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While solving some sequence and series problems I came across the following series. The question is find the value of $$\sum_{n=1}^{\infty} 4^{n-1}\frac{n!(n-1)!}{(2n+1)!!^2}.$$ I first calculated $n+1$ and $n$th terms and found that $$\frac{a_{n+1}}{a_n}=\frac{4(n+1)(n)}{(2n+3)^2}.$$ $4n^2+4n<4n^2+12n+9$ so $a_{n+1}<a_n$ thus it may converge. The proof doesn't seem to be rigorous also I have no idea how to proceed. A combinatorical proof will be more helpful. I am not certain about it so I am also including calculus as a tag.

metamorphy
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Archis Welankar
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1 Answers1

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This series converges by Stirling's asymptotics. To compute its value, let $$4S:=\sum_{n=1}^{\infty}4^n\frac{n!(n-1)!}{(2n+1)!!^2}=\sum_{n=1}^{\infty}4^n\frac{n!(n-1)!}{(2n+1)!}\frac{2^n n!}{(2n+1)!!}\\=\sum_{n=1}^{\infty}4^n\frac{n!(n-1)!}{(2n+1)!}\int_0^1\frac{x^{2n+1}\,dx}{\sqrt{1-x^2}}=\int_0^1\frac{f(x)\,dx}{\sqrt{1-x^2}},$$ where $f(x)=\displaystyle\sum_{n=1}^{\infty}4^n\frac{n!(n-1)!}{(2n+1)!}x^{2n+1}$ (and we use "beta" integrals). Now $$1+\frac{x}{2}\left(\frac{f(x)}{x}\right)'=\sum_{n=0}^{\infty}\frac{n!^2(2x)^{2n}}{(2n+1)!}=\sum_{n=0}^{\infty}(2x)^{2n}\int_0^1 y^n(1-y)^n\,dy\\=\int_0^1\frac{dy}{1-4x^2y(1-y)}=\frac{1}{2x\sqrt{1-x^2}}\left.\arctan\frac{(2y-1)x}{\sqrt{1-x^2}}\right|_{y=0}^{y=1}=\frac{\arcsin x}{x\sqrt{1-x^2}}.$$ Integrating this, we find $f(x)=2\left(x-\sqrt{1-x^2}\arcsin x\right)$ and $$S=\frac{1}{2}\int_0^1\left(\frac{x}{\sqrt{1-x^2}}-\arcsin x\right)dx=\color{blue}{1-\frac{\pi}{4}}.$$

metamorphy
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