We define these two series, for $i \neq j \in \mathbb{N}^{*^{2}}$:
$$(-1)^i \sum \limits_{n=1}^{+\infty} (-1)^{n+1} n^{-a} \cos(b \ln(n)) \ln(n)^{i} = A.$$
$$(-1)^j \sum \limits_{n=1}^{+\infty} (-1)^{n+1} n^{-a} \cos(b \ln(n)) \ln(n)^{j}= B. $$
where $a \in ]0;1[$ and $b \in \mathbb{R}$.
How to know the sign of the product AB?
The goal is to prove that the two series have the same sign.
The series converge. But if you also find short arguments to prove it, I also take it.