Searching for the radius of convergence for: $$\sum\limits_{n=1}^{+\infty}\frac{3x^n}{(2n)!}$$ leads me to the limit: $$\lim\limits_{n\to+\infty}\frac{(2n)!}{(2(n+1))!}.$$ How to simplify: $$\frac{(2n)!}{(2(n+1))!}?$$ WolframAlpha gives: $$\frac{(2n)!}{(2(n+1))!}=\frac{1}{2(n+1)(2n+1)}=\frac{1}{4n^2+6n+2},$$ but I don't see how to get it.
So, the radius of convergence is equals to $R=+\infty$, isn't it?
From the definition of the factorial, then
$$(2(n+1))!=(2n+2)!=(2n+2)(2n+1)2n!$$
Therefore
$$\frac {(2n)!}{(2(n+1))!}=\frac {(2n)!}{2(n+1)(2n+1)(2n)!}=\frac 1{2(n+1)(2n+1)}$$