Determine using the comparison test whether the series $\sum_{n=1}^\infty \frac{1}{\sqrt[3]{n^2 + 1}}$ diverges How can I determine whether or not the following series converges using the comparison test? 
$$\sum_{n=1}^\infty \frac{1}{(n^2 + 1)^\frac{1}{3}}$$
As $n$ goes to infinity, the sum is roughly equal to $\sum_{n=1}^\infty \frac{1}{(n^2)^\frac{1}{3}} = \sum_{n=1}^\infty \frac{1}{n^\frac{2}{3}}$.
I believe that $\sum_{n=1}^\infty \frac{1}{(n^2 + 1)^\frac{1}{3}} < \sum_{n=1}^\infty \frac{1}{n^\frac{2}{3}}$ for all $n$.
Using the $p$ test, it is clear that the latter sum diverges. However I cannot say that the former also diverges as the inequality sign does not satisfy the conditions of the comparison test for divergence.

Note that $n^2+1 \le 2 n^2$ so you have
${1 \over \sqrt[3]{n^2+1}} \ge {1 \over \sqrt[3]{2}} {1 \over \sqrt[3]{n^2}} $.
