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Assume $|x| \neq 1$. Find the sum of the following series:

$$\sum_{n=0}^\infty\frac{(1/2)(3/2)\cdots((2n-1)/2)}{n!}\left(\frac{2x}{1+x^2}\right)^{2n} = 1 + \frac{1}{2}\left(\frac{2x}{1+x^2}\right)^2+\frac{1}{2}\frac{3}{4}\left(\frac{2x}{1+x^2}\right)^4+\cdots$$

José Carlos Santos
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tastykakes
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  • Where is $n!$ effect in R.h.s $$\sum_{n=0}^\infty\frac{(1/2)(3/2)\cdots((2n-1)/2)}{n!}(\frac{2x}{1+x^2})^{2n} = 1 + \frac{1}{2}(\frac{2x}{1+x^2})^2+\frac{1}{2}\frac{3}{4}(\frac{2x}{1+x^2})^4+\cdots$$ ? – Khosrotash Aug 03 '17 at 09:34
  • See https://math.stackexchange.com/questions/746388/calculating-1-frac13-frac1-cdot33-cdot6-frac1-cdot3-cdot53-cdot6-cdot – lab bhattacharjee Aug 03 '17 at 09:59

1 Answers1

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Since the Taylor series of $\frac1{\sqrt{1-x}}$ is$$1+\frac12x+\frac12\times\frac34x^2+\frac12\times\frac34\times\frac56x^3+\cdots,$$and this series converges (to $\frac1{\sqrt{1-x}}$) when $|x|<1$, you have$$|x|<1\Longrightarrow\frac1{\sqrt{1-x^2}}=1+\frac12x^2+\frac12\times\frac34x^4+\frac12\times\frac34\times\frac56x^6+\cdots$$Therefore,\begin{multline*}1+\frac12\left(\frac{2x}{1+x^2}\right)^2+\frac12\times\frac34\left(\frac{2x}{1+x^2}\right)^4+\frac12\times\frac34\times\frac56\left(\frac{2x}{1+x^2}\right)^6+\cdots=\\=\frac1{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}=\frac{1+x^2}{1-x^2}.\end{multline*}

José Carlos Santos
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