# Check if the following series diverges or converges

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!

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

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.

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

• Yeah, I know that too... I just want to check If by the limit comparison what I've done is indeed correct! – Bruno Reis Feb 9 '17 at 1:03
• @BrunoReis I am not sure if I understood your question. See the edit. – Henricus V. Feb 9 '17 at 1:05
• Why the downvote? – Henricus V. Feb 9 '17 at 3:58
• 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 – Bruno Reis Feb 9 '17 at 11:43