The series of reciprocals of the primes

The infinite series $\sum_{n=1}^{\infty} \frac{1}{p_n}$ diverges.

Proof. suppose  $\sum_{n=1}^{\infty} \frac{1}{p_n}$ converges. So  $\sum_{n=k+1}^{\infty} \frac{1}{p_n} < \frac{1}{2}$. Let $Q=p_0p_1...p_k$, hence $1+nQ$ is not divisible by any $p_0,p_1,...,p_k$. So $1+nQ = p_{k+1}p_{k+2}...p_m$. By consequence $\frac{1}{1+nQ} \le \sum_{k+1}^{\infty}\frac{1}{p_i} < \frac{1}{2}$. Gives $\sum_n \frac{1}{1+nQ} \le \sum_n \left( \frac{1}{2}\right)^n$, which implies $\sum_n \frac{1}{1+nQ} $ to converge, which is false as can be shown by the integral test: Let $f(n) = \frac{1}{1+nQ}$, $f$ is positive and monotone decreasing, and $\int_1^{\infty} f(n) =  \lim_{n \rightarrow \infty} \left[ \frac{log(1+nQ)}{Q} \right]_1^{\infty} = \infty$.