convergence of the series $\sum u_n, u_n = \frac{n^n x^n}{n!}$ for $x>0$
Solution 1:
$$ n+n^2\log\frac{n}{n+1} = n-n^2\log\frac{n+1}{n} = n-n^2\log\left(1+\frac{1}{n}\right) $$ 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}
Solution 2:
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.