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Question Number 163662 by amin96 last updated on 09/Jan/22

Commented by amin96 last updated on 09/Jan/22

$$\mathbf{PROVE}\mathbf{THAT}$$

Answered by Ar Brandon last updated on 09/Jan/22

$$S=2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{n!}{(2n+1)!!}=2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{n!\times n!\times 2^n}{(2n+1)!}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{\Gamma }}^2(n+1)}{\mathrm{\Gamma }(2n+2)}\times 2^{n+1}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{2}^{n+1}\beta (n+1,n+1)=\underset{n=0}{\sum ^{\mathrm{\infty }}}{2}^{n+1}{\int }_0^1{x}^n{(1-x)}^{n}dx$$$$=2{\int }_0^1\underset{n=0}{\sum ^{\mathrm{\infty }}}{(2x-2x^2)}^{n}dx=2{\int }_0^1\frac{dx}{2x^2-2x+1}$$$$={\int }_0^1\frac{dx}{{(x-\frac{1}{2})}^2+\frac{1}{4}}=2{\left[\arctan (2x-1)\right]}_0^1$$$$=2(\frac{\pi }{4}+\frac{\pi }{4})=2\times \frac{\pi }{2}=\pi$$

Commented by amin96 last updated on 09/Jan/22

$$yessir.bravoooo$$

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