Sum of Reciprocals of Primes is Divergent/Proof 4

Theorem
The series:
 * $\displaystyle \sum_{p \mathop \in \Bbb P} \frac 1 p$

where:
 * $\Bbb P$ is the set of all prime numbers

is divergent.

Proof
Let $n \in \N$ be a natural number.

Let $p_n$ denote the $n$th prime number.

Consider the product:
 * $\displaystyle \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$

By Sum of Geometric Progression, we have: