I want to check my attempt at a proof for the divergence of
$$\sum_{n=1}^{\infty} \frac{1}{p_n} \tag{ $\star$ }.$$
We begin with assuming that $(\star)$ converges. If $(\star)$ converges, there is an integer $a$ so that, $$\sum_{j=a+1}^{\infty}\frac{1}{p_j} \lt \frac{1}{b}$$ where $b>1$. Note that given any $b$ there exists an $a$ that satisfies the above inequality. Now, we let $M = p_1\cdot\cdot\cdot p_a$ and consider the number $1+ nM$ for $n = 1,2,\dots$ Any factors of $1+nM$ are $p_i$ for $i \geq a+1,a+2,\dots$ Hence, we can write for each $g \geq 1$:
$$\sum_{n=1}^{g} \frac{1}{1+nM} \leq \sum_{x=1}^{\infty}(\sum_{j=a+1}^{\infty}\frac{1}{p_j})^x$$
But on the right hand side, we have a geometric series: $$\sum_{n=1}^{g} \frac{1}{1+nM} \leq \sum_{x=1}^{\infty}(\frac{1}{b})^x$$ Since the geometric series converges, it means that $\sum_{n=1}^{\infty} \frac{1}{1+nM}$ converges and is bounded above. But that is a contradiction because if we do the integral test on $\sum_{n=1}^{\infty} \frac{1}{1+nM}$, it diverges. Therefore, $(\star)$ diverges.