Let $P = \{p_1, p_2, \ldots, p_k\}$ be a finite set of prime numbers, and $M(P)$ be a set of natural numbers, whose prime divisors are in $P$. How can I prove that $$\sum_{n\in M(P)} \frac{1}{n}$$ converges and how can I find its sum? I would highly appreciate any help on this subject.
2 Answers
If $M(P)$ is the set of all natural numbers whose prime divisors in $P$, we can use a similar approach to the Euler product formula for the Riemann Zeta function.
Let $S = \sum_{n \in M(p)} \frac {1}{n}$. Then $\frac{1}{p_1} S$ is the sum over the numbers whose prime factora are in $P$ and the $p_1$ exponent in each of these numbers is at least 1. So $\left( 1-\frac{1}{p_1} \right) S $ removes the terms from $S$ that are divisible by $p_1$. Repeating this sieving process we get
$$ \left( 1 - \frac{1}{p_1} \right) \left( 1- \frac{1}{p_3} \right) \cdots \left( 1-\frac{1}{p_k} \right) S = 1.$$
You can finish from here.
Hint: convince yourself that
\begin{align} \sum_{n \in M(P)} \frac1n &= \prod_{j=1}^k \left(\sum_{n=0}^\infty \frac1{p_j^n}\right) \\&= \left(1 + \frac1{p_1} + \frac1{p_1^2} + \dots\right)\left(1 + \frac1{p_2} + \frac1{p_2^2} + \dots\right) \dots \left(1 + \frac1{p_k} + \frac1{p_k^2} + \dots\right) \end{align}
i.e. in the expansion of the product the reciprocal of each $n\in M(P)$ appears once and only once. (You may need to use the Fundamental Theorem of Arithmetic) Then use the sum of geometric series formula.
EDIT: since $M(P)$ may not contain all numbers with prime divisors only in $P$, the sum cannot be obtained without further information, but the upper bound still stands.
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$\begingroup$ $M(P)$ is * * a * * set whose prime divisors are in $P$, not the set of all of them. $\endgroup$– zwimCommented Mar 19, 2021 at 8:43
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$\begingroup$ Apparently this product is related to $n / \phi(n)$ for $n = p_1p_2 \dots p_k$; see artofproblemsolving.com/community/c6h1537116p9278773. $\endgroup$ Commented Mar 19, 2021 at 8:44