Check if $ \sum_{n=2}^{\infty}\frac{1}{\log(n)^2} $ diverges or converges

Question

Check if the following series diverges or converges: $$ \sum_{n=2}^{\infty}\frac{1}{\log(n)^2} $$ I know that I'm able to compute it using Integral test... But can I use Limit comparison test, with my $b_n = \log(n)^2$?

I know that the series with the sequence $b_n$ is divergent by the test of divergence ($\lim_{n\rightarrow \infty} b_n \neq 0$).

Applying the limit comparison test I'll get: $$ \lim_{n\rightarrow \infty}\frac{1}{\frac{\log(n)^2}{\log(n)^2}}\\ \lim_{n\rightarrow \infty}\frac{1}{1} = 1 $$

And because of that my first series $\sum_{n=2}^\infty \frac{1}{\log(n)^2}$ will diverge too.

Is that correct?!

Thanks!

Answer 1

You have not applied the limit comparison test correctly. It should read

$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac1{\log^2(n)}}{\log^2(n)}\lim_{n\to\infty}=\lim_{n\to\infty}\frac1{\log^4(n)}=0$$

And the limit comparison test does not work for limits that end up to be infinite or $0$.

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We have the Cauchy condensation test:

$$\sum_{n=2}^\infty\frac1{\log^2(n)}>\sum_{n=1}^\infty\frac{2^n}{\log^2(2^n)}=\frac1{\log^2(2)}\sum_{n=1}^\infty\frac{2^n}{n^2}$$

Now all you need is the term test to finish this off.

Answer 2

$(\log n)^2 \leq n$, by comparison...

Your limit comparison test doesn't look right. If you had the series $\sum_{n=1}^\infty n^{-2}$ instead, and set $b_n = n^2$, then $$ \lim_{n \to \infty} \frac{1}{\frac{n^2}{n^2}} = 1 $$ but $\sum_{n=1}^\infty n^{-2}$ is convergent.