radius of convergence $\displaystyle \sum_{n=1}^{\infty}n! x^{n!}$

Question

I just wondering radius of convergence following series $$ \sum_{n=1}^{\infty}n! x^{n!} \\ $$ My 1st attempt is 'root test' $$ \sqrt[n!]{|a_{n!} |} =\sqrt[n!]{|n! |} =\sqrt[t]{t} \rightarrow 1 $$ So, I thoungt radius of convergence is '1'

Is this right answer?

$$ $$

$$ $$

And 2nd question!!

My friend solve this problem wiht "ratio test"

He said

$$ |\frac{(n+1)!x^{(n+1)!}}{n!x^{n!}}| \le 1 \implies |(n+1)x^{n \cdot n!}|\le1 \implies |x|\le|\frac{1}{n+1}|^{\frac{1}{n\cdot n!}} \rightarrow1 $$

So, radius of convergence is '1'

Is this right answer?

$$ $$ I've solved problem of this type through $$ |\frac{a_{n+1}x^{n+1}}{a_n x^n}|\le1 $$ So, I am uncertain for my friend's solution.

BECAUSE for the series, $$ a_2=0, a_3=0, a_k=0 ( k \not=n!) $$

thus there will be $$ \frac{0}{0} $$

$$ $$

Please give me some advice.

Thanks in advance.

Answer 1

The radius of convergence is $1$, since the sum converges for $x\in(-1,1)$, sometimes both approaches (ratio and root) will work, so it is fine that there are two methods.

Your uncertainty with $\frac{0}{0}$ is okay, because if we we had some series $\sum_{n=1}^\infty x_n$, and we knew for certain terms, where $x_n=0$, say for a set $S\subset\Bbb N$, then we would just rewrite the series as:

$\sum_{n=1}^\infty x_n=\sum_{n\in \Bbb N\setminus S}x_n$, and we would assess the remaining sequence.

Example $x_n=1/n^2$ for $n$ odd, and $x_n=0$ for $n$ even, we would just have:

$\sum_{n=1}^\infty x_n=\sum_{n\in \Bbb N\setminus 2\Bbb N}^\infty x_n$

Answer 2

The power series here is $$ \sum_{k=0}^\infty a_kx^k $$ where $$ a_k=\left\{\begin{array}{} k&\text{if $k=n!$ for some $n\in\mathbb{Z}$}\\ 0&\text{otherwise} \end{array}\right. $$ where the non-zero terms are sparse. Several tests will fail because they reference consecutive terms (ratio test, alternating series, etc). The Root Test talks about $$ \limsup_{n\to\infty}|a_n|^{1/n} $$ This test does not care about the behavior of consecutive terms, so it is well suited to this type of series. The formula to determine the radius of convergence $$ r=\frac1{\limsup\limits_{n\to\infty}|a_n|^{1/n}} $$ is based on the Root Test.

The Comparison Test also works to show that the series converges for $|x|\le1$ since for this series $$ |a_k|\le k $$ and the series $$ \sum_{k=0}^\infty kx^k=\frac{x}{(1-x)^2} $$ converges for $|x|\lt1$.

The Term Test can also be used to show that the series does not converge for $|x|\ge1$.