Convergence of $\sum_{n=1}^\infty{\left(\sqrt[n]{n}-1\right)}$
Rewrite $\sqrt[n]{n}$ as $n^{1/n}=e^{(\ln n)/n}$ and use $e^x-1\sim x$ to see that $$ \sqrt[n]{n}-1\sim \frac{\ln n}{n} $$ (where I use $\sim$ to mean “is asymptotically equal to”). Now compare with the harmonic series.
A variation on Harald's answer. For all real $x>0$ and integral $n \geq 1$ $$\log(x) \leq n(x^{1/n} - 1)$$ and in particular $$ \frac{\log(n)}{n} \leq n^{1/n} - 1.$$ So it is a divergent series.