Prove that the following series is convergent for all $z\in\Bbb C$ such that $|z|<1$. $$\sum_{n=1}^\infty z^{n!}$$
Here is what I've got so far
Claim: The above series converges for $|z|<1$.
Pick $|z|<r<1$. Then for all $n$, $|z^{n!}|<=r^{n!}$.
So $\sum\limits_{n=1}^\infty r^{n!}$ is a majorant for $\sum\limits_{n=1}^\infty z^{n!}$.
$\sum\limits_{n=1}^\infty r^{n!}$ is a real series so we can test for convergence.
This is where I get stuck, I've tried the ratio test but that doesn't seem to work and I can't think of a function that would work for the comparison test.
 A: Try the root test, instead of the ratio test.
Alternately, observe that $\sum\limits_{n=0}^\infty r^{n!}$ is a subseries (in a sense) of the convergent geometric series $\sum\limits_{k=0}^\infty r^k=\frac1{1-r}$, so we can use comparison test that way. Hint: Write $\sum\limits_{n=0}^\infty r^{n!}=\sum\limits_{k=0}^\infty c_kr^k$, with the $c_k$'s defined appropriately.
A: $$\sum_{n=1}^\infty z^{n!}= \sum_{n=1}^\infty a_{k}z^{k}$$ then $a_{k}=1$ for $k=n!$ and $a_{k}=0$ $k \neq n!$ then apply root test
A: Since $|z|<1$, then
$$\frac{|z|^{(n+1)!}}{|z|^{n!}} = |z|^{n\cdot n!} \le |z| < 1$$
for $n\ge1$, so the series is absolutely convergent by the ratio test. 
A: Forget this is an entire series and define $x_n=z^{n!}$ for every $n\geqslant1$. 


*

*If $|z|\geqslant1$, then $|x_n|\geqslant1$ for every $n$ hence the sequence $(x_n)_n$ does not converge to zero hence the series $\sum\limits_nx_n$ diverges. 

*If $|z|\lt1$, then $|x_n|\leqslant|z|^n$ for every $n$ (since $n!\geqslant n$) and the series $\sum\limits_n|z|^n$ converges hence the series $\sum\limits_nx_n$ converges (absolutely).

