1
$\begingroup$

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 I`m 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?

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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, I`m not sure why this inequality holds true, can you elaborate? $\endgroup$ – Sar Jul 22 '18 at 18:10
  • 1
    $\begingroup$ @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... $\endgroup$ – ℋolo Jul 22 '18 at 18:18
  • 1
    $\begingroup$ ... $\sum_{n\in A}n|x|^{n}=\sum_{n\in A}nx^{n}$ because $n!$ is even for all $n\ge2$ $\endgroup$ – ℋolo Jul 22 '18 at 18:18
  • $\begingroup$ Thank you very much ! Indeed this is a much simpler and more elegant solution. $\endgroup$ – Sar Jul 22 '18 at 18:20
-3
$\begingroup$

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 $\sum mx^m.$

$\endgroup$
  • 1
    $\begingroup$ It coverage for $|x|\ge1$? Really? How come, $\sum m!x^{m!}\ge \sum m!$ for $|x|\ge1$ $\endgroup$ – ℋolo Jul 22 '18 at 17:54
  • $\begingroup$ 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 $\sum nx^n.$ $\endgroup$ – Aaron Meyerowitz Jul 23 '18 at 17:55

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.