Convergence of the series $\sum_{n=0}^\infty \frac{n^n z^n}{n!}$ when $e|z|=1$

complex-analysisconvergence-divergencereal-analysissequences-and-series

Using the ratio test or any other methods, one can easily find the radius of convergence of the following series to be $1/e$.
$$
\sum_{n=0}^\infty \frac{n^n z^n}{n!}
$$
When $z=1/e$, wolfram tells me it diverges, although I cannot prove why it diverges. For $z=-1/e$, it is an alternating series, so it definitely converges. It should also be convergent when $z=\pm i/e, i^2=-1$.

Let the set $S$ be the set of angles $\theta\in (-\pi,\pi]$ for which the series converges when $z=e^{i\theta-1}$. I want to find the set $S$. I am guessing that $S=\{\theta\in(-\pi,\pi]:\theta\neq 0\}$, but I don't know how to prove it.

Can this be generalised to more general series?

Best Answer

As is quite frequently the case on the boundary of convergence, this is a job for Dirichlet's test:

If

  • $(a_n)$ is a real sequence with $a_n \downarrow 0 $
  • $(b_n)$ is a complex sequence with bounded partial sums, $ \left\lvert \sum_{k=0}^n b_k \right\rvert < M$ for some fixed $M$,

then $ \sum_{n} a_n b_n $ converges.

Namely, since $$ \frac{n^n e^{-n}}{n!} \downarrow 0 , $$ (the ratio of successive terms is $ (1+1/n)^n/e < 1$, see below for a better estimate) and $$ \left\lvert \sum_{k=0}^n e^{in\theta} \right\rvert = \left\lvert \frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}} \right\rvert < \csc{(\theta/2)} $$ is bounded for $\theta \neq 0$, we do indeed have convergence for all points on the circle apart from $\theta=0$.

As for divergence for $\theta=0$, there are various ways to do this, but probably the simplest is to derive a lower bound for $n^n e^{-n}/n!$ and apply the comparison test. The graph of $\log{x}$ is concave, so the Trapezium Rule gives an underestimate of its area; this can be massaged to give the following lower bound: $$ \frac{1}{2}\log{1} + \log{n!} - \frac{1}{2}\log{n} < \int_1^n \log{x} \, dx = 1-n + n\log{n} , $$ so $$ n! < e\sqrt{n} n^n e^{-n} . $$ Therefore $$ \frac{n^n e^{-n}}{n!} > \frac{1}{e\sqrt{n}} , $$ and finally $\sum_n n^{-1/2}$ diverges by comparison with the harmonic series, for example.

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