3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $0$.


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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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.