Radius of convergence of power series $\sum_{n=0}^{\infty} n!x^{n^2}$

$\sum_{n=0}^{\infty} n!x^{n^2} = 1 + x + 2x^4 + 6x^9\ldots$

Comparing this with $\sum_{n=0}^{\infty} a_nx^n=$

$a_n= n! $ or $0$

$|a_n|^{\frac1n} ={n!}^{\frac1n} $ or $0^{\frac1n}$

We need $L= \lim \sup |a_n|^{\frac1n}$ and the radius of convergence $R$ will be $R=\frac1L$

$0^{\frac1n} \to 0 $ as $n\to \infty$

$\lim {n!}^{\frac{1}{n}}= \lim \frac{(n+1)!}{n!} \to \infty$ as $n \to \infty$

So $\lim |a_n|^{\frac1n} = 0 $ or $ \infty$

But I am not sure what $\lim \sup |a_n|^{\frac1n}$ is ?? My doubt is that : We are looking for the greatest limit point of $|a_n|^{\frac1n}. $ Clearly $\infty$ seems the correct answer but $\infty$ does not belong in R and can't be a limit point. So what is the correct answer?

  • $\begingroup$ Hi, Abhay, is this question related to CSIR preparation ? If yes, could you group study with me ? $\endgroup$ – spkakkar Feb 6 at 12:21
  • $\begingroup$ @spkakkar I am not preparing for CSIR. Right now I am still an undergraduate $\endgroup$ – Abhay Feb 6 at 12:51
  • $\begingroup$ Oh Ok. thanks for reply. $\endgroup$ – spkakkar Feb 6 at 12:51

The radius of convergence is $1$, because, if $x>0$, then$$\frac{(n+1)!x^{(n+1)^2}}{n!x^{n^2}}=(n+1)x^{2n+1}$$and the sequence $\bigl((n+1)x^{2n+1}\bigr)_{n\in\mathbb N}$ converges to $0$ if $x\in(0,1)$ and diverges if $x\geqslant1$.

  • $\begingroup$ Did you use the ratio test to find the limit of $(n+1)x^{2n+1}\; \; x\in (0,1)$? $\endgroup$ – Abhay Feb 6 at 12:54
  • 1
    $\begingroup$ That's not how I thought about it, but that's a way of doing it, yes. $\endgroup$ – José Carlos Santos Feb 6 at 13:02
  • $\begingroup$ Thank you very much, but can you tell me what's the flaw in my approach, ie, what is the problem if I simply consider the limit points of $|a_n|^{1/n}$ and then take the limit superior $\endgroup$ – Abhay Feb 6 at 13:24
  • $\begingroup$ There is nothing wrong with it. It's just that it is hard to compute $\lim_{n\to\infty}\sqrt[n]{n!}$ without Stirling's formula. $\endgroup$ – José Carlos Santos Feb 6 at 13:26
  • $\begingroup$ But I did calculate it in my post, I just used the fact that $\lim a_n ^{1/n}= \lim \frac{a_{n+1}}{a_n}$ and got the limit equal to infinity $\endgroup$ – Abhay Feb 6 at 14:19

Actually, $$a_n=\begin{cases}k!&\text{if }n=k^2\\0&\text{otherwise}\end{cases} $$ Thus $\sqrt[n]{a_n}$ is sometimes (but infinitely often) $\sqrt[k^2]{k!}$ and sometimes $0$. Therefore $$\limsup_{n\to\infty}\sqrt[n]{a_n}=\limsup_{k\to\infty}\sqrt[k^2]{k!} $$ As $\sqrt[k^2]{k!}\le \sqrt[k^2]{k^k}=\sqrt[k]k\to1$, the $\limsup$ must be $\le1$. On the other hand, $\sim \frac k2$ of the factors making up $k!$ are $\ge \frac k2$, hence $\sqrt[k^2]{k!}\ge \sqrt[k^2]{(k/2)^{k/2}}=\sqrt{\sqrt[k]{k/2}}\to1$, so the limsup is also $\ge1$.


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