Let $\omega(n)$ denote the number of distinct prime factors of n.
Does the series $$\sum_{n=1}^{\infty}\dfrac{1}{4^{\omega(n)}n}$$ converge or diverge?
From a little search I did about the $\omega$ function, I found that $$\log(\log(n))-(\log(\log(n)))^{1/2+\epsilon}<\omega(n)$$ for almost all n. So $$\sum_{n=2}^{\infty}\dfrac{1}{4^{\omega(n)}n}<\sum_{n=2}^{\infty}\dfrac{1}{4^{\log(\log(n)-\log(\log(n))^{1/2+\epsilon})}}<\sum_{n=2}^{\infty}\dfrac{1}{4^{A\log(\log(n))}n}<\infty,$$ where $$A=\min_{n\geq2}\dfrac{\log(\log (n))-(\log(\log (n)))^\epsilon}{\log(\log(n))}=\dfrac{\log(\log (2))-(\log(\log (2)))^\epsilon}{\log(\log(2))}<\dfrac{\log(\log (n))-(\log(\log (n)))^\epsilon}{\log(\log(n))},$$ since $$n\mapsto\dfrac{\log(\log (n))-(\log(\log (n)))^\epsilon}{\log(\log(n))}$$ is increasing (positive derivative).
Now we have to consider two sets of natural numbers, one which $$\log(\log(n))-(\log(\log(n)))^{1/2+\epsilon}<\omega(n)$$ and one which $$\log(\log(n))-(\log(\log(n)))^{1/2+\epsilon}>\omega(n).$$
And split the series $$\sum_{n=1}^{\infty}\dfrac{1}{4^{\omega(n)}n}$$ over these two sets. Lets call the first one A and the second one B. Since se inequality holds for almost all n, B should be "small" in some sense (?). The terms of the series are positive so we can rearrange them as we please
$$\sum_{n=1}^{\infty}\dfrac{1}{4^{\omega(n)}n}=\sum_{n\in A}^{\infty}\dfrac{1}{4^{\omega(n)}n}+\sum_{n\in B}^{\infty}\dfrac{1}{4^{\omega(n)}n}.$$
Now, the inequalities are valid for the first sum, over A, what can I do with the sum over B? How can I control it?
