# Convergence radius of a power series

I came across the following question :

Find the convergence radius for the series $\sum_{n=1}^\infty n!x^{n!}$.

My initial intuition led me to believe it should converge for $x\in(-1,1)$.

I would really appreciate if anyone could review my proof and point out if Im missing anything.

The proof goes as following :

Let $a_n$ be the sequence of coefficients of $x^n$. For every $n_m=k!$ for some $k\in \mathbb{N}$, $a_{n_m}=n_m$.

By Cauchy-Hadamard theorem, $\limsup (|a_n|^\frac 1 n)=\frac 1R$, and by D'Alembert criterion, if the limit $\frac {a_{n_m+1}} {a_{n_m}}$ exists , it equals to $\lim (|a_{n_m}|^\frac 1 {n_m})$ , and indeed $\lim \frac {a_{n_m+1}} {a_{n_m}}=\frac {{n_m+1}} {{n_m}}=1$. Therefore we have $1$ as a partial limit of $|a_n|^{\frac 1 n}$. In addition, for every $n\neq n_m$, $a_n = 0$, meaning $1$ is the limit supremum. (The partial limits are either 1 or 0)

From here we can deduce R=1.

Divergence for $x=1$ is obvious, since $n!$ doesn't converge to $0$, and divergence for $x=-1$ for the same reason (since $n!$ are all even after $n=2$, we get the same series as $x=1$)

Is my proof correct?

Note that $$\sum_{n=1}^M n!x^{n!}\le \sum_{n=1}^{M!} n|x|^{n}$$

So what can we say about the limit case?

• The weak inequality holds in the limit case, and the second limit is the derivative of a geometric sum, which converges for $x\in(-1,1)$, but would that mean that the series we're interested does NOT converge for $|x|\geq 1$ ?(Or do we need to address this case seperately ?) In addition, Im not sure why this inequality holds true, can you elaborate? – Sar Jul 22 '18 at 18:10
• @Sar This does only show that the series converges for $x\in(-1,1)$, to show that otherwise the series diverge just notice that for $|x|\ge 1$ we have $\sum m!x^{m!}\ge \sum m!$ which you showed is diverge. The inequality holds because take the set $A=\{n!\mid n\le M\}$ and the set $B=\{n\mid n\le M!\}$, notice that $A\subseteq B$, so we have $\sum_{n=1}^M n!x^{n!}=\sum_{n\in A}nx^{n}$, with that we can notice that $\sum_{n=1}^{M!} n|x|^{n}=\sum_{n\in B} n|x|^{n}=\sum_{n\in A}n|x|^{n}+\mbox{positive things}$, like you said in the post we have... – ℋolo Jul 22 '18 at 18:18
• ... $\sum_{n\in A}n|x|^{n}=\sum_{n\in A}nx^{n}$ because $n!$ is even for all $n\ge2$ – ℋolo Jul 22 '18 at 18:18
• Thank you very much ! Indeed this is a much simpler and more elegant solution. – Sar Jul 22 '18 at 18:20

It certainly does NOT converge for $|x| \geq 1.$ The interval of convergence is some $(-R,R).$ It is bounded above (for $x \gt 0$) by $\sum mx^m.$

• It coverage for $|x|\ge1$? Really? How come, $\sum m!x^{m!}\ge \sum m!$ for $|x|\ge1$ – ℋolo Jul 22 '18 at 17:54
• Sorry, left out a word. My point was that , while the ratio and root tests will certainly give convergence, easiest is to say that it is absolutely convergent on $(-1,1)$ by comparison to $\sum nx^n.$ – Aaron Meyerowitz Jul 23 '18 at 17:55