# Convergence of $\sum_{n=1}^\infty\frac{n!\, i^n}{n^n}$

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.

• Stirling implies $n! \sim \sqrt{2\pi n} \frac{n^n}{e^n}$ – Alvin Lepik Sep 1 '19 at 5:15
• ikr but I can't figure out from where the $\sqrt{2\pi \: n}$ comes from. – holahola Sep 1 '19 at 5:19
• As was said, you don't need to hunt a fly with a cannon. Absolute convergence implies convergence. As for Stirling: I don't know where it comes from either, but work through the proof and you'll see that's the case. – Alvin Lepik Sep 1 '19 at 5:21
• @AlvinLepik the factor $\sqrt{2\pi n}$ follows from ${2n \choose n} \sim \frac{4^n}{\sqrt{\pi n}}$, which follows from CLT. – mathworker21 Sep 1 '19 at 5:27
• @mathworker21 oh, that's cool. When we proved the formula, we took it as a given and worked with $\varepsilon - \delta$. CLT was something off-limits back then. – Alvin Lepik Sep 1 '19 at 5:33

The series is absolutely convergent. You don't need Striling's approximation. Just apply ratio test: $$\frac {|a_{n+1}|} {|a_n|} \to \frac 1 e$$.
• @StefanQuandt $n!, n^n$ are too large for computers – mathworker21 Sep 1 '19 at 5:28
• That's a good thing to consider for the future. Thanks! Also, when I'm evaluating the ratio, I get 1. Could you show it a little bit more explicitly? $$lim_{n\rightarrow \infty} \left|\frac{(n+1)!i^{n+1}n^n}{(n+1)^{(n+1)} n! i^n}\right| = \left| \frac{n+1}{(n+1)^1}\right|$$ – holahola Sep 1 '19 at 5:32
• @StefanQuandt You should get $\frac {n^{n}} {(n+1)^{n}}=\frac 1 {(1+\frac 1 n)^{n}}$ – Kavi Rama Murthy Sep 1 '19 at 5:43