TITLE: Prove interval of convergence for the series $\sum_{k=1}^\infty \frac{(k-1)!}{k^k} x^{k-1}$
QUESTION [1 upvotes]: Consider the following Taylor series:
$$\sum_{k=1}^\infty \frac{(k-1)!}{k^k} x^{k-1}$$
Numerical experiments suggest that the series converges on the interval:
$$x\in (-e,e)$$
Disclaimer
I'm not sure how to prove this.
I guessed the following integral representation:
$$\sum_{k=1}^\infty \frac{(k-1)!}{k^k} x^{k-1}=\int_0^\infty \frac{dt}{e^t-xt}$$
Now we can kind of 'prove' the convergence interval.
$$e^t>xt$$
$$x<\frac{e^t}{t}$$
$$\min \frac{e^t}{t}=e$$
It's pretty obvious that the integral converges for all $x<e$ and diverges for $x=e$.

How to prove that the series diverges for $x>e$ ?

REPLY [2 votes]: Look at the ratio test:  if $a_n$ is the $n$th term of the series, then
$$
\begin{align*}
\lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_n}\right\rvert&=\lim_{n\to\infty}\frac{n!}{(n-1)!}\cdot\frac{n^n}{(n+1)^{n+1}}\cdot\lvert x\rvert\\
&=\lim_{n\to\infty}n\cdot\frac{1}{n+1}\cdot\left(1+\frac{1}{n}\right)^{-n}\cdot\lvert x\rvert\\
&=e^{-1}\lvert x\rvert.
\end{align*}
$$
So, by the Ratio test, this series converges when $\lvert x\rvert<e$, diverges when $\lvert x\rvert>e$, and the behavior is indeterminant when $\lvert x\rvert=e$.
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