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In the course of trying to answer Closed form of a series involving the squared Beta function I came up with this expresion, which can be proven in a round-about manner via the Beta function: $$ \pi = 2+\frac23 \sum_{n=1}^\infty \prod_{k=1}^{n-1} \frac{4k^2}{(2k+3)(2k+1)} $$ I'm wondering if this is a known expression, and is there a straightforward proof?

It converges fairly slowly, since the term ratio is $$\frac{n^2}{(n+\frac12)(n+\frac32)}$$ so people doing digits-of-pi work would not be interested.

I have searched the web and could not find this, nor the trivially equivalent relation $$\pi = 2+\frac{1}{2}\sum_{n=1}^\infty\frac{2^{2n}(n-1)!)^2}{(2n+1)((2n-1)!!)^2}$$ In particular, MathWorld and Wikipedia (as well as references like Abramowitz and Stegun and Rektorys) don't have it.

On the other hand, full Mathematica and even Wolfram alpha does get the right answer.

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Mark Fischler
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  • Summing your "trivially equivalent relation" does not yield $2\pi$. Did you make an error, or did I? https://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D1%7D%5E%7B1000%7D+%5Cfrac%7B2%5E%7B2n%7D((n-1)!)%5E2%7D%7B(2n%2B1)((2n-1)!!)%5E2%7D – Brevan Ellefsen Feb 02 '17 at 21:27
  • Cut and paste error. Correcting it now... I have put back the $2$ and the summation sign. – Mark Fischler Feb 02 '17 at 21:46
  • Note that conventionally, an empty product has value $1$, not $0$. Thus the $n=1$ term has value $\frac23$ in both the original form and the equivalent pure sum. – Mark Fischler Feb 02 '17 at 21:53

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Such identities for $\pi$ are simple consequences of the Taylor series of $\arcsin(x)$ and $\arcsin^2(x)$.
For instance, $$ \arcsin^2(x) = \frac{1}{2}\sum_{n\geq 1}\frac{(4x^2)^n}{n^2\binom{2n}{n}} \tag{1}$$ is proved through different techniques in this historical thread and at pages $19,20$ of my course notes (sorry, they are in Italian). Of course, if we differentiate both sides of $(1)$ we get: $$\frac{\arcsin(x)}{2x\sqrt{1-x^2}}=\sum_{n\geq 1}\frac{(4x^2)^{n-1}}{n\binom{2n}{n}}\tag{2} $$ that is enough to recover the wanted result as a simple integral, $$ \int_{0}^{1}\frac{\arcsin(\sqrt{x})}{\sqrt{x}}\,dx = \pi-2\tag{3} $$ through the following steps:

  • In $(2)$, replace $x^2$ with $x$;
  • multiply both sides by $\sqrt{1-x}$;
  • perform termwise integration over $(0,1)$.
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Jack D'Aurizio
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