I have this goofy series $\sum \limits_{n=2}^\infty \frac{ \log_2 \left[ n \log_2^2 n \right]}{n \log_2^2 n}$ 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 $\lim \limits_{x \rightarrow \infty} \frac{\log_a x}{x^b} \rightarrow 0 \;\forall b > 0$.

So I'm looking at things like $\frac{\log_2 \left[ n^3 \right]}{n \log_2^2 n}$ (larger, unfortunately), $\frac{\log_2 \left[ n^3 \right]}{n^2 \log_2 n}$ (converges), and $\frac{\log_2 \left[ n^2 \right]}{n \log_2^2 n}$ (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?


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

$$n \log_2^2 n \ge n$$

by a little bit, so let's just test the series with $n$, rather than the complicated argument inside the logarithm. This leads to

$$\sum_n \frac{\log_2 [n \log_2^2 n]}{n \log_2^2 n} \ge \sum_n \frac{\log_2 n}{n \log_2^2 n} = \sum_{n} \frac{1}{n \log_2 n} = \infty$$

Another thing that suggests this approach is to rewrite the numerator as

$$\log_2 n + \log_2 \Big(\log_2^2 n\Big) = \log_2 n + 2 \log_2 \log_2 n$$

If $\log_2 n$ is small relative to $n$, then $\log_2 \log_2 n$ is tiny in comparison. This strongly suggests comparison to the series without this term.

  • $\begingroup$ Geez, I should have seen that. My head was still spinning from the amount of algebra it took to get everything else out of the way. Thanks. $\endgroup$ – bright-star Feb 3 '14 at 6:03

Hope it's not too far off topic but would like to follow up @T.Bongers' answer by sharing my favourite comparison of logarithms and powers. (And it's too long to fit in a comment.)

Consider the graph of $y=x^{1/10}$. You can calculate that if $x=1000000$ then $y$ is a bit less than $4$. Imagine drawing the graph on a scale of $1$ unit equals $1$ millimetre. Then although the graph is always increasing, it's like walking up a hill (if that's the right word) which climbs $4$ millimetres in a kilometre. This is clearly a climb which you would not even notice - you would think it was totally flat.

So, $y=x^{1/10}$ increases incredibly, almost unimaginably, slowly...

...but $y=\ln x$ increases even more slowly than that!! Certainly $\ln x$ is larger than $x^{1/10}$ for small values of $x$. If you draw both graphs on the same axis for say $1\le x\le1000$, you will get a very misleading impression - eventually, $x^{1/10}$ takes over and is larger than $\ln x$.

How long does it take for this to happen? Well, with a bit of trial and error you can find that the crossover point is approximately $x=4\times10^{15}$. If you (try to) draw this at a scale of $10$ centimetres for a unit - suitable for a blackboard in class - you find that the crossover point is about $3000$ times further away than the sun!

Hope this helps to illustrate how unimaginably slowly (I didn't say "almost" this time) the logarithm function grows.

  • $\begingroup$ That's a good one. Closer to home for me is visual/auditory senses, which give you great sensitivity at low ranges without causing a seizure when a truck passes :) $\endgroup$ – bright-star Feb 3 '14 at 23:20

Lemma $\displaystyle \sum_{n=2}^{\infty} \dfrac{\log_2 n}{n}$ diverges.

Proof By comparison to $\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n}$.

For $n\geq 2$, $\log_2 n \geq 1$. So $\dfrac{\log_2 n}{n} \geq \dfrac{1}{n}$. Since $\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n}$ diverges, $\displaystyle \sum_{n=2}^{\infty} \dfrac{\log_2 n}{n}$ diverges.


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.