Please help me find the sum of this series:
1+23⋅12+23⋅56⋅122+23⋅56⋅89⋅123+⋯
All I could figure out was to find the nth term as:
an=2⋅(2+3)⋯(2+3(n−1))3⋅6⋅9⋯3(n−1)⋅12n−1
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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