 # What's a good class of functions for bounding/comparing ratios of complicated logarithms? I have th fabios3 2022-06-25 Answered
What's a good class of functions for bounding/comparing ratios of complicated logarithms?
I have this goofy series $\sum _{n=2}^{\mathrm{\infty }}\frac{{\mathrm{log}}_{2}\left[n{\mathrm{log}}_{2}^{2}n\right]}{n{\mathrm{log}}_{2}^{2}n}$ that Wolfram Alpha tells me diverges by the comparison test (and indeed, in the larger problem I'm working on I need to prove that the expression containing it diverges), but I'm struggling to find a good divergent lower bound.
I can't just throw away all the inner logarithms--then I run into the theorem $\underset{x\to \mathrm{\infty }}{lim}\frac{{\mathrm{log}}_{a}x}{{x}^{b}}\to 0\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall }b>0$
So I'm looking at things like $\frac{{\mathrm{log}}_{2}\left[{n}^{3}\right]}{n{\mathrm{log}}_{2}^{2}n}$ (larger, unfortunately), $\frac{{\mathrm{log}}_{2}\left[{n}^{3}\right]}{{n}^{2}{\mathrm{log}}_{2}n}$ (converges), and $\frac{{\mathrm{log}}_{2}\left[{n}^{2}\right]}{n{\mathrm{log}}_{2}^{2}n}$ (still too large).
Is there a more general class or form I can use to find a simple divergent lower bound, instead of stabbing in the dark?
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Logarithms are really small, compared to any polynomial (or any positive power of $n$). So one very frequently useful trick is to try to ignore the logarithm, since it's not going to have a huge affect on the convergence of the series. More precisely, we have that
$n{\mathrm{log}}_{2}^{2}n\ge n$
$n{\mathrm{log}}_{2}^{2}n\ge n$
by a little bit, so let's just test the series with $n$, rather than the complicated argument inside the logarithm. This leads to
$\sum _{n}\frac{{\mathrm{log}}_{2}\left[n{\mathrm{log}}_{2}^{2}n\right]}{n{\mathrm{log}}_{2}^{2}n}\ge \sum _{n}\frac{{\mathrm{log}}_{2}n}{n{\mathrm{log}}_{2}^{2}n}=\sum _{n}\frac{1}{n{\mathrm{log}}_{2}n}=\mathrm{\infty }$
Another thing that suggests this approach is to rewrite the numerator as
${\mathrm{log}}_{2}n+{\mathrm{log}}_{2}\left({\mathrm{log}}_{2}^{2}n\right)={\mathrm{log}}_{2}n+2{\mathrm{log}}_{2}{\mathrm{log}}_{2}n$
If ${\mathrm{log}}_{2}n$ is small relative to $n$, then ${\mathrm{log}}_{2}{\mathrm{log}}_{2}n$ is tiny in comparison. This strongly suggests comparison to the series without this term.

We have step-by-step solutions for your answer! Abram Boyd
Lemma $\sum _{n=2}^{\mathrm{\infty }}\frac{{\mathrm{log}}_{2}n}{n}$ diverges.
Proof By comparison to $\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n}$
For $n\ge 2$, ${\mathrm{log}}_{2}n\ge 1$. So $\frac{{\mathrm{log}}_{2}n}{n}\ge \frac{1}{n}$. Since $\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n}$ diverges, $\sum _{n=2}^{\mathrm{\infty }}\frac{{\mathrm{log}}_{2}n}{n}$ diverges.

We have step-by-step solutions for your answer!