I need to find the $x$ values for which the series $\sum_{n=1}^{\infty}\frac{x^{n}n!}{n^{n}}$ converges (when $\left|x\right|\neq e$).
I have tried to use the ratio test as follows:
$$
\left|\frac{\frac{x^{n+1}\left(n+1\right)!}{\left(n+1\right)^{n+1}}}{\frac{x^{n}n!}{n^{n}}}\right|=\left|\frac{x\cdot\left(n+1\right)\cdot n^{n}}{\left(n+1\right)^{n+1}}\right|=\left|x\cdot\left(\frac{n}{n+1}\right)^{n}\right|=\left|x\right|\cdot\left(\left(\frac{n+1}{n}\right)^{-1}\right)^{n}=
$$
$$
=\left|x\right|\cdot\left(\left(\frac{n+1}{n}\right)^{-1}\right)^{n}=\left|x\right|\cdot\left(\left(\frac{n+1}{n}\right)^{n}\right)^{-1}\xrightarrow[n\rightarrow\infty]{}\left|x\right|\cdot\frac{1}{e}
$$
So that the series converges absolutely for $\left|x\right|<e$ and it doesn't converges absolutely for $\left|x\right|>e$.
However, I got stuck on finding the values for which the series is conditionally convergent (if there are any).
How can I move on with this?
Best Answer
The series converges absolutely for $|x|<\mathrm e$ and diverges grossly for $|x|>\mathrm e$. Assuming $x$ is real, $x=\pm\mathrm e$ are the only two values for which the series is conditionally convergent.
Let $a_n:=\frac{\mathrm e^n\,n!}{n^n}$. Your computation shows that $$\frac{a_{n+1}}{a_n}=\mathrm e\cdot\left(1+\frac1n\right)^{\!-n}\ge1$$ for all $n$, so $(a_n)_{n\ge1}$ is non-decreasing. Since it is not eventually zero, this implies that $(a_n)_{n\ge1}$ cannot converge to $0$. Thus $\sum a_n$ and $\sum(-1)^na_n$ also diverge grossly.