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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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n is always > 1 and this allows us to use AM-GM inequality,

$^3\sqrt\frac 1{(n-\frac 1{\sqrt{2}})(n+\frac 1{\sqrt{2}})} > \ ^3\sqrt{\frac{1}{n^2}}$ (Which surprisingly is the same as using common sense; Nevermind)

$$ \sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n^2-\frac{1}{2}}} > \sum_1^\infty \frac{1}{n^{2/3}}$$

if n>1,

$n^{2/3}<n$ Which is a consequence of the Trivial Inequality*,

$$ \sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n^2-\frac{1}{2}}} > \sum_1^\infty \frac{1}{n}$$

Which diverges.

There you go.

If you need more comfort with handling comparisons I suggest going through Inequalities like the Trivial Inequality, RMS-AM-GM-HM inequality and the Cauchy-Schwarz Inequality although these might not be that useful.

(*)$n^2>n \rightarrow n^3>n^2 \rightarrow n>n^{2/3}$ since we are taking principle roots and Real n>1

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