$ f(s) = 1 + \sum_{n=2}^{\infty} \frac{n^{-s}}{p_n}= 0$?

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Let $p_n$ be the $n$ th prime number. Let $f(s)$ be a Dirichlet series defined on the complex plane as :

$$f(s) = 1 + \sum_{n=2}^{\infty} \frac{n^{-s}}{p_n}= 1 + \frac{2^{-s}}{2}+ \frac{3^{-s}}{3} + \frac{4^{-s}}{5} + \frac{5^{-s}}{7} + \dots$$

This function is well defined for $0 \leq \Re (s)$ and probably(?) has analytic continuation to the entire complex plane. I would like to see its continuation, but that is not the (main) question here.

Consider solving $f(s)=0$.

It appears all but a small finite amount of zero's have $\Re (s)\leq -\frac{1}{2}$.

I tried the following

Let $\Re(s) > -1$ and

$$ f_k(s) =1 + \sum_{n=2}^{k} \frac{n^{-s}}{p_n}$$

And then solving

$$f_k(s) = 0$$

and assuming

$$ \lim_k f_k(s) = f(s) = 0$$ for the complex values $s$ ($\Re(s) > -1$) that give zero's.

And then numerically solve for that.

Not sure if that is a good and valid way.

Is that true and how to prove or disprove those assumed properties ?