[tex]\displaystyle\sum_{k\ge1}\frac{\sqrt[3]{k^2+1}}{\sqrt{k^3+2}}\sim\sum_{k\ge1}\frac{k^{2/3}}{k^{3/2}}=\sum_{k\ge1}\frac1{k^{5/6}}[/tex]
Recall that [tex]\displaystyle\sum_{k\ge1}\frac1{k^p}[/tex] diverges for [tex]0\le p\le1[/tex]. 5/6 falls in this interval, so this series must diverge, and so must the original one by comparison.
