Radius of Convergence of the series $\sum_{n=1}^{\infty}x^{n!}$ I was trying to find the Radius of Convergence of the following series: $\sum_{n=1}^{\infty}x^{n!}$. I wrote the sum as $\sum_{n=1}^{\infty}x^{((n-1)!)^n}$ and replaced $x^{(n-1)!}$ with $y$. Then the series, $\sum_{n=1}^{\infty}y^n$, converges when $|y|<1$. Now if $y=1$, then $x$ must be $0$ because $\exists$ no $n\in \mathbb{N}$: $x^{(n-1)!}=0$. So the radius of convergence must be $0$. Am I on the right track?
 A: This is an example
of a Lacunary function.
See
https://en.wikipedia.org/wiki/Lacunary_function
The radius of convergence is $1$,
but,
unlike functions such as
$\sum_{n=0}^{\infty} x^n
=\dfrac1{1-x}$,
which can be continued
beyond the unit circle,
this function has essential singularities
at all points
on the unit circle
$e^{2\pi ir}$
where $r$ is rational.
To see this,
note that,
if $r = a/b$,
then
$\begin{array}\\
(e^{2\pi ir})^{n!}
&=(e^{2\pi i a/b})^{n!}\\
&=e^{2\pi i n! a/b}\\
&=1
\qquad\text{if } n \ge b\\
\end{array}
$
so that all terms after the first $b$
are $1$.
A: Why not using the root test? Your series is


*

*$\sum_{k=1}^{\infty}a_kx^k$ with 


$$a_k = \begin{cases}
0 & k \neq n! & (n \in \mathbb{N})\\
1 & k = n! &(n \in \mathbb{N})
\end{cases}$$
Now, calculate $\limsup_{k \to \infty}\sqrt[k]{a_k}$. The only relevant subsequence $a_{k_n}$ of $a_k$ which can produce a limes different from $0$ is $a_{k_n} = a_{n!}= 1$. Let $\rho$ denote the radius of convergence. Hence
$$\limsup_{k \to \infty}\sqrt[k]{a_k} = \lim_{n \to \infty}\sqrt[k_n]{a_{k_n}}= \lim_{n \to \infty}\sqrt[n!]{1} = 1 \Rightarrow \rho = \frac{1}{\limsup_{k \to \infty}\sqrt[k]{a_k}} = \frac{1}{1} = 1$$
A: Your substitution isn't valid, because you're letting the single variable $y$ stand in for $x^{(n-1)!},$ which varies with both $x$ and $n.$
Instead, we can see by comparison with $\sum_{n=1}^\infty |x|^n$ that it converges whenever $|x|<1.$ If I recall correctly, it converges only when $|x|<1.$
