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How to find sum of the following series

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

Please give me some hints. Thanks in advance.

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marked as duplicate by José Carlos Santos, Did real-analysis Jan 27 '18 at 20:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The $.$ denotes multiplication? You can use $5 \cdot 8$ for $5\cdot 8$. $\endgroup$ – martini Sep 18 '13 at 14:09
  • $\begingroup$ 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? $\endgroup$ – abiessu Sep 18 '13 at 14:14
  • $\begingroup$ @abiessu this is not correct. $\endgroup$ – Siddhant Trivedi Sep 18 '13 at 14:16
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    $\begingroup$ $(2^{2/3}-1)/2$; $\endgroup$ – GEdgar Sep 18 '13 at 14:22
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    $\begingroup$ Try to see what binomial coefficient is represented by your terms. In the end, see how it is related to the series $(1-x)^{-2/3}$. $\endgroup$ – GEdgar Sep 18 '13 at 14:42
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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.

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