calculus - Convergence of $sum_{n=1}^inftyfrac{n!, i^n}{n^n}$

I need to study this sum: $\sum_{n=1}^{\infty} \frac{n!\, i^n}{n^n}$. Taking: $$\lim_{n\rightarrow \infty} \; \left|\left(\frac{n!\: i^n}{n^n}\right)^{\frac{1}{2}}\right|$$
$$\rightarrow \lim_{n\rightarrow \infty} \left|\left(\frac{n!}{n^n}\right)^{\frac{1}{2}}\right|$$

Using Stirling approximation considering the limit:

$$\ln n! \approx n \ln\:n-n$$

then $n! \approx (\frac{n}{e})^n$ (is this correct?):

$$\rightarrow \lim_{n\rightarrow \infty} \left|\frac{n}{n\, e}\right| = \frac{1}{e}$$ This doesn't make too much sense because I know the series diverges.

I think I'm missing a $\sqrt{2\pi n}$ in Stirling approximation but I don't understand why that pops out taking the $e^{(\;)}$ from the first expression.

Answer

The series is absolutely convergent. You don't need Striling's approximation. Just apply ratio test: $\frac {|a_{n+1}|} {|a_n|} \to \frac 1 e$.