Variable Transform of Power Series

Question

I realize that this question is a here, but I have a different question. I am asked to find the radius of convergence $R$ of

$\sum_{n=1}^\infty z^{n!}$

I want to put it in the form $\sum_{k=0}^\infty c_k z^k$, as this is the form demanded by all the theorems I can use (note, I do not have the ratio test available to calculate $R$). I though to choose $k = n!$, and then $c_k = 1$ for all $k$. However, Daniel Fischer's comment says otherwise. So my question is, why is

$c_k = \begin{cases} 1, ~if~k=n! ~for~some ~n>1 \\ 2, ~if~k=1 \\ 0, ~otherwise \end{cases}$

true? (Note,I am not sure if I faithfully reproduced his comment)

Also, I was given this problem, and I am wondering if $n=1$ is a mistake. Does the problem make sense if $n$ begins at $1$, rather than $0$?

Answer 1

Okay, I figured out the answer to my question. In order the calculate the radius of convergence $R$, which my book defines as

$\frac{1}{R} = \lim \sup |c_k|^{1/k}$,

the given power series has to be of the form $\sum_{k=0}^\infty c_k (z-z_0)^k$. Thus, I want to find $c_k$ for which

$\sum_{n=0}^\infty z^{n!} = \sum_{k=0}^\infty c_k z^k$

holds, as then I will be in a position to calculate $R$ If one expands both sides of the equation, it becomes clear what we need $c_k$ to be in order for this to hold:

$2z + z^{2!} + z^{3!} + ... = c_0 + c_1 z + c_2 z^2 + ...$

For instance, we need the $z^3$ term to die, and we need $c_1 = 2$, so we need the definition of $c_k$ to be such that $c_3 = 0$ and $c_1 = 2$. And that's how one gets the definition that Daniel Fischer got.

Most may have seen that as obvious, but that's what was confusing. Hopefully it helps others.