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!

  • 5
    $\begingroup$ You have $\sqrt[3]{n^2 - \frac{1}{2}} \leq \sqrt[3]{n^2}$. $\endgroup$ – Joe Johnson 126 May 17 '17 at 12:08

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.