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Here is the problem:

Find the radius and the interval of convergence of $$\sum_{n\geq 1}\frac{2\cdot 4\cdot 6\cdots (2n)\,x^{n}}{1\cdot 3\cdot 5\cdots (2n-1)}.$$

I thought that I could write the numerator as $n!2^n$, and the denominator as $\frac{(2n)!}{(n!2^n)}$, but I am not sure that either of these are correct. If you have any advice on how to solve the problem, it would be much appreciated! :) Thanks!

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  • $\begingroup$ Your expressions for the numerator and denominator are both correct. $\endgroup$ – Ant Jan 8 '18 at 2:32
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By the way: Mathematica recognizes your function as $${x\over 1-x}+{\sqrt{x}\arcsin\bigl(\sqrt{x}\bigr)\over(1-x)^{3/2}}\ .$$

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Rewrite the sum like you suggested: $$\sum_{n=1}^{\infty}{\left(\frac{2\cdot4\cdot6\cdot\cdots\left(2n\right)}{1\cdot3\cdot5\cdot\cdots\left(2n-1\right)}x^n\right)}=\sum_{n=1}^{\infty}{\left(\frac{n!\cdot2^n}{\frac{\left(2n\right)!}{n!\cdot2^n}}x^n\right)}=\sum_{n=1}^{\infty}{\left(\frac{{\left(n!\right)}^2\cdot2^{2n}}{\left(2n\right)!}x^n\right)}$$ Now, the ratio test says that this sum converges if the ratio between successive terms: $$\frac{a_{n+1}}{a_n}=\frac{\frac{{\left(\left(n+1\right)!\right)}^2\cdot2^{2\left(n+1\right)}}{\left(2\left(n+1\right)\right)!}x^{n+1}}{\frac{{\left(n!\right)}^2\cdot2^{2n}}{\left(2n\right)!}x^n}$$ is less than one. You can check that this fraction reduces to: $$\frac{2n+2}{2n+1}x$$ so the radius of convergence is $1$ (and the interval of convergence is $-1<x<1$).

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