n-0-2n-1-8-n-n-2-Help-please-

Question Number 150432 by Jamshidbek last updated on 12/Aug/21

\underset{n = 0}{\sum^{\infty}} \frac{(2 n + 1)!}{8^{n} \cdot {(n!)}^{2}} = ? Help please

Answered by Olaf_Thorendsen last updated on 12/Aug/21

f (x) = \frac{1}{{(1 - x)}^{3 / 2}} = {(1 - x)}^{- \frac{3}{2}}

f^{'} (x) = \frac{1.3}{2} {(1 - x)}^{- \frac{5}{2}}

f^{″} (x) = \frac{1.3.5}{2^{2}} {(1 - x)}^{- \frac{7}{2}}

f^{‴} (x) = \frac{1.3.5.7}{2^{3}} {(1 - x)}^{- \frac{9}{2}}

\dots

f^{(n)} (x) = \frac{1.3.5.7 \dots (2 n + 1)}{2^{n}} {(1 - x)}^{- \frac{(2 n + 3)}{2}}

f^{(n)} (x) = \frac{(2 n + 1)!}{2^{n} . 2.4.6 \dots 2 n} {(1 - x)}^{- \frac{(2 n + 3)}{2}}

f^{(n)} (x) = \frac{(2 n + 1)!}{2^{2 n} n!} {(1 - x)}^{- \frac{(2 n + 3)}{2}}

f^{(n)} (0) = \frac{(2 n + 1)!}{2^{2 n} n!}

f (x) = \underset{n = 0}{\sum^{\infty}} \frac{f^{(n)} (0)}{n!} x^{n}

f (\frac{1}{2}) = \frac{1}{{(1 - \frac{1}{2})}^{3 / 2}} = \underset{n = 0}{\sum^{\infty}} \frac{(2 n + 1)!}{2^{2 n} n!^{2}} . {(\frac{1}{2})}^{n}

\Rightarrow \underset{n = 0}{\sum^{\infty}} \frac{(2 n + 1)!}{2^{3 n} n!^{2}} = \underset{n = 0}{\sum^{\infty}} \frac{(2 n + 1)!}{8^{n} . n!^{2}} = 2 \sqrt{2}

Commented by Tawa11 last updated on 12/Aug/21

Great sir .

Commented by Tawa11 last updated on 12/Aug/21

Sir help Q 150462

Commented by Ar Brandon last updated on 12/Aug/21

Super!

Commented by amin96 last updated on 12/Aug/21

V e r y n i c e

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