prove-that-n-1-2n-1-2n-1-2-n-2n-1-2-pi-2-4-1-

Question Number 163045 by mnjuly1970 last updated on 03/Jan/22

p r o v e t h a t

\underset{n = 1}{\sum^{\infty}} \frac{(2 n + 1)!!}{(2 n)!!} \frac{1}{2^{n} {(2 n + 1)}^{2}} = \frac{π \sqrt{2}}{4} - 1

Answered by qaz last updated on 03/Jan/22

\underset{n = 1}{\sum^{\infty}} \frac{(2 n + 1)!!}{(2 n)!!} \cdot \frac{1}{2^{n} {(2 n + 1)}^{2}}

= \underset{n = 1}{\sum^{\infty}} \frac{(2 n + 1)!}{{(2^{n} n!)}^{2}} \cdot \frac{1}{2^{n} {(2 n + 1)}^{2}}

= \underset{n = 1}{\sum^{\infty}} (\begin{matrix} 2 n \\ n \end{matrix}) \frac{1}{2^{3 n}} \int_{0}^{1} x^{2 n} dx

= \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} (\begin{matrix} - 1 / 2 \\ n \end{matrix}) {(- \frac{1}{2} x^{2})}^{n} dx

= \int_{0}^{1} ({(1 - \frac{1}{2} x^{2})}^{- 1 / 2} - 1) dx

= \frac{π \sqrt{2}}{4} - 1

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