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

calculate Σ_(n=0) ^∞  ((n!^2 )/((2n)!))

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

Answered by qaz last updated on 24/Jul/21

Σ_(n=0) ^∞ (((n!)^2 )/((2n)!))  =Σ_(n=0) ^∞ (2n+2)∫_0 ^1 (x−x^2 )^n dx  =∫_0 ^1 (2yD+2)∣_(y=x−x^2 ) Σ_(n=0) ^∞ y^n dx  =∫_0 ^1 (2yD+2)∣_(y=x−x^2 ) (1/(1−y))dx  =∫_0 ^1 (2/((1−y)^2 ))∣_(y=x−x^2 ) dx  =∫_0 ^1 (2/((1−x+x^2 )^2 ))dx  =(((4/3)x−(2/3))/(1−x+x^2 ))∣_0 ^1 +∫_0 ^1 ((4/3)/(1−x+x^2 ))dx  =(4/3)+((8(√3)π)/( 27))

$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{n}!\right)^{\mathrm{2}} }{\left(\mathrm{2n}\right)!} \\ $$$$=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2n}+\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{x}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2yD}+\mathrm{2}\right)\mid_{\mathrm{y}=\mathrm{x}−\mathrm{x}^{\mathrm{2}} } \underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{y}^{\mathrm{n}} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2yD}+\mathrm{2}\right)\mid_{\mathrm{y}=\mathrm{x}−\mathrm{x}^{\mathrm{2}} } \frac{\mathrm{1}}{\mathrm{1}−\mathrm{y}}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}}{\left(\mathrm{1}−\mathrm{y}\right)^{\mathrm{2}} }\mid_{\mathrm{y}=\mathrm{x}−\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}}{\left(\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\frac{\mathrm{4}}{\mathrm{3}}\mathrm{x}−\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} }\mid_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{\mathrm{4}}{\mathrm{3}}}{\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{8}\sqrt{\mathrm{3}}\pi}{\:\mathrm{27}} \\ $$

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