# Convergence of the series $\sum_{n=1}^{\infty}\frac{1}{4^{\omega(n)}n}$

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?

• Please excuse me if I'm missing something rather trivial... but the divergence of this series seems to just be a consequence of the divergence of the reciprocals of the primes.
– user296602
Feb 9, 2018 at 0:53
• It is possible to write down an asymptotic formula for the partial sum $\sum_{n\le x} \frac1{4^{w(n)}n}$. Feb 9, 2018 at 1:32

$$\sum_{n \ \text{prime}} \frac{1}{4^{\omega(n)}n} = \frac{1}{4} \sum_{n \ \text{prime}} \frac{1}{n}$$