# Sum of infinite series [duplicate]

How to find sum of the following series

$$\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\cdots={1\over 6} + \sum_{n=1}^\infty{\Pi_{i=1}^n{3i+2}\over (n+1)!6^{n+1}}$$

## marked as duplicate by José Carlos Santos, Did real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 27 '18 at 20:20

• The $.$ denotes multiplication? You can use $5 \cdot 8$ for $5\cdot 8$. – martini Sep 18 '13 at 14:09
• This looks like ${1\over 6} + \sum_{n=1}^\infty{\Pi_{i=1}^n{3i+2}\over (n+1)!6^{n+1}}$, is that correct? – abiessu Sep 18 '13 at 14:14
• $(2^{2/3}-1)/2$; – GEdgar Sep 18 '13 at 14:22
• Try to see what binomial coefficient is represented by your terms. In the end, see how it is related to the series $(1-x)^{-2/3}$. – GEdgar Sep 18 '13 at 14:42
The general term of the series reads: $$\frac{1}{2}\frac{\prod_{k=1}^n (3k-1)}{\prod_{k=1}^n 6 k} = \frac{1}{2^{n+1}} \prod_{k=1}^n \left(1-\frac{1}{3k}\right) = \frac{1}{2^{n+1}} \frac{\left(2/3\right)_n}{n!} = \frac{1}{2^{n+1}} \frac{\Gamma\left(n+2/3\right)}{n! \cdot \Gamma\left(2/3\right)}$$ The series thus reads: $$\mathcal{S}= \sum_{n=1}^\infty \frac{1}{2^{n+1}} \frac{\left(2/3\right)_n}{n!} = \frac{1}{\Gamma(2/3)} \sum_{n=1}^\infty \frac{1}{2^{n+1}} \frac{\Gamma\left(n+2/3\right)}{n!}$$ Using Euler's integral: $$\Gamma\left(n+2/3\right) = \int_0^\infty t^{n-1/3} \mathrm{e}^{-t} \mathrm{d}t$$ and interchanging the summation and the integration warranted by Tonelli's theorem: $$\begin{eqnarray} \mathcal{S} &=& \frac{1}{\Gamma(2/3)} \sum_{n=1}^\infty \frac{1}{2^{n+1}} \frac{\Gamma\left(n+2/3\right)}{n!} = \frac{1}{\Gamma(2/3)} \int_0^\infty \left(\sum_{n=1}^\infty \frac{t^{n-1/3}}{2^{n+1} n!} \right) \mathrm{e}^{-t} \mathrm{d}t \\ &=& \frac{1}{\Gamma(2/3)} \int_0^\infty \left(\frac{\exp\left(t/2\right)-1}{2 t^{1/3}} \right) \mathrm{e}^{-t} \mathrm{d}t \\ &=& \frac{1}{2\Gamma(2/3)} \left( \int_0^\infty t^{-1/3} \mathrm{e}^{-t/2} \mathrm{d}t - \int_0^\infty t^{-1/3} \mathrm{e}^{-t} \mathrm{d}t \right) \\ &=& \frac{1}{2\Gamma(2/3)} \left( 2^{2/3} \Gamma\left(2/3\right) - \Gamma\left(2/3\right) \int_0^\infty t^{-1/3} \mathrm{e}^{-t} \mathrm{d}t \right) = \frac{2^{2/3}-1}{2} \end{eqnarray}$$ Alternatively, you might note that the series term can be written in terms of a binomial: $$\frac{\left(2/3\right)_n}{n!} = \binom{2/3}{n}$$ and hence: $$\mathcal{S} = \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \binom{2/3}{n} = \frac{1}{2} \left( \sum_{n=0}^\infty \frac{1}{2^n} \binom{2/3}{n} -1 \right)$$ this can now be finished with the Newton's generalized binomial theorem.