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Question Number 131103 by mnjuly1970 last updated on 01/Feb/21

$$\dots nicecalculus\dots$$$$provethat::$$$$\sum _{n\in \mathbb{N}}\left(\frac{{\mathrm{\Gamma }}^2\left(n\right)}{2^{-n}(2n-1)!}\right)=\pi$$$$$$

Answered by Ar Brandon last updated on 01/Feb/21

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

Commented by mnjuly1970 last updated on 01/Feb/21

$$excellentmrbrandon$$$$extraordinary..$$

Answered by Dwaipayan Shikari last updated on 01/Feb/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{\Gamma }}^2\left(n\right)}{2^{-n}\mathrm{\Gamma }\left(2n\right)}={\int }_0^1\underset{n=1}{\sum ^{\mathrm{\infty }}}{2}^n{x}^{n-1}{(1-x)}^{n-1}dx$$$$=2{\int }_0^1\frac{1}{1-2x(1-x)}dx={\int }_0^1\frac{1}{x^2-x+\frac{1}{2}}dx=2{\int }_0^1\frac{1}{{(x-\frac{1}{2})}^2+\frac{1}{4}}dx$$$$=2\left(\frac{\pi }{2}\right)=\pi$$

Commented by mnjuly1970 last updated on 01/Feb/21

$$verynice.thankyousomuch=.$$

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