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?!


  • $\begingroup$ You'll want to start the sum from $n=2$. $\endgroup$ – Simply Beautiful Art Feb 9 '17 at 1:01
  • $\begingroup$ @SimplyBeautifulArt Yeah, I'll correct that. Sorry. But is what I've done correct? $\endgroup$ – Bruno Reis Feb 9 '17 at 1:02
  • $\begingroup$ $(\log n)^2$ and $\log(n^2)$ are both unambiguous as to what is being squared. $\log(n)^2$ may be problematic. $\endgroup$ – Michael Hardy Feb 9 '17 at 1:20

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


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

We have the Cauchy condensation test:


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

  • $\begingroup$ Thanks. O just want to check if that was possible.. what you've just said makes a lot of sense. TY $\endgroup$ – Bruno Reis Feb 9 '17 at 1:10
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    $\begingroup$ No problem :-) (also, I love using Cauchy condensation test for any series involving logarithms. Best way to deal with them IMO) $\endgroup$ – Simply Beautiful Art Feb 9 '17 at 1:11
  • $\begingroup$ Thank you for that mate! I'll read about it... $\endgroup$ – Bruno Reis Feb 9 '17 at 2:31

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

  • $\begingroup$ Yeah, I know that too... I just want to check If by the limit comparison what I've done is indeed correct! $\endgroup$ – Bruno Reis Feb 9 '17 at 1:03
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    $\begingroup$ @BrunoReis I am not sure if I understood your question. See the edit. $\endgroup$ – Henricus V. Feb 9 '17 at 1:05
  • $\begingroup$ Why the downvote? $\endgroup$ – Henricus V. Feb 9 '17 at 3:58
  • $\begingroup$ Oh mate, if I did It wasn't on purpous! Your explanation is good. It's just because the other one is a little bit more clear. I'll upvote urs. I don't think that I've downvoted yours! Thanks $\endgroup$ – Bruno Reis Feb 9 '17 at 11:43

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