Radius of Convergence of $\sum\ z^{n!}$ Does anyone know how to find the radius of convergence of the series $\sum\ z^{n!}$, where $z$ is a complex number?
I tried to use the definition: $$\frac{1}{R}=\limsup\left|\frac{a_{n+1}}{a_{n}}\right|$$ but I wasn't successful. 
 A: Using the ratio test,
$$\Bigl|\frac{z^{(n+1)!}}{z^{n!}}\Bigl|=|z|^{(n+1)!-n!}=|z|^{n(n!)}\ .$$
If $n\to\infty$ then this last expression tends to infinity if $|z|>1$, or to zero if $|z|<1$.  So the series converges for $|z|<1$, diverges for $|z|>1$, and the radius of convergence is $1$.
The ratio test in the format you used, where $a_k$ is the coefficient of $z^k$, does not work well because lots of the $a_k$ are zero and so the required limit does not exist.
A: $$\sum_{n = 0}^{\infty} z^{n!} = \sum_{j = 0}^{\infty} a_{j} z^{j}$$
where
$$
a_{j} =
\begin{cases}
1, \ if \ j = m! \ for \ some \ m \in \mathbb{N},\\
0, \ otherwise.
\end{cases}
$$
Then,
$$\frac{1}{R} := \lim \sup |a_{n}|^{\frac{1}{n}} = 1$$.
A: For any series
$S(z)
=\sum_{n=1}^{\infty} z^{a_n}
$
where $a_n$ is a increasing sequence
of positive integers,
the radius of convergence
is one,
since it converges
for $|z| < 1$
since
$|z|^{a_n}
\le |z|^n
$
since
$a_n \ge n$
and diverges for
$|z| > 1$
since
$|z|^{a_n} \ge |z|^n$
for the same reason.
At $|z| = 1$,
things can get complicated.
However,
it certainly diverges
for $z = 1$.
