# Testing Boundary Points Of $\sum_{n = 1}^{\infty} \frac{n!z^n}{n^n}$

I'm having some trouble testing the series indicated in the title at its boundary points. I'll sketch the preliminary work, then arrive at the problem. It is clear that the series converges absolutely for all complex $z$ satisfying $|z| < e$ and diverges for all such $z$ satisfying $|z| > e$, since

$$\lim_{n \rightarrow \infty} \left|\frac{(n + 1)!z^{n + 1}}{(n + 1)^{n + 1}}\cdot\frac{n^n}{n!z^n}\right|~=~|z|\lim_{n \rightarrow \infty} \left(\frac{n}{n + 1}\right)^n~=~\frac{|z|}{e}$$

and $|z|/e < 1$ iff $|z| < e$. In the case where $|z| = e$, I expanded the exponential and resultant logarithms, then applied Euler's formula, to obtain the following expression for the series:

$$\sum_{n = 1}^{\infty} n!e^{n(1+ \log n)}(\cos n\theta + i\sin n \theta) \quad \text{where} \quad \theta = \arg \frac{z}{n}$$

I want to show that the series diverges when $|z| = e$. Assuming the series converges, it follows that

$$\sum_{n = 1}^{\infty}n!e^{n(1 + \log n)} \cos n \theta \quad \text{and} \quad \sum_{n = 1}^{\infty}n!e^{n(1 + \log n)} \sin n \theta$$

both converge. I wanted to show that the former series converges if and only if the latter series diverges, but my reasoning isn't rigorous. Now the former series converges whenever $\theta$ is an integer multiple of $\pi$, in which case the latter series diverges. And the latter series converges whenever $n\theta$ is an odd integer multiple of $\pi/2$, in which case the former series diverges. I wanted to strengthen these implications to necessary conditions for convergence, but immediately realized something was amiss when I couldn't prove them. The rest of the story is a messy attempt to apply every other sufficiently strong convergence test in my textbook, and isn't going well. So I'd like some help showing that the series diverges when $|z| = e$, or perhaps a useful hint of some kind.

Your ideas are good, but overkill. Stirling's formula gives the next few asymptotics of $n!$:

$$n! = \left(\frac n e \right)^n \sqrt{2\pi n} \left( 1 + O\left( \frac1n\right)\right)$$

as $n\to \infty$.

In particular,

$$\frac{n!z^n}{e^n} = \left(\frac z e\right)^n \sqrt{2\pi n} \left( 1 + O\left( \frac1n\right)\right)$$

and when $|z| = e$, grows in absolute value like $\sqrt{2\pi n}$.

Therefore nowhere on the circle of radius $e$ do your summands converge to $0$, and so the series diverges.

• Thanks for looking at the problem. Stirling's formula actually hasn't been covered in my textbook (Apostol's Calculus), but looking at the asymptotics of the factorial is a good idea. – portin.daniel Dec 20 '13 at 0:03