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Question Number 189323 by mnjuly1970 last updated on 14/Mar/23

Answered by witcher3 last updated on 14/Mar/23

$$\iff \mathrm{\Sigma }\frac{\mathrm{\Gamma }\left(\mathrm{n}\right)\mathrm{\Gamma }\left(\mathrm{n}\right)}{\mathrm{\Gamma }\left(2\mathrm{n}\right).2^{-\mathrm{n}}}=\sum _{\mathrm{n}⩾1}\frac{\beta (\mathrm{n},\mathrm{n})}{2^{-\mathrm{n}}}$$$$=\sum _{\mathrm{n}⩾1}{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}-1}{(1-\mathrm{x})}^{\mathrm{n}-1}}{2^{-\mathrm{n}}}$$$$=2{\int }_0^1\sum _{\mathrm{n}⩾0}{\left(2\mathrm{x}(1-\mathrm{x})\right)}^{\mathrm{n}}$$$$2\mathrm{x}(1-\mathrm{x})<\frac{1}{2}$$$$=2{\int }_0^1\frac{1}{1-2\mathrm{x}(1-\mathrm{x})}$$$$=2{\int }_0^1\frac{\mathrm{dx}}{{2\mathrm{x}}^2-2\mathrm{x}+1}$$$$=2{\int }_0^1\frac{\mathrm{dx}}{2((\mathrm{x}-\frac{1}{2})+\frac{1}{4})}=4{\int }_0^1\frac{\mathrm{dx}}{{(2\mathrm{x}-1)}^2+1}$$$$={4\tan }^{-1}\left(1\right)=\pi$$

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