Prove that series $\sum_{n\in M(P)} \frac{1}{n}$ converges and find its sum

Question

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.

Answer 1

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.

Answer 2

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.