Compute the radius of convergence for $\sum_{k=0}^{\infty}k!z^{k!}$ Just working on some practice questions and I'm not too sure what to do with this one. I've never encountered the $n!$ in the exponent of $z$ in these types of questions before.
Computing the radius of convergence in the usual way (LimSup$\frac{||a_k||}{||a_{k+1}||}$) I get a radius of convergence of $0$. But I'm quite certain I need to account for the $n!$ in the exponent of $z$, but have no idea what to do.
Any help would be greatly appreciated!
 A: You cannot use the ratio test in this case in the "straightforward" way by considering
$$\left|\frac{a_{k+1}z^{k+1}}{a_kz^k}\right|$$
as $k\to\infty$, because the sequence $a_{k+1}/a_k$ oscillates wildly - and, even worse, is undefined for many $k$.
But rather than using a "technique" to compute the radius of convergence, remember what it actually means: a series in powers of $z$ has radius of convergence $R$ if it diverges for $|z|>R$ and converges for $|z|<R$.  So, apply the ratio test to the terms which are "actually there":
$$\left|\frac{(k+1)!\,z^{(k+1)!}}{k!\,z^{k!}}\right|
  =(k+1)|z|^{(k+1)!-k!}=(k+1)|z|^{k(k!)}\ .$$
If $|z|>1$ then this tends to $\infty$ as $k\to\infty$, so the series diverges by the ratio test.  If $|z|<1$ it tends to $0$ and the series converges; so the radius of convergence is $1$.
A: Hint:
$a_n$ in your case is not $n!$. $a_n$ is the number that is next to $z^n$, meaning
$$a_n=\begin{cases}n&\text{if }\exists k: n=k!\\0&\text{otherwise}\end{cases}$$
A: Compare the series with $\sum_{k=0}^\infty k z^k$ which has the convergence radius $1$.
Because your series has a lot of dropped coefficients, the quotient definition is not the best idea. Use a definition that uses a single term:
$$r=\frac{1}{\lim\operatorname{sup}_{n\to\infty} \sqrt[n]{|a_n|}} $$
Because of that $\operatorname{sup}$ in there, it doesn't matter if most of the terms are zero. The ones that survive are the same as in the comparison series $\sum k z^k$, so the radius of convergence is the same.
A: $$u_n=n!z^{n!}$$
so
$$\frac{u_{n+1}}{u_n}=(n+1)z^{n(n!)}=\exp(ln(n+1)+n(n!)lnz)$$
From there, you can see two cases :
z<1, so $\ln(z)<0$, hence $$n\to \infty \implies \frac{u_{n+1}}{u_n}\to 0$$
 since $$n\to \infty \implies \frac{n!n}{ln(n+1)}\to \infty$$
z$\geq$1,so $\ln(z)\geq 0$, hence $$n\to \infty \implies \frac{u_{n+1}}{u_n}\to \infty$$
Therefore, $R=1$.  
