Can anyone explain why $\zeta(1+it)= \sum_{n=1}^{\infty} \frac{1}{n^{1+it}}=0$ will imply $\sum_{\text{p prime}} \frac{1-\Re(-p^{-it})}{p}< \infty?$
Here is what I have so far:
We know $\zeta(s)$ does not have any zeros for $\Re(s) \geq 1.$ So $\zeta(1+it) \not=0.$
Thus, we have,
$\sum_{\text{p prime}}\frac{1-\Re(-p^{-it})}{p}= \sum_{\text{p prime}} \frac{1}{p}- \sum_{\text{p prime}}\frac{\Re(-p^{-it})}{p}$.
The first sum is known to be infinite and the second one ressembles $\zeta(1+it).$ However, I dont know how to conclude the whole sum is infinite.