Power Series interval of convergence - testing the end points Calculating interval of convergence for the power series $\sum_{n=1}^{\infty}{\frac{n^n}{n!}}x^n$
will give |x| < 1/e.
Testing the negative end point (-1/e) for inclusion is straightforward, but is there a simple way to determine inclusion of the positive end point 1/e without appealing to Stirling’s approximation?
 A: For every $n\geqslant1$, let $a_n=\sqrt{n}\cdot n^n/(n!\mathrm e^n)$, then 
$$
\frac{a_{n+1}}{a_n}=\left(1+\frac1n\right)^{n+1/2}\frac1{\mathrm e},
$$ 
hence 
$$
\log\left(\frac{a_{n+1}}{a_n}\right)=\left(n+\frac12\right)\log\left(1+\frac1n\right)-1=\left(n+\frac12\right)\cdot u\left(\frac1n\right),
$$
where the function $u$ is defined by
$$
u(t)=\log(1+t)-\frac{2t}{2+t}.
$$
The derivative
$$
u'(t)=\frac1{1+t}-\frac4{(2+t)^2}=\frac{t^2}{(1+t)(2+t)^2}
$$
is positive hence $u(t)\gt u(0)=0$ for every positive $t$. 
This shows that $\log(a_{n+1}/a_n)\gt0$ for every $n$, hence the sequence $(a_n)$ is increasing, in particular $a_n\geqslant a_1=1/\mathrm e$ for every $n\geqslant1$. Finally, the series
$$
\sum_{n\geqslant1}\frac{n^n}{n!}\left(\frac1{\mathrm e}\right)^n=\sum_{n\geqslant1}\frac{a_n}{\sqrt{n}}\geqslant\sum_{n\geqslant1}\frac1{\mathrm e\sqrt{n}},
$$
diverges. (Note that the seemingly crude estimate $a_n\geqslant1/\mathrm e\gt0.367$ is actually rather tight since $a_n\leqslant1/\sqrt{2\pi}\lt0.399$ for every $n\geqslant1$ with $a_n\to1/\sqrt{2\pi}$ when $n\to\infty$.)
A: I don't know if it's inherently different than using Stirling's approximation, but what about
$$n!\leq e\left(\frac{n+1}{e}\right)^{n+1}?$$
