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!


1 Answer 1


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 $$\sum_{n\le X} C^{\omega(n)} n^{-1}\sim a_C \log^{C-1} X$$

  • $\begingroup$ 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 $\sum_{n } C^{\omega(n)} n^{-s} \ll \sum_{n} \tau_m(n) n^{-s}$? $\endgroup$
    – Johnny T.
    Commented Feb 6, 2020 at 11:51
  • 1
    $\begingroup$ I'm not, $\sum_n C^{\omega(n)} n^{-s}\le \zeta(s)^m$ is a termwise bound. $\endgroup$
    – reuns
    Commented Feb 6, 2020 at 14:08

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