# Upper bound on the sum $\sum_{n \leq X} C^{\omega(n)}/n$

I would like to obtain an upper bound for the sum $$\sum_{n \leq X} C^{\omega(n)}/n$$, where $$C> 0$$ and $$\omega(n)$$ is the number of distinct prime divisors on $$n$$. I was wondering if I could obtain an upper bound that is at at most a power of $$\log X$$.. Any suggestion is appreciated!

For $$C\le 1$$ then $$\sum_{n \leq X} C^{\omega(n)}n^{-1}\le \sum_{n\le X} n^{-1} \le 1+\log X$$.
For $$C > 1$$ then $$\sum_n C^{\omega(n)} n^{-s}=\prod_p (1+C\sum_{k\ge 1}p^{-sk})\le \zeta(s)^m=\sum_n \tau_m(n)n^{-s},\qquad m=\lceil C\rceil$$ by induction on $$m$$ $$\sum_{n\le X}\tau_{m+1}(n)=\sum_{d\le X} \sum_{n\le X/d}\tau_m(n)= O(\sum_{n\le X/d} X/d \log^{m-1}( X/d))=O(X \log^m X)$$ thus by partial summation $$\sum_{n\le X} C^{\omega(n)} n^{-1}=O(\sum_{n\le X}\tau_m(n)n^{-1})=O(\log^m X)$$ A more precise argument will give $$\sum_{n\le X} C^{\omega(n)} n^{-1}\sim a_C \log^{C-1} X$$
• Thank you for your answer. How do you deduce $\sum_{n \leq X} C^{\omega(n)} n^{-1} \ll \sum_{n \leq X} \tau_m(n) n^{-1}$ from $\sum_{n } C^{\omega(n)} n^{-s} \ll \sum_{n} \tau_m(n) n^{-s}$? Commented Feb 6, 2020 at 11:51
• I'm not, $\sum_n C^{\omega(n)} n^{-s}\le \zeta(s)^m$ is a termwise bound. Commented Feb 6, 2020 at 14:08