Wednesday, 31 August 2016

summation - Evaluate the sum of this series





Please help me find the sum of this series:



1+2312+2356122+235689123+



All I could figure out was to find the nth term as:



an=2(2+3)(2+3(n1))3693(n1)12n1



What To do ahead of it. I don't know. Please help.


Answer




Let S denote the sum. We write each term (with indices starting at 0) as



\left( \prod_{k=1}^{n} \frac{3k-1}{3k} \right) \frac{1}{2^n} = \frac{\prod_{k=0}^{n-1} (-\frac{2}{3}-k)}{n!} \left(-\frac{1}{2}\right)^n = \binom{-2/3}{n} \left(-\frac{1}{2}\right)^n.



Then we easily recognize S as a bionmial series and hence



S = \sum_{n=0}^{\infty}\binom{-2/3}{n} \left(-\frac{1}{2}\right)^n = \left(1 - \frac{1}{2}\right)^{-2/3} = 2^{2/3}.


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