Convergence of $\sum_{n=1}^{\infty}{\sqrt[n]{n}-1 \over n}$

Since $n=\sqrt n\cdot\sqrt n\cdot1\cdot1\cdots1$, the Arithmetic-Geometric Mean inequality tells us

$$\sqrt[n]n\le{\sqrt n+\sqrt n+1+1+\cdots+1\over n}={2\sqrt n+(n-2)\over n}\lt1+{2\over\sqrt n}$$

It follows that

$$\sum_{n=1}^\infty{\sqrt[n]n-1\over n}\lt\sum_{n=1}^\infty{2\over n^{3/2}}$$

Hint. One may notice that, as $n \to \infty$, $$ \sqrt[n]{n}=n^{1/n}=e^{\frac{\ln n}n}=1+\frac{\ln n}n+\mathcal O\left(\frac{\ln^2 n}{n^2}\right) $$ giving $$ \frac{\sqrt[n]{n}-1}n=\frac{\ln n}{n^2}+\mathcal O\left(\frac{\ln^2 n}{n^3}\right) $$ and thus the convergence of the given series using Bertrand's result.