Does the sum of reciprocals of the harmonic divisor numbers converge?

Define the following:

Harmonic divisor number - $n$ such that $\sigma(n) \mid n\sigma_0(n)$. Equivalently, the harmonic mean of divisors of $n$ is an integer.

Deficient, perfect, or abundant number - $n$ such that $2n$ is greater than, equal to, or less than, respectively, $\sigma(n)$, the sum of the divisors of $n$. Every positive integer falls into one of these three categories.

Semiperfect number = $n$ such that $n$ is a sum of distinct proper divisors of $n$.

Practical number - $n$ such that every $k \in [1,n]$ is a sum of distinct divisors of $n$.

We should of course consider whether any known subsets of the harmonic divisor numbers have a divergent sum of reciprocals or if any known supersets have a convergent sum of reciprocals. The perfect numbers are a subset of the harmonic divisor numbers, but have a convergent sum of reciprocals. I don't know of any other subsets to consider. As for supersets, I'm pretty sure every harmonic divisor number is non-deficient, but I can't prove this and it wouldn't be any help anyways because every multiple of an abundant number is also abundant and the sum of reciprocals of the first $r$ multiples of any integer is just a constant times the $r$th harmonic number, thus divergent. Numerical evidence supports the stronger statement that the harmonic divisor numbers are also a subset of the intersection of the practical numbers and semiperfect numbers (same as the practical numbers that aren't powers of two), but the sum of reciprocals of practical semiperfect numbers is divergent still, so this wouldn't help either.

I don't know what other approaches there are to this problem. The harmonic divisor numbers are known to have natural density $0$, so the sum of reciprocals may or may not converge.

  • $\begingroup$ Who showed that the harmonic divisor numbers have density 0? Did they give a quantitative bound on the number of such integers less than $x$? $\endgroup$ Oct 5, 2013 at 4:06
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    $\begingroup$ @GregMartin G. Cohen says in "Superharmonic numbers" that this was proved by H. J. Kanold, but I haven't been able to find a translation of the paper. $\endgroup$ Oct 5, 2013 at 4:10
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    $\begingroup$ seem to be only linear growth in the count between $10^n$ and $10^{n+1} - 1.$ That is, less than $$ \sum \frac{n^2}{10^n} $$ which converges rapidly. $\endgroup$
    – Will Jagy
    Oct 5, 2013 at 5:47
  • $\begingroup$ @WillJagy Thanks. It's at least .21836 (from oeis list), so likely with $7$ in place of $10$, but I get your point. $\endgroup$ Oct 5, 2013 at 6:02
  • $\begingroup$ try this list ma.noda.tus.ac.jp/u/tg/files/list4 $\endgroup$
    – Will Jagy
    Oct 5, 2013 at 6:06

1 Answer 1


Yes, that sum of reciprocals does converge.

Kanold proved that for any $c>1/2$, the number of harmonic (divisor) numbers up to $x$ is eventually less than $x^c$. Any set of numbers with this property (for some $c<1$) has a convergent reciprocal sum. (To show this, bound the contribution to the reciprocal sum of the numbers in the set between $2^{k-1}$ and $2^k$, then sum over $k$.)

In fact, Pollack and Pomerance have given a better bound for the number of harmonic numbers up to $x$; see Section 5.2 of this paper1, where Kanold's result is cited. (And I learned of that paper because it cites Cohen's "Superharmonic numbers", so thank you for starting the detective trail!)

1Pollack, Paul; Pomerance, Carl, Prime-perfect numbers, Integers 12, No. 6, 1417-1437, A14 (2012). ZBL1271.11006.


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