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Question Number 144924 by qaz last updated on 30/Jun/21

$$\mathrm{S}\left(\mathrm{x}\right)=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\left(2\mathrm{n}\right)!!}{(2\mathrm{n}+1)!!}{\mathrm{x}}^{2\mathrm{n}}=?........(\mid \mathrm{x}\mid <1)$$

Answered by Ar Brandon last updated on 30/Jun/21

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

Commented byqaz last updated on 30/Jun/21

$$\mathrm{i}\mathrm{found}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{complicate}\mathrm{to}\mathrm{use}\mathrm{DE}\mathrm{to}\mathrm{solve}\mathrm{it}....$$ $$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented byAr Brandon last updated on 30/Jun/21

Je vous en prie !

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