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).



Series of reciprocals of smooth numbers
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.


