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Question Number 163045 by mnjuly1970 last updated on 03/Jan/22
       prove that    Σ_(n=1) ^∞ (( ( 2n+1 )!!)/((2n )!!)) (1/(2^( n) (2n +1)^( 2) )) =((π(√2))/4)−1
$$ \\ $$$$\:\:\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(\:\mathrm{2}{n}+\mathrm{1}\:\right)!!}{\left(\mathrm{2}{n}\:\right)!!}\:\frac{\mathrm{1}}{\mathrm{2}^{\:{n}} \left(\mathrm{2}{n}\:+\mathrm{1}\right)^{\:\mathrm{2}} }\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}−\mathrm{1} \\ $$
Answered by qaz last updated on 03/Jan/22
Σ_(n=1) ^∞ (((2n+1)!!)/((2n)!!))∙(1/(2^n (2n+1)^2 ))  =Σ_(n=1) ^∞ (((2n+1)!)/((2^n n!)^2 ))∙(1/(2^n (2n+1)^2 ))  =Σ_(n=1) ^∞  (((2n)),(( n)) )(1/2^(3n) )∫_0 ^1 x^(2n) dx  =∫_0 ^1 Σ_(n=1) ^∞  (((−1/2)),((     n)) )(−(1/2)x^2 )^n dx  =∫_0 ^1 ((1−(1/2)x^2 )^(−1/2) −1)dx  =((π(√2))/4)−1
$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2n}+\mathrm{1}\right)!!}{\left(\mathrm{2n}\right)!!}\centerdot\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} \left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2n}+\mathrm{1}\right)!}{\left(\mathrm{2}^{\mathrm{n}} \mathrm{n}!\right)^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} \left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2n}}\\{\:\mathrm{n}}\end{pmatrix}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3n}} }\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{2n}} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\begin{pmatrix}{−\mathrm{1}/\mathrm{2}}\\{\:\:\:\:\:\mathrm{n}}\end{pmatrix}\left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right)^{−\mathrm{1}/\mathrm{2}} −\mathrm{1}\right)\mathrm{dx} \\ $$$$=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}−\mathrm{1} \\ $$

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