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) = limsup_{nrightarrowinfty}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 $g(k)=(log k)^alpha (loglog k)^beta$ 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 $sigma_c(F)leqRe(s_0)+epsilon$ 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.