Convergence/Divergence of $\sum_{n=1}^{\infty}\frac{n+n^2+\cdots+n^n}{n^{n+2}}$ 
$$\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}$$

$$\sum_{n=1}^\infty \frac{1}{n^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{n^{n+2}}=\sum_{n=1}^\infty \frac{n+n+\cdots+n}{n^{n+2}}\leq\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq \sum_{n=1}^{\infty}\frac{n^n+n^n+\cdots+n^n}{n^{n+2}}=\sum_{n=1}^\infty \frac{n^{n+1}}{n^{n+2}}=\sum_{n=1}^\infty \frac{1}{n}$$
But is still does not help to conclude about converges/diverges
 A: You can make your approach work if you do the following:
$$\sum_{n=1}^\infty \frac{1}{n^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{n^{n+2}}=\sum_{n=1}^\infty \frac{n+n+\cdots+n}{n^{n+2}}\leq\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq \sum_{n=1}^{\infty}\frac{n^{\color{red}{n-1}}+n^{\color{red}{n-1}}+\cdots+n^{\color{red}{n-1}}+n^n}{n^{n+2}}=\sum_{n=1}^\infty \frac{n^{n-1}(n-1)+n^n}{n^{n+2}}$$
$$=\sum_{n=1}^\infty \frac{n^{n}+n^n-n^{n-1}}{n^{n+2}}=\sum_{n=1}^\infty \frac{2}{n^2}-\sum_{n=1}^\infty \frac{1}{n^3}$$
A: We have: $n+n^2+\cdots + n^n = n(1+n+n^2+\cdots + n^{n-1})= \dfrac{n(n^n-1)}{n-1}\implies \displaystyle \sum_{n=1}^\infty \dfrac{n+n^2+\cdots n^n}{n^{n+2}} < \displaystyle \sum_{n=2}^\infty \dfrac{1}{n(n-1)}$, and this one converges .
A: If $n>1$ we have $n+n^2+\dots + n^n=\frac{n^{n+1}-1}{n-1}-1\leq\frac{n^{n+1}}{n-1}\leq 2n^{n}$
So $\frac{n+n^2+\dots + n^n}{n^{n+2}}\leq n^{-2}$
Therefore $\sum_{n=2}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq\sum_{n=2}^\infty n^{-2}<\infty$
A: Easy with equivalents (this requires the general term to have a constant sign) and some computation:
$$n+n^2+\dots+n^n=\frac{n(n^n-1)}{n-1}\sim_\infty n^n$$
so
$$\frac{n+n^2+\cdots+n^n}{n^{n+2}}\sim_\infty\frac{n^n}{n^{n+2}}=\frac1{n^2},$$
which converges.
