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As in this previous question, let a $k-$almost prime be a positive integer having exactly $k$ prime factors, not necessarily distinct. Let $\mathbb{P}_k$ be the set of the $k$-almost primes and let $$ \rho_k(n):=\sum\limits_{\substack{q\in \mathbb{P_k}\\q\le n}}\frac1q. $$ The answer to that question states that (I presume for fixed $k$) the asymptotic estimate for $\rho_k(n)$ has leading term $$ \rho_k(n) \asymp \frac{1}{k!}(\log \log n)^k, $$ and error term of $O((\log \log n)^{k-1}).$

How would a rigorous proof showing the error term proceed?

Is there a reference to such a result in the literature?

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I'm just focusing on the error term since the main term was already discussed in your previous question. You know that $$ \sum_{p\leq n}\frac1{p} = \log\log n + O(1)\;; $$ then $$ \rho_k(n) \ll \left(\sum_{p\leq n}\frac1{p}\right)^k = O((\log\log n)^k) + O((\log\log n)^{k-1}) \;. $$

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  • $\begingroup$ The exponent should be $k$ not n in your first equation after "then"? How do you rigorously argue the error term? $\endgroup$
    – kodlu
    Feb 28, 2016 at 20:46
  • $\begingroup$ Sorry for the typo, of course the exponent is $k$. Concerning your second question, you have something like $(\log\log n + O(1))^k$, the main term is the product of $k$ $\log\log n$, the main error comes from $k$ product like $O((log\log n)^{k-1})O(1)=O((log\log n)^{k-1})$. $\endgroup$
    – PITTALUGA
    Feb 29, 2016 at 15:53
  • $\begingroup$ Thanks. Can you confirm that with the linked question's answer and your answer, it has been established that $$\rho_k(n) \asymp ( \log \log n)^k$$ $\endgroup$
    – kodlu
    Feb 29, 2016 at 21:28
  • $\begingroup$ @kodlu Confirmed! $\endgroup$
    – PITTALUGA
    Mar 1, 2016 at 10:39

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