A question about reciprocal sums and growth

For integers nZn \in \mathbb{Z}, we know that the partial sum of their reciprocals grows asymptotically as n=1x1nlnx \displaystyle \sum_{n=1}^x \frac{1}{n} \simeq \ln x

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

p=1x1pln(lnx) \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 A\mathbb{A} of Z\mathbb{Z} such that, for aAa \in \mathbb{A} , we have

a=?x1aln(ln(lnx)) \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, 6 months 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), 3(2), 5(3), 11(5), 17(7), 31(11), 41(13), \ldots

There are NlnN \frac {N}{\ln N} primes up to NZ N \in \mathbb{Z} and NlnNln(NlnN)=N(lnN)2(lnN)(ln(lnN)) \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.


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

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

Henry U - 2 years, 6 months ago

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

I also found that the series

x=11xlnxlnlnx \displaystyle \sum_{x=1}^\infty \frac{1}{x\cdot lnx \cdot lnlnx}

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

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

x=11pxln(px/x) \displaystyle \sum_{x=1}^\infty \frac{1}{p_x \cdot ln(p_x / x)}

This roughly corresponds to the sequence generated by pxln(px/x)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, [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, 326,334,346,358,362,382,386,394,398,422,446,454,458,466,478,482,502,514,526,538,542,554,562,566,586,614,622,626,634,662,674,694,326, 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]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 ]

Levi Walker - 2 years, 6 months ago

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The formula

x=11xlnxln(lnx) \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(lnx))) \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...

Henry U - 2 years, 6 months ago

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@Henry U Yeah, it seems like n=1x1ni=1kln(k)(n)ln(k+1)(x) \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>0k>0. I think the k=0k=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.

Levi Walker - 2 years, 6 months ago

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@Levi Walker 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 lnx \ln x already grows very slowly.

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

Henry U - 2 years, 6 months ago

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