I need some help with the next series. I want to prove that the power series diverges if $x=8$ or $x=-8$, but, I don't know how. I really appreciate any hint/help. Thanks.
First, the power series is $\displaystyle\sum_{n=1}^{\infty} \displaystyle\frac{(n!)^2 x^n}{2^n ((2n)!)}$. We claim that the series converges if $|x|<8$. If we use the ratio test, then, we need to compute $$\lim\limits_{n\to\infty} \displaystyle\frac{\frac{(n!)^2}{2^n (2n!)}}{\frac{((n+1)!)^2}{2^{n+1} (2n+2)!}}=\lim\limits_{n\to\infty}\displaystyle\frac{(n!)^2 2^{n+1} (2n+2)!}{2^n (2n)!((n+1)!)^2}=\lim\limits_{n\to\infty} \displaystyle\frac{2(2n+1)(2n+2)}{(n+1)(n+1)}=\lim\limits_{n\to\infty}\displaystyle\frac{8n^3+12n+4}{n^2+2n+1}=8$$Thus, the radius of convergence is $8$. Then, the power series converges in $(-8,8)$. But, what happens in $8$ and $-8$? Wolfram says that the series diverges, but, I don't know how to prove that claim. Both, the ratio and the root test have failed. I thought in the comparison test (by inequalities) but, seems so hard. Probably the solution is very easy, but, really I can't see it.
You can prove (using induction or other means, like complex analysis) that $$ {2\,n\choose n}=\frac{(2\,n)!}{(n!)^2}=\frac{(2\,n)(2\,n-1)\cdots(n+1)}{n(n-1)\cdots1}\le4^n. $$ Then, $$ \frac{(n!)^2\, 8^n}{2^n (2\,n)!}=\frac{(n!)^2\, 4^n}{(2\,n)!}\ge1. $$ This implies that the general term of the series does not converge to $0$, and that the series does not converge.