# A question about reciprocal sums and growth

For integers $n \in \mathbb{Z}$, we know that the partial sum of their reciprocals grows asymptotically as $\displaystyle \sum_{n=1}^x \frac{1}{n} \simeq \ln x$

Likewise, we know that an analogous result holds for prime numbers $p \in \mathbb{Z}$:

$\displaystyle \sum_{p=1}^x \frac{1}{p} \simeq \ln(\ln x)$

I'm curious to know if we can go any deeper. I'm looking for a subset $\mathbb{A}$ of $\mathbb{Z}$ such that, for $a \in \mathbb{A}$, we have

$\displaystyle \sum_{a=?}^x \frac{1}{a} \simeq \ln(\ln(\ln x))$

Does a set like this exist? Is it a subset of the prime numbers, or are there elements of this set which are not primes? What special properties do these numbers have? Note by Levi Walker
2 years ago

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Maybe the primes whose index is also prime, so:

$3(2), 5(3), 11(5), 17(7), 31(11), 41(13), \ldots$

There are $\frac {N}{\ln N}$ primes up to $N \in \mathbb{Z}$ and $\frac {\frac {N}{\ln N}} {\ln \left( \frac {N}{\ln N} \right)} = \frac {N} {(\ln N)^2 - (\ln N) \cdot (\ln (\ln N))}$ of them have a prime index, maybe this helps, but I have no idea how to prove something like that.

Edit:

$\frac 1{\ln N}$ of the integers up to $N$ are primes and $\frac 1 {\ln N}$ of them have a prime index, so there are $\frac 1 {(\ln N)^2}$ primes with prime index up to $N$.

How does this fit with $\frac {N} {(\ln N)^2 - (\ln N) \cdot (\ln (\ln N))}$?

- 2 years ago

I think your asymptotic formulas are correct - I found this paper which corroborates what you did. It looks like the $-ln \: N \cdot ln(ln \: N))$ term is a higher-order correction.

I also found that the series

$\displaystyle \sum_{x=1}^\infty \frac{1}{x\cdot lnx \cdot lnlnx}$

Will asymptotically grow on the order of $lnlnlnx$. I'm not sure if the same will hold for the prime-primes, though.

Assuming $p_n \simeq n \cdot ln \: n$, maybe we can change the sum above to

$\displaystyle \sum_{x=1}^\infty \frac{1}{p_x \cdot ln(p_x / x)}$

This roughly corresponds to the sequence generated by $p_x \cdot \lceil ln(p_x / x) \rceil$, which I found (using Python) to be

$[2, 3, 5, 7, 11, 13, 17, 19, 23, 58, 62, 74, 82, 86, 94, 106, 118, 122, 134, 142, 146, 158, 166, 178, 194, 202, 206, 214, 218, 226, 254, 262, 274, 278, 298, 302, 314,$ $326, 334, 346, 358, 362, 382, 386, 394, 398, 422, 446, 454, 458, 466, 478, 482, 502, 514, 526, 538, 542, 554, 562, 566, 586, 614, 622, 626, 634, 662, 674, 694,$ $698, 706, 718, 734, 746, 758, 766, 778, 794, 802, 818, 838, 842, 862, 866, 878, 886, 898, 914, 922, 926, 934, 958, 974, 982, 998 ]$

- 2 years ago

The formula

$\displaystyle \sum_{x=1}^{\infty} \frac 1 {x \cdot \ln x \cdot \ln(\ln x)}$

looks like it could be generalized to give an asymptotic growth of

$\ln (\ln (\ln (\ldots \ln x \ldots)))$.

About the prime-primes, it was just the first thing that came to my mind and a simple application of the prime number theorem, but of course the growth of prime-primes does not directly connect to the growth of the sum of their reciprocals.

Or does it? Maybe there is a formula that "converts" between these two growth rates...

- 2 years ago

Yeah, it seems like $\displaystyle \sum_{n=1}^x \frac{1}{n \prod_{i=1}^k ln^{(k)}(n)} \simeq ln^{(k+1)}(x)$

Should hold for any integer $k>0$. I think the $k=0$ case just returns us to the harmonic series.

There might be some way to find the asymptotic density from those growth rates, which will give us information about whether a set is "dense" or not. I'm not entirely sure how to go about that though.

- 2 years ago

This is interesting since you can let these sums grow arbitrarily slow but they still converge. It can probably be one of the slowest reciprocal sums since $\ln x$ already grows very slowly.

I think the best thing is that the formula is so simple, just composed logarithms multiplied.

- 2 years ago