Convergence of a complex power series Consider the series: 
$$\sum_{n=1}^{\infty} \frac{n!}{n^n}z^n$$

I found radius of convergence using almbert $R=e$

I know that this series converges for $|z|&lt;e$ , does it converge for |z|=e
Letting $z = ee^{i\theta}$, your sum becomes

$$\sum_{n=1}^{\infty} \left(\frac{e}{n}\right)^nn!\,e^{in\theta}.$$

In order for the sum to converge, we need that the summand goes to zero as $n$ tends to infinity. Does that happen here?
Let's use Stirling's formula: $n! \sim \sqrt{2\pi n} (n/e)^n$, so you have $$\sum_{n=1}^\infty \frac{n!}{n^n} e^n e^{int} \sim \sum_{n=1}^\infty \sqrt{2\pi n}  e^{int}$$ if $|z|=e$. But $\lim\limits_{n\to\infty}|\sqrt{2\pi n}  e^{int}| = \infty$, and hence еhis series diverges.