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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!

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

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