Find the radius of convergence of the power series $J_{o}(x)= \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2^{2n}(n!)^2}$ Find the radius of convergence of the power series $$J_{o}(x)= \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2^{2n}(n!)^2}$$
Should I separate this into the product of two limits, namely $\frac{(-1)^n}{(n!)^2}$ and $\frac{x^{2n}}{2^{2n}}$ so that the first gets dominated in the denominator by the factorial?
 A: When looking for the radius of convergence for the series $\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2^{2n}(n!)^2}$,  you have to find half of the number of integer values of $x$ that would make the series to convergent. The way to find this is to start with the ratio test, which states that a series $\sum_{n=0}^{\infty} a_n$ is convergent when $ \lim_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n}\right\vert < 1$.
$$\lim_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n}\right\vert = \lim_{n\to\infty} \left\vert\frac{x^{2n+2}}{2^{2n+2}(n+1)!^2} * \frac{2^{2n}(n!)^2} {x^{2n}} \right\vert =\lim_{n\to\infty} \left\vert \frac{x^2}{4(n+1)^2} \right\vert =0 < 1 \text{ For all values of $x$} $$
Since the ratio test states that the series converges whenever $\lim_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n}\right\vert < 1$ and in the case of this series, $ \lim_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n}\right\vert = \lim_{n\to\infty} \left\vert \frac{x^2}{4(n+1)^2} \right\vert =0 <1$ for all values of $x$, so the series is convergent for all values of $x$. Since there are an infinite number of integer values of $x$, and all of them would make the series convergent, half of the number of values that make the series convergent is still infinite. Therefore the radius of convergence is $\infty$, which makes the interval of convergence $(-\infty, \infty)$. 
