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Q: Calculate

$$S=1+{3\over4}+{3\times5\over4\times8}+{3\times5\times7\over4\times8\times12}+\dots$$

I calculated that the $n$th term is given by $$ \begin{align} A_n&={3\times5\times7\times\dots\times(2n+1)\over4\times8\times12\times\dots\times4n}\\ &={{(2n+1)!\over2\times4\times8\dots\times2n}\over4^nn!}\\ &={(2n+1)!\over8^n(n!)^2} \end{align}$$ Hence $$S=\sum_{n=0}^{\infty}{(2n+1)!\over8^n(n!)^2} =\sum_{n=0}^{\infty}{\binom{2n}{n}(2n+1){1\over8^n}}$$

I'm not sure how to proceed

I feel like generating function for $\sum{\binom{2n}{n}(2n+1)}$ would be helpful but don't know how to derieve it.

Robert Z
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Anvit
  • 3,379

2 Answers2

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Hint. Note that $$A_n={3\times5\times7\times\dots\times(2n+1)\over4\times8\times12\times\dots\times4n}=\frac{(2n+1)!!}{n! 4^n} =\frac{(-1/2)^n}{n!}\left(-\frac32\right)\ldots\left(-\frac{2n+1}2\right).$$

Robert Z
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This looks like the binomial series for $$(1+x)^{-3/2}$$ evaluated at $x=-1/2$.

Angina Seng
  • 158,341
  • Yeah the answer's $\sqrt8$ but how to reach the conclusion without already knowing it? If I knew $(1+x)^{(-3/2)}$ was an option i would just expand – Anvit Apr 14 '18 at 06:02