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I'm trying to determine whether the following statement is true or false:

Let $b_n$ be the largest prime factor of the positive integer $n$, and $\{a_n\}$ be a increasing positive sequence such that $\displaystyle\sum_{n=1}^\infty\frac1{na_n}$ converges, then $\displaystyle\sum_{n=1}^\infty\frac1{na_{b_n}}$ is also convergent.

My friend gave me this question. It is one of the problems in a non-official math contest held several years ago.

To be honest, I have no idea on how to start, it seems that we can not write an explicit formula for $b_n$, so I'm stuck. Clearly, $b_n\le n$, and thus $a_{b_n}\le a_n$ and $$\frac1{na_n}\leq\frac1{na_{b_n}},$$ which is not conclusive.

Any help would be appreciated.

Feng
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    Possibly helpful: https://math.stackexchange.com/q/105903/42969 (it answers your question for the case $a_n = n$). – Martin R Jul 11 '22 at 07:21
  • @MartinR Thank you! It seems that we need a lot of knowlegde beyond real-analysis to solve this problem... That really surprises me! – Feng Jul 11 '22 at 07:32
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    @Feng The question is intricately linked with the exact evolution of $b_n$, so the fact that analytical number theory is needed (or at the very least that it's the most natural way to approach the problem) seems rather obvious to me. – Arthur Jul 11 '22 at 07:34
  • This is not my area of expertise, so I have no idea how complicated it is. I just found the related Q&A with Approach0 – Martin R Jul 11 '22 at 07:34
  • @MartinR I already used Approach0 to find some related posts, but I didn't find anything useful. Clearly, your searching skills are much better! – Feng Jul 11 '22 at 07:46
  • @Arthur I haven't learned about analytical number theory. So this question is hard to me. :-( Let's wait for some wonderful answers! – Feng Jul 11 '22 at 07:48
  • @Feng As current question is broader than the one referenced by Martin R, it is unlikely that the answer will require less analytical number theory. – mathcounterexamples.net Jul 11 '22 at 07:56
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    It seems to be impossible to work on this problem without the knowledge of Mertens' theorems. See my answer – TravorLZH Jul 11 '22 at 11:42

1 Answers1

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$$ S(x)=\sum_{n\le x}{1\over na_{b_n}}=\sum_{p\le x}{1\over a_p}\sum_{\substack{n\le x\\b_n=p}}\frac1n=\sum_{p\le x}{1\over pa_p}\color{blue}{\sum_{\substack{m\le x/p\\b_m\le p}}\frac1m}. $$

To continue, we simply drop the requirement that $m\le x/p$ so that it follows from Mertens' 3rd theorem that $$ \color{blue}{\sum_{\substack{m\le x/p\\b_m\le p}}\frac1m}\le\sum_{\substack{m\ge1\\p'|m\Rightarrow p\le p}}\frac1m=\prod_{p'\le p}\left(1-\frac1p\right)^{-1}\ll\log p. $$ As a consequence, it suffices to prove that the sum $$ S_1(x)=\sum_{p\le x}{\log p\over pa_p} $$ is convergent. To continue, we define $\vartheta(x)=\sum_{p\le x}\log p$ so that when $x$ is an integer, \begin{aligned} S_1(x) &=\sum_{n\le x}{\vartheta(n)-\vartheta(n-1)\over na_n}=\sum_{n\le x}{\vartheta(n)\over na_n}-\sum_{n\le x-1}{\vartheta(n)\over (n+1)a_{n+1}} \\ &={\vartheta(x)\over xa_x}-\sum_{n\le x-1}\vartheta(n)\left[{1\over(n+1)a_{n+1}}-{1\over na_n}\right]. \end{aligned} To continue, we apply Chebyshev's upper bound $\vartheta(n)=O(n)$ so that $$ S_1(x)\ll{1\over a_x}+\color{purple}{\sum_{n\le x}n\left[{1\over na_n}-{1\over(n+1)a_{n+1}}\right]}, $$ and applying summation by parts again to the purple sum indicates that $S_1(x)$ converges.

Feng
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TravorLZH
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  • Thanks for the fantastic answer! It seems to me that this answer is very similar to that of this question found by Martin R. I've learned a lot from this answer! In order to let more people join in the discussion of this problem, I'll wait for several hours to accept your answer. Thanks again! – Feng Jul 11 '22 at 13:24