Does this prime sum converge?

List the primes in ascending order $p_1,p_2,p_3,...$ . Then any integer positive integer $N \ge 2$ can certainly be factorised in terms of the first $N$ (at most) primes, and hence $\sum_{n=1}^N \frac{1}{n} \; \le \; \prod_{j=1}^N \left(1 - \tfrac{1}{p_j}\right)^{-1}$ and hence $\begin{aligned} \ln\left( \sum_{n=1}^N \frac{1}{n} \right) & \le \; \sum_{j=1}^N \ln\left(1 - \frac{1}{p_j}\right)^{-1} \; = \; -\sum_{j=1}^N \ln\left(1 - \frac{1}{p_j}\right) \; = \; \sum_{j=1}^N \sum_{k=1}^\infty \frac{1}{kp_j^k} \; = \; \sum_{j=1}^N \frac{1}{p_j} + \sum_{j=1}^N \sum_{k=2}^\infty \frac{1}{kp_j^k} \\ & \le \; \sum_{j=1}^N \frac{1}{p_j} + \sum_{j=1}^N \sum_{k=2}^\infty \frac{1}{p_j^2} \times \frac{1}{k2^{k-2}} \; \le \; \sum_{j=1}^N \frac{1}{p_j} + \sum_{n=1}^\infty \frac{1}{n^2} \times \sum_{k=2}^\infty \frac{1}{2^{k-1}} \; = \; \sum_{j=1}^N \frac{1}{p_j} + \tfrac16\pi^2 \end{aligned}$ Since $\sum_n n^{-1}$ diverges, we deduce that $\sum_p p^{-1}$ diverges.