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Question Number 147863 by mathmax by abdo last updated on 24/Jul/21

$$\mathrm{calculate}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{\mathrm{n}{!}^2}{\left(2\mathrm{n}\right)!}$$$$$$

Answered by qaz last updated on 24/Jul/21

$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(\mathrm{n}!)}^2}{\left(2\mathrm{n}\right)!}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}(2\mathrm{n}+2){\int }_0^1{(\mathrm{x}-{\mathrm{x}}^2)}^{\mathrm{n}}\mathrm{dx}$$$$={\int }_0^1(2\mathrm{yD}+2){\mid }_{\mathrm{y}=\mathrm{x}-{\mathrm{x}}^2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{y}}^{\mathrm{n}}\mathrm{dx}$$$$={\int }_0^1(2\mathrm{yD}+2){\mid }_{\mathrm{y}=\mathrm{x}-{\mathrm{x}}^2}\frac{1}{1-\mathrm{y}}\mathrm{dx}$$$$={\int }_0^1\frac{2}{{(1-\mathrm{y})}^2}{\mid }_{\mathrm{y}=\mathrm{x}-{\mathrm{x}}^2}\mathrm{dx}$$$$={\int }_0^1\frac{2}{{(1-\mathrm{x}+{\mathrm{x}}^2)}^2}\mathrm{dx}$$$$=\frac{\frac{4}{3}\mathrm{x}-\frac{2}{3}}{1-\mathrm{x}+{\mathrm{x}}^2}{\mid }_0^1+{\int }_0^1\frac{\frac{4}{3}}{1-\mathrm{x}+{\mathrm{x}}^2}\mathrm{dx}$$$$=\frac{4}{3}+\frac{8\sqrt{3}\pi }{27}$$

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