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

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?

• Hi, Abhay, is this question related to CSIR preparation ? If yes, could you group study with me ? Feb 6 '19 at 12:21
• @spkakkar I am not preparing for CSIR. Right now I am still an undergraduate Feb 6 '19 at 12:51
• Oh Ok. thanks for reply. Feb 6 '19 at 12:51
• Does this answer your question? Radius of convergence of $\sum_{n=0}^\infty n!x^{n^2}$ Nov 9 '20 at 3:43

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 $$x\geqslant1$$.
• Did you use the ratio test to find the limit of $(n+1)x^{2n+1}\; \; x\in (0,1)$? Feb 6 '19 at 12:54
• 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 Feb 6 '19 at 13:24
• 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. Feb 6 '19 at 13:26
• 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 Feb 6 '19 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 $$\ge1$$.