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Question Number 144925 by qaz last updated on 30/Jun/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\left(2\mathrm{n}\right)!!}{2^{\mathrm{n}}\cdot (\mathrm{n}+1)\cdot (2\mathrm{n}+1)!!}=?$$

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

$$\left(2\mathrm{n}\right)!!=2\mathrm{n}(2\mathrm{n}-2)(2\mathrm{n}-4)...=2^{\mathrm{n}}\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)...=2^{\mathrm{n}}\mathrm{n}!$$$$(2\mathrm{n}+1)!!=(2\mathrm{n}+1)(2\mathrm{n}-1)...=\frac{(2\mathrm{n}+1)\left(2\mathrm{n}\right)(2\mathrm{n}-1)(2\mathrm{n}-2)...}{\left(2\mathrm{n}\right)(2\mathrm{n}-2)...}=\frac{(2\mathrm{n}+1)!}{2^{\mathrm{n}}\mathrm{n}!}$$$$\mathrm{S}\left(\mathrm{n}\right)=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\left(2\mathrm{n}\right)!!}{2^{\mathrm{n}}(\mathrm{n}+1)(2\mathrm{n}+1)!!}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(2^{\mathrm{n}}\mathrm{n}!)}^2}{2^{\mathrm{n}}(\mathrm{n}+1)(2\mathrm{n}+1)!}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{2^{\mathrm{n}}{(\mathrm{n}!)}^2}{(\mathrm{n}+1)(2\mathrm{n}+1)!}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{2^{\mathrm{n}}}{\mathrm{n}+1}\beta (\mathrm{n}+1,\mathrm{n}+1)$$$$={\int }_0^1\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{2^{\mathrm{n}}}{\mathrm{n}+1}{\mathrm{x}}^{\mathrm{n}}{(1-\mathrm{x})}^{\mathrm{n}}\mathrm{dx}={\int }_0^1\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(2\mathrm{x}-{2\mathrm{x}}^2)}^{\mathrm{n}}}{\mathrm{n}+1}\mathrm{dx}-1$$$$\mathrm{y}\left(\mathrm{t}\right)=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{t}}^{\mathrm{n}}=\frac{1}{1-\mathrm{t}}\Rightarrow {\int }_0^{\mathrm{t}}\mathrm{y}\left(\mathrm{t}\right)\mathrm{dt}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{t}}^{\mathrm{n}+1}}{\mathrm{n}+1}={\int }_0^{\mathrm{t}}\frac{1}{1-\mathrm{t}}\mathrm{dt}=\ln \left(\frac{1}{1-\mathrm{t}}\right)$$$$\mathrm{S}\left(\mathrm{n}\right)={\int }_0^1\frac{\ln ({2\mathrm{x}}^2-2\mathrm{x}+1)}{{2\mathrm{x}}^2-2\mathrm{x}}\mathrm{dx}-1$$

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