Find out the answer to "How to prove:π2=18∑n=0∞n!n!(2n+2)!"

ymhonnolfq

Answered question

2022-02-27

How to prove:

π^{2} = 18 \sum_{n = 0}^{\infty} \frac{n! n!}{(2 n + 2)!}

Pooja Copeland

Beginner2022-02-28Added 8 answers

One has

S = \sum_{n = 0}^{\infty} \frac{n! n!}{(2 n + 2)!} = \sum_{n = 0}^{\infty} \frac{1}{(2 n + 2) (2 n + 1) (\begin{matrix} 2 n \\ n \end{matrix})}

How you have the following Maclaurin expansion:

\frac{\arcsin x}{\sqrt{1 - x^{2}}} = \sum_{n = 0}^{\infty} \frac{2^{2 n} x^{2 n + 1}}{(2 n + 1) (\begin{matrix} 2 n \\ n \end{matrix})}

S = 2 \int_{0}^{1} \frac{\arcsin (\frac{x}{2})}{\sqrt{1 - {(\frac{x}{2})}^{2}}} d x = 2 {[\arcsin^{2} (\frac{x}{2})]}_{0}^{1} = 2 \times {(\frac{π}{6})}^{2} = \frac{π^{2}}{18}

and you are done.

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