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Be a non-empty set of primes $A $. Let us define $A^{\otimes}$ as the set of numbers smooth over $A$, that are the naturals having all their prime divisors in $A$ (where $1$ is arbitrarily considered as smooth over any set).

Using an elementary proof, I have established that the following sums $$ \sum_{p \in A} \frac{1}{p} \quad \quad \mathsf{and} \quad \quad \sum_{n \in A^{\otimes}} \frac{1}{n} $$ are of the same nature, i.e. either they both converge or they both diverge; also, if these sums converge, then $$ \sum_{p \in A} \frac{1}{p} \; < \; \log \left( \sum_{n \in A^{\otimes}} \frac{1}{n} \right) \; < \; 2 \sum_{p \in A} \frac{1}{p} $$ I have been told$-$quite tersely$-$by an expert in number theory that proving these results is "a very simple exercise". Now, as I consider my proof as non-obvious, would anyone be kind enough

  • either to provide some references, should these results be well-known,
  • or to give some clues that these are indeed "simple" to prove, possibly using common non-elementary techniques?

PS: Note that the former proposition has several interesting corollaries, e.g. deriving from Brun's theorem that the sum of the reciprocals of numbers smooth over twin primes is convergent (this was incidentally my initial motivating case study).

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    $\begingroup$ One formally has $\sum_{n\in A^\otimes} \tfrac{1}{n} = \prod_{p\in A} (1 - \tfrac{1}{p})^{-1}$ by same argument as the Euler product of the zeta function, from which it follows that the convergence of the sum over primes is equivalent to convergence of the full sum over $A$-smooth numbers by standard facts about the convergence of infinite sums and products. Presumably this is the "very simple exercise" the expert you mentioned meant. This is quite similar in spirit to Euler's proof of Euclid's theorem. en.wikipedia.org/wiki/Euclid%27s_theorem#Euler's_proof $\endgroup$ May 23, 2022 at 20:38
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    $\begingroup$ @Anurag Sahay. Thanks for your prompt comment! Actually, I was well aware of the Euler product formula and the equality that you state is textually in my proof. Now, the inequalities and convergence arguments were less clear for me but the proof given below clarifies the point. $\endgroup$ May 23, 2022 at 21:19

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These are series of positive numbers, so they can be rearranged without affecting their values, whether or not they converge (look up Riemann rearrangement theorem).

Since $A^\otimes$ is the set of positive integers whose prime factorization is built from primes in $A$, $$ \sum_{n \in A^\otimes} \frac{1}{n} = \prod_{p \in A} \frac{1}{1-1/p}. $$ Why? Intuitively, expand each $1/(1-1/p)$ into a geometric series and multiply all these series together to get the series on the left. The rigorous justification of this intuitive idea is the same as the proof that the Riemann zeta-function $\sum_{n \geq 1} 1/n^s$ for ${\rm Re}(s) > 1$ can be expressed as an Euler product, and the reason can be described as follows: if $\{z_n\}$ are complex numbers where $|z_n| \leq 1 - \varepsilon$ for some $\varepsilon \in (0,1)$ and $\sum |z_n|$ is convergent (such as the sequence $1/p$ for $p\in A$ if $\sum_{p \in A} 1/p$ converges, using $\varepsilon = 1/2$) then the infinite product $\prod 1/(1-z_n)$ can be expressed as a series by writing each $1/(1-z_n)$ as a geometric series $\sum_{j \geq 0} z_n^j$ and multiplying all these series together finitely many terms at a time. That may seem like something tedious, but it's a basic kind of argument in the subject that you learn once and then refer to it any time you ever need to do something like this again.

Returning to the displayed series and product above, take logarithms of both sides: $$ \log\left(\sum_{n \in A^\otimes} \frac{1}{n}\right) = \sum_{p \in A} -\log\left(1 - \frac{1}{p}\right). $$ For prime $p$, $0 < 1/p \leq 1/2$. When $0 < x < 1$, $$ -\log(1 - x) = \sum_{k \geq 1} \frac{x^k}{k} = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots, $$ where all terms in the series are positive, so $-\log(1-x) > x$, and when $0 < x \leq 1/2$ we have $$ -\log(1-x) < \sum_{k \geq 1} x^k = \frac{x}{1-x} \leq \frac{x}{1-1/2} = 2x. $$ Taking $x = 1/p$ here, we get $$ \frac{1}{p} < -\log\left(1 - \frac{1}{p}\right) < \frac{2}{p}. $$ Sum all terms in this inequality over all $p$ and we get the inequalities that you were tersely told are "a very simple exercise".

I don't know how this argument compares with the way you had worked it out, but the conversion of the initial series over $n$ in $A^\otimes$ into a product over primes in $A$ and the bounds $x < -\log(1-x) \leq 2x$ for $0 < x \leq 1/2$ are well-known ideas by those who have worked with Dirichlet series and Euler products.

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  • $\begingroup$ Many thanks for this quick and precise answer! Actually, I was well aware of the Euler product formula and the first equality that you state is textually in my proof (admittedly, the "easy" part!). For deriving the inequalities, my proof follows a more convoluted path, so your approach is clearly better (and very informative for me!). $\endgroup$ May 23, 2022 at 21:14

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