Part a) of a question required showing that the radius of convergence of the power series $\sum_{} \frac{n!}{n^n}x^n$ and $\sum_{} \frac{n^n}{n!}x^n$ are e and 1/e respectively. This was fairly simple.
Part b) of the same question required showing that $a(n)=(1+\frac{1}{n})^n$ is a strictly increasing sequence (whose limit is e). I was also able to get this out.
But part c) now states: Deduce from your working in a) and b) that $\sum_{} \frac{n!}{n^n}x^n$ diverges for x=e and x=-e.
I know that I am meant to substitute the values of the endpoints e and -e into the sequence. However, i have tried numerous tests - (ratio test which gave r=1) and have also been unable to get anywhere with the integral, comparison and kth term tests for divergence.
Any help will be greatly appreciated. Thank you
For $n\geq2,$ wwe have $$\log{n!}=\sum_{k=2}^n{\log{k}}>\sum_{k=2}^n{\int_{k-1}^k\log{x}\mathrm{dx}}=(x\log{x}-x)|_1^n=n\log{n}-n+1$$
(The inequality $\log{k}>\int_{k-1}^k\log{x}\mathrm{dx}$ is true because $\log{x}$ is an increasing function, so on the interval $(k-1,k),\ \log{k}>\log{x}.)$
Therefore,$$ \log{n!e^n\over n^n}=\log{n!}+n-n\log{n}>(n\log{n}-n+1)+n-n\log{n}=1$$so that $${n!e^n\over n^n}>e,$$
and $\sum {n!e^n\over n^n}$ diverges, since the $n$th term doesn't go to $0$.
Your question says that you are supposed to answer part c) based on your workings of parts a) and b), and you haven't shown those, so I don't know if you can use this answer directly, but perhaps it will help.
A more careful estimate of the integral in the first line is an essential part of the derivation of Stirling's approximation.