OIPLZLqm6

2022-02-26

I want to know how to compute the closed form for the series
$\sum _{n=0}^{\mathrm{\infty }}\frac{\left(2n\right)!}{{\left(n!\right)}^{2}}{\left(\frac{|A|}{2}\right)}^{2n},\phantom{\rule{1em}{0ex}}|A|<1,$

### Answer & Explanation

Pregazzix2a

Applying the Taylor development to the function $f\left(A\right)=\frac{1}{\sqrt{1-{A}^{2}}}$
$f\left(A\right)={\left(1-{A}^{2}\right)}^{-\frac{1}{2}}$
$=1+\frac{1}{1!}\left(-\frac{1}{2}\right)\left(-{A}^{2}\right)+\frac{1}{2!}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right){\left(-{A}^{2}\right)}^{2}+\dots +\frac{1}{n!}\left(\prod _{i=0}^{n-1}\left(-\frac{1}{2}-i\right)\right){\left(-{A}^{2}\right)}^{n}+\dots$
$=1+\sum _{n=1}^{+\mathrm{\infty }}\left(\frac{1}{n!}{\left(-{A}^{2}\right)}^{n}\prod _{i=0}^{n-1}\left(-\frac{1}{2}-i\right)\right)$
$=1+\sum _{n=1}^{+\mathrm{\infty }}\frac{\left(2n-1\right)!!}{{2}^{n}n!}{\left({A}^{2}\right)}^{n}$
$=1+\sum _{n=1}^{+\mathrm{\infty }}\left(\frac{\left(2n-1\right)!!}{{2}^{n}n!}×\frac{\left(2n\right)!!}{{2}^{n}n!}\right){\left({A}^{2}\right)}^{n}$
$=1+\sum _{n=1}^{+\mathrm{\infty }}\frac{\left(2n\right)!}{{\left({2}^{n}\right)}^{2}{\left(n!\right)}^{2}}{\left({A}^{2}\right)}^{n}$
$=\sum _{n=0}^{+\mathrm{\infty }}\frac{\left(2n\right)!}{{\left(n!\right)}^{2}}{\left(\frac{{A}^{2}}{2}\right)}^{n}$
Q.E.D

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