}{\pars{2/3}_{n} \over 2^{n}\,\, n!} \\[5mm] = &\ 1 + \sum_{n = 1}^{\infty} {\Gamma\pars{2/3 + n}/\Gamma\pars{2/3} \over 2^{n}\,\, \Gamma\pars{n + 1}} = 1 + {1 \over \Gamma\pars{1/3}\Gamma\pars{2/3}}\sum_{n = 1}^{\infty} {\Gamma\pars{2/3 + n}\Gamma\pars{1/3} \over 2^{n}\,\, \Gamma\pars{n + 1}} \\[5mm] = &\ 1 + {\sin\pars{\pi/3} \over \pi}\sum_{n = 1}^{\infty} {1 \over 2^{n}}\int_{0}^{1}x^{-1/3 +n}\,\,\,\pars{1 - x}^{-2/3}\,\,\dd x \\[5mm] = &\ 1 + {\root{3} \over 2\pi}\int_{0}^{1}x^{-1/3}\,\,\pars{1 - x}^{-2/3}\,\, \sum_{n = 1}^{\infty}\pars{x \over 2}^{n}\,\dd x \\[5mm] = &\ 1 + {\root{3} \over 2\pi}\int_{0}^{1}x^{-1/3}\,\,\pars{1 - x}^{-2/3}\,\, {x/2 \over 1 - x/2}\,\dd x = 1 + {\root{3} \over 2\pi}\int_{0}^{1}\pars{x \over 1 - x}^{2/3}\,\, \,{\dd x \over 2 - x} \end{align}

With the substitution $\ds{x = {t \over 1 + t}}$: \begin{align} &\color{#f00}{1 + \sum_{n = 1}^{\infty} {\prod_{k = 1}^{n}\pars{3k - 1} \over 6^{n}\,\, n!}} = 1 + {\root{3} \over 2\pi}\ \underbrace{\int_{0}^{\infty} {t^{2/3} \over \pars{t + 1}\pars{t + 2}}\,\dd t} _{\ds{{2\pi \over \root{3}}\,\pars{2^{2/3} - 1}}}\ =\ \color{#f00}{2^{2/3}} \approx 1.5874\label{1}\tag{1} \end{align}

The integral, in \eqref{1}, is trivially evaluated by an integration over a suitable contour in the complex plane.