If $\sum_{n=1}^\infty\frac1{na_n}$ converges, then $\sum_{n=1}^\infty\frac1{na_{b_n}}$ is also convergent 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.
 A: $$
S(x)=\sum_{n\le x}{1\over na_{b_n}}=\sum_{p\le x}{1\over a_p}\sum_{\substack{n\le x\\b_n=p}}\frac1n=\sum_{p\le x}{1\over pa_p}\color{blue}{\sum_{\substack{m\le x/p\\b_m\le p}}\frac1m}.
$$
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.
