# convergence of the series $\sum u_n, u_n = \frac{n^n x^n}{n!}$ for $x>0$

While checking the convergence of the series $$\sum u_n, u_n = \frac{n^n x^n}{n!}$$ for $$x>0$$, we used ratio test to say that for $$0< x < \frac1e$$ $$\sum u_n$$ is convergent and for $$\frac1e < x<\infty$$ $$\sum u_n$$ is divergent.

We use Logarithimic Test for the case $$x = \frac1e$$, where we came across computing the limit $$\lim_{n \to \infty} n+n^2 \log \frac{n}{n+1}$$ but I am stuck in finding the limit.

$$n+n^2\log\frac{n}{n+1} = n-n^2\log\frac{n+1}{n} = n-n^2\log\left(1+\frac{1}{n}\right)$$ So do some asymptotics. As $$n \to +\infty$$, we have \begin{align} \log\left(1+\frac{1}{n}\right) &= \frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right) \\ n^2\log\left(1+\frac{1}{n}\right) &= n-\frac{1}{2}+o(1) \\ n-n^2\log\left(1+\frac{1}{n}\right) &= \frac{1}{2}+o(1) \end{align}
It follows from Stirling's approximation that$$\frac{n^n\left(\frac1e\right)^n}{n!}=\frac{n^n}{e^nn!}\geqslant\sqrt{2\pi n}.$$Therefore, the series diverges.