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Question Number 144322 by qaz last updated on 24/Jun/21

\underset{n = 0}{\sum^{\infty}} \frac{(2 n)!!}{(2 n + 1)!! (n + 1)} x^{2 n + 2} = ? \dots \dots \dots . ∣ x ∣⩽ 1

Answered by mindispower last updated on 25/Jun/21

(2 n)!! = 2^{n} . n!

(2 n + 1)!! = \frac{(2 n + 1)!}{2^{n} . n!}

\Leftrightarrow \sum_{n ⩾ 0} \frac{2^{n} . n!}{\frac{(2 n + 1)!}{2^{n} n!} (n + 1)} x^{2 n + 2}

= \sum_{n ⩾ 0} \frac{2^{2 n} . {(n!)}^{2} x^{2 n + 2}}{(2 n + 1)! . (n + 1)}

= \frac{1}{2} \sum_{n ⩾ 0} \frac{2^{2 (n + 1)} . {(n! (n + 1))}^{2}}{(2 n + 2)! {(n + 1)}^{2}} . x^{2 (n + 1)}

= \frac{1}{2} . \sum_{n ⩾ 1} \frac{{(2^{n})}^{2} . {(n!)}^{2}}{(2 n)! n^{2}} . x^{2 n}

= \frac{1}{2} \sum_{n ⩾ 1} \frac{{(2 x)}^{2 n}}{n^{2} . C_{2 n}^{n}} = {a r c s i n}^{2} (x)