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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?

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3 Answers 3

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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

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

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.

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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. –  Trevor Alexander Feb 3 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.

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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 :) –  Trevor Alexander Feb 3 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.

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