Radius of convergence of $\sum_{n=1}^\infty n!\left(\frac{z^n}{n}\right)^n$ I'm currently studying for my exam next month and I'm struggling to calculate the radius of convergence of
$$\sum_{n=1}^\infty n!\left ( \frac{z^n}{n} \right )^n$$
In class we always had easier ones like
$$\sum_{n=0}^\infty n!z^n$$ where we could simply use the following formula: 
$R=\frac{1}{\varlimsup |a_n|^{(1/n)}}$
So how could I caluclate the radius of convergence of my original series?
Thanks!
 A: If you rewrite it as:
$$\sum a_kz^k$$
with $$a_k=\begin{cases}\frac{n!}{n^n}&k=n^2\\0&k\text{ is not a perfect square}\end{cases}$$
Then:
$$\limsup_{k\to\infty} |a_k|^{1/k}=\limsup_{n\to\infty} |a_{n^2}|^{1/n^2}$$
And we have:
$$\frac{1}{n^n}\leq a_{n^2}\leq 1$$
So:
$$\frac{1}{n^{1/n}}\leq |a_{n^2}|^{1/n^2}\leq 1$$
By the squeeze theorem, since $\lim_{n\to\infty} \frac{1}{n^{1/n}}=1$, we get $\lim_{n\to\infty}  |a_{n^2}|^{1/n^2}=1$.
So we get a radius of convergence of $1$.
A: Since$$\frac{(n+1)!\left(\frac{z^{n+1}}{n+1}\right)^{n+1}}{n!\left(\frac{z^n}n\right)^n}=n\frac{\frac{z^{(n+1)^2}}{(n+1)^{n+1}}}{\frac{z^{n^2}}{n^n}}=\frac n{n+1}z^{2n+1}\left(\frac n{n+1}\right)^n$$and since$$\lim_{n\to\infty}\frac n{n+1}\left(\frac n{n+1}\right)^n=\frac1e,$$the series converges absolutely if $|z|<1$ and diverges if $|z|>1$. Therefore, the radius of convergence is $1$.
A: The radius of convergence $\rho$ of a power series of the form $\sum a_n z^{n^2}$ can be evaluated by
$$\rho=\frac1{\limsup|a_n|^{1/n^2}}$$
because the other coefficients are zero, so they doesnt add anything to the limit superior. In this case we have
$$\rho=\frac1{\limsup|a_n|^{1/n^2}}=\frac1{\lim (n!/n^n)^{1/n^2}}=\frac1{\lim( \sqrt{2\pi n}e^{-n})^{1/n^2}}=1$$
were we used the Stirling asymptotic $n! \sim \sqrt{2\pi n} (\tfrac ne)^n$ when $n$ goes to infinity and the known fact that $\lim C^{1/n}=\lim n^{1/n}=1$ for any $C>0$.
