Radius of convergence of $\sum_{n=1}^\infty n!\left(\frac{z^n}{n}\right)^n$

Question

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!

Answer 1

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 $1$.

Answer 2

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 $C>0$.

Answer 3

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 $1$.