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Here I have this sum, $\displaystyle \sum_{n=1}^\infty \displaystyle \prod_{k=1}^n \dfrac{2k+1}{4k}$.

I have no idea how to sum this up. Any help would be appreciated!

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closed as off-topic by Namaste, Trevor Gunn, Jack D'Aurizio Sep 17 '18 at 16:21

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    $\begingroup$ Start by simplifying the summand by writing the product in terms of factorials and try to notice the connection of the summand to ${-3/2 \choose n}$. Then consider the power-series of $(1 - 4 x)^{-3/2} - 1$. $\endgroup$ – Winther Sep 17 '18 at 15:07
  • $\begingroup$ Please avoid "no-clue" questions. They do not make good quality questions and are likely to be removed. $\endgroup$ – Trevor Gunn Sep 17 '18 at 15:23
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Is this true: $$\prod_{k=1}^n\frac{2k+1}{4k}=\frac{\prod_{k=1}^n2k+1}{\prod_{k=1}^n4k}=\frac{\prod_{k=1}^n2k+1}{4^n\prod_{k=1}^nk}=\frac{\Gamma(2n+1)}{4^n\Gamma(n)}$$

EDIT as pointed out by another user: $$\frac{\prod_{k=1}^n2k+1}{4^n\prod_{k=1}^nk}=\frac{(2n+1)!!}{4^nn!}$$

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  • $\begingroup$ I'm not yet familiar with Gamma function. $\endgroup$ – user591656 Sep 17 '18 at 15:11
  • $\begingroup$ I am familiar with it but not in this context, so I need someone else to verify that this is true $\endgroup$ – Henry Lee Sep 17 '18 at 15:12
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    $\begingroup$ $\prod_{k=1}^n (2k + 1) = (2n + 1)(2n - 1) \cdots 3 \cdot 1 = (2n + 1)!! = \frac{(2n + 2)!}{2^{n + 1}(n + 1)!}$ and $\prod_{k = 1}^n k = n! = \Gamma(n + 1)$. $\endgroup$ – Trevor Gunn Sep 17 '18 at 15:12

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