1

I'm trying to understand why the series $\sum_{n=2}^\infty \frac{1}{n \ln n}$ diverges. It's clear that for any $p > 1$, the series $\sum_{n=1}^\infty \frac{1}{n^p}$ (the p-series) converges. Furthermore, we have the inequality:

$$\frac{1}{n \ln n} < \frac{1}{n}$$

for all sufficiently large $n$.

Intuitively, it seems the series $\frac{1}{n \ln n}$ should also converge since its terms are smaller than those of the smallest divergent p-series. However, the standard tests (comparison, ratio, root) don't seem to help directly.

Could someone please provide:

  • Insights into why the intuitive comparison to p-series doesn't work in this case.
  • References (literature or online resources) with a perhaps more intuitive approach to understanding series' and why my way of thinking is incorrect.
Martin Sleziak
  • 56,060
Marko Pekas
  • 11
  • 3
    integral test ... $$\int_2^\infty\frac{dx}{x} = \left[\ln x\right]{x=2}^\infty=\infty\ \int_2^\infty\frac{dx}{x\ln x} =\left[\ln \ln x\right]{x=2}^\infty =\infty$$ – GEdgar Jun 15 '24 at 20:15
  • Hopefully this also helps to appreciate why $\Theta(n\ln(n))$ runtime is better than $\Theta(n^{1.00001})$ runtime for an algorithm! – LSpice Jun 15 '24 at 20:32
  • 4
    Perhaps the Cauchy Condensation test https://en.wikipedia.org/wiki/Cauchy_condensation_test may be helpful for you. $\sum_{n=2}^\infty \frac{1}{n \log n}$ essentially condenses to the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$, whilst $\sum_{n=2}^\infty \frac{1}{n^p}$ condenses to the geometric series $\sum_{n=1}^\infty \frac{1}{2^{(p-1)n}}$, and the harmonic series condenses to the series $\sum_{n=1}^\infty 1$. Try mapping your intuition over to the condensed case: for instance, the harmonic series is smaller than the smallest divergent geometric series, yet still diverges. –  Jun 15 '24 at 20:44
  • @GEdgar yes, thank you. I know it diverges. It was more a question of intuitive understanding of why it converges. – Marko Pekas Jun 15 '24 at 22:49
  • 1
    @LSpice that makes sense! Thank you. – Marko Pekas Jun 15 '24 at 22:52
  • 1
    @Terry Tao - amazing, thank you. That's something I didn't see in my university literature. It helps a lot. It seems my reasoning for why something converges/diverges was flawed – Marko Pekas Jun 15 '24 at 22:55
  • See also Infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$ and other questions linked there. – Martin Sleziak Jun 16 '24 at 06:31

2 Answers2

2

It seems that Terry Tao is not on MSE. Since his answer is too good to be left in the comments, I will post it here, made CW to avoid reputation. If I am wrong and he is after all a member here, then I will be happy to delete this answer so that he can post it himself.

Perhaps the Cauchy Condensation test may be helpful for you. $\sum_{n = 2}^\infty \frac1{n\log n}$ essentially condenses to the harmonic series $\sum_{n = 1}^\infty \frac1 n$, whilst $\sum_{n = 2}^\infty \frac1{n^p}$ condenses to the geometric series $\sum_{n = 1}^\infty \frac1{2^{(p - 1)n}}$, and the harmonic series condenses to the series $\sum_{n = 1}^\infty 1$. Try mapping your intuition over to the condensed case: for instance, the harmonic series is smaller than the smallest divergent geometric series, yet still diverges.

LSpice
  • 2,891
1

Intuitively, it seems the series $\frac{1}{n \ln n}$ should also converge since its terms are smaller than those of the smallest divergent $p$-series.

This simply doesn't follow. All you can conclude is that if $\sum \frac{1}{n \ln n}$ diverges then it does so more slowly than the harmonic series, which is true.

In fact we can keep going: the series $\sum \frac{1}{n \ln n \ln^2 n}$ (here $\ln^2 n = \ln (\ln n)$, and we need to omit the first few terms of the series so that this is defined) also diverges (by two applications of Cauchy condensation, for example, as mentioned in the comments) but its terms are even smaller and so it diverges even more slowly. And we can keep going with $\ln^3$ and so forth. It's even the case that for any divergent series you care to write down you can write down another divergent series whose terms are even smaller and which diverges even more slowly.

Qiaochu Yuan
  • 468,795
  • Thank you. It seems my reasoning for why something converges/diverges is flawed. Perhaps the actual value of the terms is less relevant than the way a function "acts". Which makes sense since you can multiply any series by a constant and get a "smaller" series that keeps its divergence. – Marko Pekas Jun 15 '24 at 22:57
  • @Marko: I enthusiastically encourage you to write out your reasoning in more detail. Maybe you will be able to see the error then, and if not you can ask a follow-up question about it. – Qiaochu Yuan Jun 15 '24 at 22:59