Define $f(s)$ as
$$ f(s) = 1 + \sum_{n=2}^{\infty} \frac{n^{-s}}{\ln(n)}$$
where we take the upper complex plane as everywhere analytic.
Notice this is an antiderivative of the Riemann Zeta function, hence there is a log branch. But like i said we take the upper complex plane as everywhere analytic, in other words we take the branch cut at the reals $<1$.
Now it appears that
$$f(s) = 0$$
implies that the real part of $s$ is smaller or equal than $1/2$ :
$Re(s) \leq \frac{1}{2}$
Is this true ?
I tried the following
Let $Re(s) > 0$ and
$$ f_k(s) = 1 + \sum_{n=2}^{k} \frac{n^{-s}}{\ln(n)}$$
And then solving
$$f_k(s) = 0$$
and assuming
$$ \lim_k f_k(s) = f(s) = 0$$ for the complex values $s$ ($Re(s) > 0$) that give zero's.
And then numerically solve for that.
Not sure if that is a good and valid way.
MAIN question :
$$ 0 = 1 + \sum_{n=2}^{\infty} \frac{n^{-s}}{\ln(n)} \implies^{??} Re(s) \leq \frac{1}{2}$$