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!