How to approach this series using comparison test?

I'm wondering if someone could give me a direction to go off of with the following series:

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

I'm wondering how I can show this series diverges/converges using the comparison test for series; this series kind of resembles $1/n$ which diverges, hence why I'm thinking something like this.

I'm honestly stuck, and would really appreciate a hint!

• You have $\sqrt[3]{n^2 - \frac{1}{2}} \leq \sqrt[3]{n^2}$. – Joe Johnson 126 May 17 '17 at 12:08