Let

$$F(s)=\sum_{k=1}^\infty \frac{f(k)}{k^s} \mbox{ and }F^*(s)=\sum_{k=2}^\infty \frac{f(k)g(k)}{k^s},$$ where the infinite sum $\sum f(k)$ diverges, $f(k)$ and $g(k)$ are real numbers, $s$ a complex number, and $\sigma_c(F)$, the abscissa of convergence, is given by the well-known formula $$\sigma_c(F) = \lim\sup_{n\rightarrow\infty}\frac{\log|A(n)|}{\log n}, \mbox{ with } A(n)=\sum_{k=1}^n f(k).$$

I am wondering what functions $g(k)$ preserves the abscissa of convergence, that is, what functions $g(k)$ result in $\sigma_c(F^*)=\sigma_c(F)$. I am particularly interested in the case where $f(k)=\lambda(k)$ is the Liouville function, resulting in $F(s)=\zeta(2s)/\zeta(s)$, thus having the same roots as $\zeta$ for $\Re(s)>\sigma_c(F)$.

Obviously, the abscissa of convergence is preserved if $g(k)=(\log k)^\alpha$ and $\alpha$ is any positive integer: in that case, $F^*(s)$ is the $\alpha$-th derivative of $F(s)$, and theorem 11.12 in Apostol's book on number theory essentially states that $\sigma_c(F)=\sigma_c(F^*)$ in that case. By reversing integration and derivation, it seems obvious that it must also be true if $\alpha$ is any negative integer. And if it is true for (say) $\alpha=2$ and $\alpha=3$, there is no reason to believe it does not work with (say) $\alpha=2.81$. So this has to be true for any real $\alpha$. One would also easily imagine that it must work too if $g(k)=(\log k)^\alpha (\log\log k)^\beta$ for any real numbers $\alpha,\beta$.

My question:

Is there a reference (or can you prove / disprove) that $\sigma_c(F)=\sigma_c(F^*)$ assuming $A(n)$ diverges and $g(k)=(\log k)^\alpha$, for any real number $\alpha$, or at least if $\alpha$ is a negative integer?

Other questions and remarks

Let's say $g(k)=\lambda(k)$ is the Liouville function. I am wondering if $A(n)$ diverges, I am sure it does, but can't remember seeing a proof. Also the distribution of $+1$ and $-1$ in the $(\lambda(k))$ sequence is 50/50. Is that a consequence of the prime number theorem? I think I read someone saying this.

Finally, if $F(s)$ convergences for some $s=s_0$, then it is known that $F(s)$ converges for $\Re(s)>\Re(s_0)$, and thus $\sigma_c(F)\leq\Re(s_0)+\epsilon$ for any $\epsilon>0$. In the case $f(k)=\lambda(k)$, proving that $F(s)$ converges at $s=0.9$ (a real number) would imply that $\zeta(s)$ has no zero in $0.9 < \Re(s) < 1$. This is impossible to prove yet. When I made my computations, it really seemed to converge, and what's more, to the correct value computed by Mathematica (Mathematica is based on the analytic continuation of $\zeta$, my computations are based on the series $F(s)$). Maybe it might be easier to prove the convergence of $F^*(0.9)$ by choosing (say) $\alpha=-1$. It would have the same exact implications.