It is common knowledge that

$$\sum_{\text{Integer}}^\infty \frac{1}{n} \sim \log(n),$$

and that

$$\sum_{\text{Prime}}^\infty \frac{1}{p} \sim \log(\log(n)).$$

I am looking for another subseries of the harmonic series that diverges with some number of iterated logarithms like

$$\sum_{??}^\infty \frac{1}{q} \sim \log(\log(\log(n))).$$

For convenience this will be called "third order" divergence. I am specifically looking for a series that is not trivial. A trivial example would be a series that is constructed so that terms are only added when the cumulative sum is less than $\log(\log(\log(\log (n)))$. The more "natural" the better (although I realize that this is subjective). Any answers could include the reciprocal of primes of the form $4n+1$, all twin primes, all odd perfect numbers, etc. The only other thing needed would be to show the order of divergence.

I am also looking for verification/disproof of the current conjectured answer.

Thank you.

Bounty rules: I am looking for a proof or significant demonstration of a series that exhibits this behavior. It is not only limited to "third order" logarithms and can be any "order" greater than $2$.

I have decided to accept and bounty Winther's answer even though it would not be considered "natural". I believe that the proof is correct and it generalizes to any order of logarithm. It also provides a good explanation as to why the prime series diverges by second order logarithms.

Even though the question is answered I would still accept the submission of other "more natural" series if any of them happen to pop up.

  • $\begingroup$ @BabyDragon Unfortunately the series of reciprocal "super-primes" is known to converge. $\endgroup$ – user7530 Jun 15 '14 at 20:32
  • $\begingroup$ The notation is not right ($n$ in both sides are not related), the first one should be $\sum_{k=1}^n 1/k \approx \log n$ $\endgroup$ – leonbloy Jun 16 '14 at 1:08
  • $\begingroup$ @user7530 Do you have a link to something about the "super-primes"? Looking at that analysis could help answer this question. $\endgroup$ – Brad Jun 16 '14 at 13:38
  • $\begingroup$ @Brad there is this: cs.uwaterloo.ca/journals/JIS/VOL12/Broughan/broughan16.pdf $\endgroup$ – user7530 Jun 16 '14 at 15:09

The idea here is basically in line with user7530's answer, but allows for a simple proof. It might not satisfy your 'natural' condition, but the terms in the sequence I construct can be easily computed and it also generalizes to find a sequence with a $k$'th order divergence so it might be of interest.

We start by considering

$$ \sum^N\frac{1}{n(\log n)(\log\log n)\ldots\log^{(k)}n}$$

where $\log^{(k)}x = \log(\log(\log(\ldots(\log x))))$. By the integral test the sum diverges as $\log^{(k+1)} N$.

Now consider $a_n = n\lfloor \log n\ldots \log^{(k)} n\rfloor$ which is an stricktly increasing sequence of natural numbers (for $n> e\uparrow\uparrow k$ ; using power-tower notation). To prove the order of the divergence of the sum $1/a_n$ notice that

$$\sum^N \frac{1}{a_n} > \sum^N \frac{1}{n \log n\ldots \log^{(k)} n} \sim \log^{(k+1)}N$$

and for any $\epsilon >0$ we have

$$\sum^N \frac{1}{a_n} < \sum^N \frac{1}{n(\log n\ldots \log^{(k)} n) - n} < \mathcal{O}_{\epsilon}(1) + \frac{1}{1-\epsilon}\sum^N \frac{1}{n(\log n\ldots \log^{(k)} n)} \sim \frac{\log^{(k+1)}N}{1-\epsilon}$$

where $\mathcal{O}_{\epsilon}(1)$ is a constant that depends only on $\epsilon$. Thus the series $\sum \frac{1}{a_n}$ has a $(k+1)$'th order divergence.


I don't have a formal proof of order of divergence, but I expect

$$\sum_{n=1}^{\infty} \frac{1}{n p(\lceil p(n)/n\rceil)}$$

will do the trick, where $p(n)$ is the $n$th prime. The intuition is that since $p(n)\sim n\log n$, the denominator scales like $n\log n\log \log n$, and the series approximates the Riemann sum of $\frac{d}{dx}\log\log\log x$.


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