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Is there a closed form for the expression $$\sum_{n=1}^{\infty}\left(n+\frac12\right)B^2\left(n,\frac32\right)$$ where $B(x,y)$ is the Beta function? Here is what I've done.

I wrote:

$$\sum_{n=1}^{\infty}\left(n+\frac12\right)B^2\left(n,\frac32\right)=\sum_{n=1}^{\infty}\left(n+\frac12\right)\int_0^1\int_0^1(xy)^n\sqrt{(1-x)(1-y)}dxdy$$ which gives $$\int_0^1\int_0^1\frac{(3-xy)xy}{(xy-1)^2}\sqrt{(1-x)(1-y)}dxdy$$ And unfortunatly I stuck here.

song01
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  • For the OP: many users here tend to downvote so-called PSQs (Problem Statement Questions) like yours, and answers, too (some god will forgive them, but the poor human Jack won't). So it is in your (and community's) best interest to improve your actual question by adding some context (your attempts, why this question is relevant to you, something along these lines). Cheers. – Jack D'Aurizio Feb 02 '17 at 19:16
  • Anyway, yes: $$\large{\pi-2}$$ – Jack D'Aurizio Feb 02 '17 at 19:18
  • That is related with the convolution formula for central binomial coefficients. – Jack D'Aurizio Feb 02 '17 at 19:21
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    Is $\pi-2$ closed form enough for you? BTW, I don't dislike PSQs as long as the problem is clean and interesting. – Mark Fischler Feb 02 '17 at 19:22
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    Thanks Jack for your answer. I didn't know that I should also write what I've done and where I stuck. I edited my message so to show my attempt. – song01 Feb 02 '17 at 19:22
  • This famous question (http://math.stackexchange.com/questions/878477/a-closed-form-of-sum-k-0-infty-frac-1k1k-gamma2-left-frack2) deals with a similar series. You may easily adapt the shown techniques to your problem. – Jack D'Aurizio Feb 02 '17 at 19:31

2 Answers2

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You may exploit: $$ \sum_{n\geq 1}(n+1/2) B(n,3/2) z^{n-1} = \frac{\arcsin(\sqrt{z})}{\sqrt{z(1-z)}}\tag{1} $$ to get that your series is simply related with: $$ \int_{0}^{1}\frac{\arcsin(\sqrt{z})}{\sqrt{z}}\,dz \tag{2}$$ that is not a difficult integral to compute (through the substitution $z=\sin^2\theta$).

Community
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Jack D'Aurizio
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Yes, the sum is $\pi-2$. The method shown by @Jack Aurizio shows that.

BTW, your integral has a mistake in it; the integral you present has value $\frac{176}{9}-6\pi$.

Start from the definition of the beta function $$ (n+\frac12)B^2(n,\frac32) = (n+\frac12) \frac{\Gamma^2(n)\Gamma^2(\frac32)}{\Gamma^2(n+\frac32)}= (n+\frac12) \frac{(n-1)!)^2\frac{\pi}{4}}{(n+\frac12)^2\Gamma^2(n+\frac12)}= \frac{\pi(n-1)!)^2}{4(n+\frac12)\Gamma^2(n+\frac12)} $$ Now apply the formula $$\Gamma(n+\frac12) = \frac{(2n)!}{4^n n!}\sqrt{\pi}$$ to obtain $$ \frac{\pi(n-1)!)^2}{4(n+\frac12)\Gamma^2(n+\frac12)} = \frac{1}{2}\frac{2^{4n}(n!)^2(n-1)!)^2}{(2n+1)(2n)!(2n)!} $$ Next use $\frac{(2n)!}{2^n}= n!(2n-1)!!$ to get $$ \frac{\pi(n-1)!)^2}{4(n+\frac12)\Gamma^2(n+\frac12)} = \frac{1}{2}\frac{2^{2n}(n!)^2(n-1)!)^2}{(2n+1)(n!)^2((2n-1)!!)^2} = \frac{1}{2}\frac{2^{2n}(n-1)!)^2}{(2n+1)((2n-1)!!)^2} $$ This is a fairly simple seires: The first term is $\frac23$ and the term ratio for $n\geq 1$ is $$ \frac{a_{n+1}}{a_n} = \frac{4n^2}{(2n+3)(2n+1)} $$ So now we know that $$ \pi = 2+\frac23 \sum_{n=1}^\infty \prod_{k=1}^{n-1} \frac{4k^2}{(2k+3)(2k+1)} $$ This is not among the relations I can find in Mathworld, Abramowitz and Stegun, Rektorys, or various lists of series for $\pi$ I can find on the web.

And now we have this (round-about) way of proving it.

Mark Fischler
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