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My task is this:

Find the convergence radius of$$\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n.$$

My work so far:

By ratio test we get that$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left|\frac{(n+1)^2x}{4(n+1)(n+1/2)}\right|=\lim_{n\to\infty}\left|\frac{x}{4}\right|<1\implies x\in(-4,4).$$

Now for the endpoints my first approach since the ratio test is inconclusive, was to try factor out something and then compare it. We notice that$$\frac{(\pm4)^n(n!)^2}{(2n)!}=\frac{(\pm4)^nn!n!}{2n(2n-1)\ldots n!}= \frac{(\pm4)^nn!}{2n(2n-1)\ldots(n+1)}.$$

Now if we could compare it to $$e^4=\sum_{n=0}^\infty \frac{4^n}{n!}$$ in some way or something similar, the job would be done. I just can't see it right now and need some help to finish this one off.

Thanks in advance!

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From your calculation, we have that when $x=\pm 4$, $$\left|\frac{a_{n+1}}{a_n}\right| =\frac{n+1}{n+1/2}\gt 1.$$

So when $x=\pm 4$, the terms are increasing in absolute value. As a consequence, they do not have limit $0$, and therefore the series diverges.

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  • $\begingroup$ I don't think this claim holds true and for a counter example consider $a_n = 1/n\to \left| \frac{a_{n+1}}{a_n}\right| = \frac{n}{n+1} <1$, but $1/n$ diverges for all n. $\endgroup$
    – Thomas
    Commented May 16, 2016 at 10:19
  • $\begingroup$ @Thomas: The claim is that if the terms do not have limit $0$, then the series diverges. The converse is (as you point out) not true. Terms going to $0$ does not imply convergence. But in the answer we are using the fact that the terms do not go to $0$. $\endgroup$
    – André Nicolas
    Commented May 16, 2016 at 11:39
  • $\begingroup$ Forgive me, but I must ask you to clarify a bit more or provide a link to this test. I'm familiar with the divergence test which states that if $\lim_{n\to\infty}a_n\neq 0$ then the series diverges, but what you state here seems like the ratio test which states that if $\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = a \in \mathbb{R}$ then $a_n$ konverges for $a<1$, diverges for $a>1$ and test inconclusive for $a=1$. Thanks in advance. $\endgroup$
    – Thomas
    Commented May 16, 2016 at 13:17
  • $\begingroup$ I am using the divergence test, but got the information from your ratio calculation, which tells us that for example if we use $x=4$, then $a_{n+1}/a_n\gt 1$. So the $a_k$ are increasing, and therefore cannot have limit $0$. The series therefore diverges by the divergence test. Let us calculate the first few terms of the series explicitly when $x=4$. We get $1+2+\frac{8}{3}+\frac{16}{5}+\cdots$. The terms are increasing. In the answer I used your ratio calculation to show that (for $x=\pm 4$) the terms are alway increasing in absolute value. $\endgroup$
    – André Nicolas
    Commented May 16, 2016 at 13:37
  • $\begingroup$ About your previous comment: I said absolutely nothing about the limit of the ratio. Of course at $x=\pm 4$ the limit is $1$. I worked with the ratio, not the limit of the ratio. Because $|a_{n+1}/a_n|$ is greater than $1$, $|a_{n+1}|\gt |a_n|$, so the terms are increasing in absolute value. As a consequence, they cannot have limit $0$. $\endgroup$
    – André Nicolas
    Commented May 16, 2016 at 16:00

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