By the $p$-series test, $\sum \dfrac{1}{n^p}$ converges iff $p > 1$. Here: $\sum 1/n^2$ has $p=2>1$ — converges; $\sum 1/\sqrt{n}$ has $p=1/2<1$ — diverges; $\sum 1/n$ has $p=1$ — diverges (harmonic series); $\sum 1/n^{1/3}$ has $p=1/3<1$ — diverges.