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First of all, I'm a novice in math and English is not my native language, so I apologize in advance for any incorrect wording, etc.

Definitions and specific example
Let's define the set $A_4 = \{2,3,5,7\}$, with the first 4 prime numbers. And $r(A_4) = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} +\frac{1}{7} = \frac{247}{210}$, as the sum of the reciprocals of the elements of $A_4$.

Let's now define the set $B_4 = \{6,10,14,15,21,35\}$, with the squarefree semiprimes with prime factors $p$ such that $p \in A_4$. Now, $r(B_4) = \frac{1}{6} + \frac{1}{10} + \frac{1}{14} +\frac{1}{15} +\frac{1}{21} +\frac{1}{35} = \frac{101}{210}$. And in a similar way, $C_4 = \{30,42,70,105\}$, with the squarefree 3-almost primes and $r(C_4) = \frac{1}{30} + \frac{1}{42} + \frac{1}{70} +\frac{1}{105} = \frac{17}{210}$. And finally, $D_4 = \{210\}$, with a square free 4-almost prime and $r(D_4) = \frac{1}{210}$.

Observation
We see: $$r(A_4) > r(B_4) > r(C_4) > r(D_4)$$

Generalization
Now change $A_4$ into $A_n$, where $n$ is a positive integer. I assume my observation holds for other - higher - values of $n$. So that: $ r(A_n) > r(B_n) > r(C_n) > \dots > r(Z_n)$, where Z doesn't mean the 26th term of the sequence, but in general 'the last' or n-th term. (I was running out of capitals... )

Question
Does my observation holds for all values of $n$? Who can I prove it (Euler probably already did it for me...)? Useful links to literature are welcome.

EDIT: counter example
Following Greg Martin's thorough answer (thx!), I make a cautious attempt to formulate a 'counter argument'.

Let's compare the first and second to last terms of the series based on $A_5 = \{a,b,c,d,e\}$. Both terms consist of the same amount of sub-terms.

first term: $$r(A_5) = u_1 = \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}$$

second to last term: $$u_{n-1}=\frac{1}{acde}+\frac{1}{bcde}+\frac{1}{bcde}+\frac{1}{bcde}+\frac{1}{bcde}$$

It's clear that the first sub-term in $u_{n-1}$ is smaller than the first of $u_{1}$. The same holds for the other sub-terms. So:

$$u_{1} > u_{n-1}$$

This is true for all values of $n$. It gives me the intuitive idea that the '<' and '>' signs 'flips' at some point. So, let me restate my question. Does my observation holds for all sufficiently large values of $n$, starting from a certain term $N$? So that: $$r(A_n) < r(B_n) < r(C_n) < \dots < r(N-1_n) < r(N_n) > r(N+1_n) \dots r(Z_n)$$

Pim Dumans
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1 Answers1

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It turns out that the answer is no, and indeed that each of these inequalities are reversed for $n$ large enough! Let's consider the first inequality as an example, noticing the identity $$ r(A_n)^2 = \sum_{1\le i,j\le n} \frac1{p_ip_j} = 2r(B_n) + \sum_{i=1}^n \frac1{p_i^2}. $$ In particular, $$ r(B_n) > \frac12r(A_n)^2 - C \quad\text{where}\quad C = \frac12\sum_p \frac1{p^2} \approx 0.2261237; $$ therefore $r(B_n)$ will ultimately be far larger than $r(A_n)$ since the latter goes to infinity. It turns out that the crossover happens at $n=176$, for which $p_n = 1049$.

In general, the reciprocal sum of the first $n$ $k$-almost primes is going to grow like a constant depending on $k$ times $(\log\log n)^k$, so each of the proposed inequalities will ultimately be reversed.

You might be interested in the 2023 paper of Gorodetsky, Lichtman, and Wong, which responds to a conjecture I made—very similar to yours—that they completely disproved :D

Greg Martin
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