1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Stirling's formula? $\endgroup$ – Angina Seng Jan 12 '18 at 7:47
  • $\begingroup$ I don't know that formula. In my course we didn't study it. $\endgroup$ – Carlos Jiménez Jan 12 '18 at 7:56
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.