# Does the sum of the reciprocal highly composite numbers converge or diverge?

Does the sum $$1/1 + 1/2 + 1/4 + 1/6 +\dots$$ converge or diverge, given it's the sum of the reciprocals of the Highly composite numbers?

List of highly composite numbers: https://oeis.org/A002182 . . . I've tried to do some approximations, but I'm unsure, considering there is a relatively small amount of highly composites even up to millions of digits.

I'm assuming that its probably pretty hard to conclusively show this series converges or diverges - and am mostly looking for anyone's thoughts on if it probably converges or diverges.

• Your OEIS link references Erdos, who references Ramanujan proving that the ratio test applied to your series is inconclusive. And Erdos's result gives the same result (although with an effective error term describing how rapidly the ratio approaches $1$ (not rapidly)). So you might want to read through the references there to see how close to a result using a convergence test you know you are able to get. Commented Aug 7 at 5:11
• HCN has an asymptotic growth formula so I think we can use $\int e^{-\sqrt{x}}dx$ to dominate the series. Commented Aug 7 at 5:12

Let $$a_n$$ denote the $$n$$-th highly composite number (A002182). You observed that “... there is a relatively small amount of highly composites even up to millions of digits.” That is because the sequence $$(a_n)$$ grows “fast” and that makes it likely that $$\sum 1/a_n$$ is convergent.

In order to make this rigorous one can try to find a lower bound for $$a_n$$. There are surely more precise estimates of the asymptotics, but the following can be obtained with elementary means and is good enough to prove the convergence:

$$a_n$$ has at least $$n$$ divisors. On the other hand, the “number of divisors function” $$d(n)$$ satisfies $$d(n) \le 2 \sqrt n$$, see for example Upper limit for the Divisor function.

Therefore $$n \le d(a_n) \le 2 \sqrt{a_n} \implies a_n \ge \frac 14 n^2$$ so that $$\sum_{n=1}^\infty \frac{1}{a_n}$$ is convergent.

See also A352418 for the decimal expansion of the sum of the reciprocals of the highly composite numbers: $$\sum_{n=1}^\infty \frac{1}{a_n} = 2.13287282352581201034387841183051877907288806895933...$$

Given $$d \ge 1$$, find $$a(d)$$ which is the smallest number with $$d$$ divisors. For every $$d$$, $$a(d)$$ is the only number that might be a highly composite number with $$d$$ divisors (for example $$a(3) = 8$$ is not highly composite).

You can calculate that $$a(d)$$ grows quite fast with $$d$$, so the sum of $$1 / a(d)$$ is bounded, and the sum of the reciprocals of highly composite numbers is less.