Radius of Convergence of the series $\sum_{n=1}^{\infty}x^{n!}$

Question

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?

Answer 1

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 $|x|<1.$

Answer 2

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 $1$.

Answer 3

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 $$\limsup_{k \to \infty}\sqrt[k]{a_k} = \lim_{n \to \infty}\sqrt[k_n]{a_{k_n}}= \lim_{n \to \infty}\sqrt[n!]{1} = 1 \Rightarrow \rho = \frac{1}{\limsup_{k \to \infty}\sqrt[k]{a_k}} = \frac{1}{1} = 1$$