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Question Number 138621 by mnjuly1970 last updated on 15/Apr/21

Answered by Dwaipayan Shikari last updated on 15/Apr/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{\Gamma }}^2(n+1)}{\mathrm{\Gamma }(2n+2)}={\int }_0^1\mathrm{\Sigma }{x}^n{(1-x)}^{n}dx$$$$={\int }_0^1\frac{1}{1-x+x^2}dx={\int }_0^1\frac{1}{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx=\frac{2}{\sqrt{3}}{\left[{tan}^{-1}\frac{2x-1}{\sqrt{3}}\right]}_0^1$$$$=\frac{2}{\sqrt{3}}(\frac{\pi }{6}+\frac{\pi }{6})=\frac{2\pi }{3\sqrt{3}}$$

Commented by mnjuly1970 last updated on 16/Apr/21

$$thankyousomuch...$$

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