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From an urn and balls problem, I end up with the need to compute the following sum $$S = \sum_{n\geq 1} \frac{1}{n}\binom{2n}{n+1}2^{-2n}$$ Using Maple I discovered $S=1$. Starting with some basics transformations, I get $$S= \sum_{n=1}^{\infty} \frac{2^{-n}(2n-1)(2n-3)\ldots 3}{(n+1)!}$$

Therefore I can write $S'$ using hypergeometric function taken at point $z=1$, $$S+1 = {}_2F_1(\frac{1}{2},1,2,1)$$

Then using Gauss Hypergeometric theorem $$S+1 = {}_2F_1(\frac{1}{2},1,2,1) = \frac{\Gamma(2)\Gamma(1/2)}{\Gamma(1)\Gamma(3/2)}=\frac{\Gamma(1/2)}{\Gamma(3/2)}$$

And using $\Gamma(z+1)=z\Gamma(z)$ I get $S=1$.

My question relates to Gauss Hypergeometric theorem. I couldn't find an online proof, or an explanation. And I was wandering if I could find a more direct approach for ${}_2F_1(\frac{1}{2},1,2,1) = 2$ and directly for my sum $S$. Gauss Theorem is quite generic, and it feels kinda using a bazooka to kill a bird. With the specific value I have, there might be an easier approach.

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  • $\begingroup$ WA gives the following answer:$$\frac{2^{-2 m-1} \left(-m \binom{2 (m+1)}{m+2}-2 \binom{2 (m+1)}{m+2}+2^{2 m+1} m+2^{2 m+1}\right)}{m+1}$$ $\endgroup$ Oct 30, 2019 at 13:07
  • $\begingroup$ @Dr.SonnhardGraubner on what input? $\endgroup$ Oct 30, 2019 at 13:14
  • $\begingroup$ The finite sum above! $\endgroup$ Oct 30, 2019 at 13:16
  • $\begingroup$ There are no $m$ parameter... the sum is infinite on $n$, so the result should not have any parameter $\endgroup$ Oct 30, 2019 at 13:17
  • $\begingroup$ My input was the finite sum! $\endgroup$ Oct 30, 2019 at 13:19

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The given series is telescopic. If we set $a_n=\frac{1}{4^n}\binom{2n}{n}$ we have $$ \frac{a_{n+1}}{a_n} = \frac{(2n+2)(2n+1)}{4(n+1)^2} = \frac{2n+1}{2n+2}=1-\frac{1}{2n+2}$$ hence $$ a_n-a_{n+1} = \frac{1}{2n+2}\binom{2n}{n}2^{-2n}= \frac{1}{2}\cdot\frac{1}{n}\binom{2n}{n+1}2^{-2n}$$ and since $a_n\to 0$ we have $$ S=\sum_{n\geq 1}\frac{1}{n}\binom{2n}{n+1}2^{-2n}=2a_1 = \color{red}{1}.$$ As an alternative approach, by exploiting $a_n=\frac{2}{\pi}\int_{0}^{\pi/2}(\cos\theta)^{2n}\,d\theta $ we have $$S=\frac{2}{\pi}\sum_{n\geq 1}\frac{1}{n+1}\int_{0}^{\pi/2}(\cos\theta)^{2n}\,d\theta=-\frac{2}{\pi}\int_{0}^{\pi/2}\frac{2\log\sin\theta+\cos^2\theta}{\cos^2\theta}\,d\theta$$ and it is not difficult to compute the last integral.

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One can recognize an instance of the binomial series: $$\frac{1}{n}\binom{2n}{n+1}2^{-2n}=2(-1)^n\binom{1/2}{n+1},$$ hence $S=\displaystyle\lim_{x\to1^-}S(x)$ where $S(x)=\color{red}{-}2\displaystyle\sum_{n\geq 1}\binom{1/2}{n+1}(-x)^{n\color{red}{+1}}=2(1-\sqrt{1-x})-x$.

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  • $\begingroup$ Thanks! I had a feeling there was something related, but I could find it myself. $\endgroup$ Oct 30, 2019 at 13:36

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