power series $\large{\Sigma_{n=0}^{\infty}} \frac{(n!)^2 x^n}{(2n)!}$, Radius of convergence

Question

I have the power series $$\large{\Sigma_{n=0}^{\infty}} \frac{(n!)^2 x^n}{(2n)!}.$$ I found the radius of convergence to be $(-4,4)$ using d'Alembert rule. Now I am trying to find what is happening on the edges but I found it really difficult to solve because for $x=-4$ we get $\large{\Sigma_{n=0}^{\infty}} \frac{(-1)^n(n!)^2 4^n}{(2n)!}$ I am not sure if it Leibniz series though I don't know how to calculate the limit of $\frac{(n!)^2 4^n}{(2n)!}$.

The same thing happens for $x=4$, I get a d'Alembert ratio limit equal to which doesn't allow to conclude whether the series converge or not.

Answer 1

Note that by the binomial theorem, $$ 4^n = (1 + 1)^{2n} = 1^{2n} + \cdots +\binom{2n}{n}1^n 1^n +\cdots + 1^{2n} >\binom{2n}{n} = \frac{{(2n)!}}{{n!^2 }} $$ for all $n\geq 1$.

Answer 2

Denoting $u_n = \frac{(n!)^2 4^n}{(2n)!}$ you have

$$\frac{u_{n+1}}{u_n} = 4\frac{(n+1)^2}{(2n+2)(2n+1)} = 1 + \frac{1}{2n+1}>1$$

Hence the absolute value of the general term of the series which is increasing can't have zero for limit and the series diverges at $x=-4, 4$.