If $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$, $\lim_{n\to\infty}\frac{a_{2n}}{a_n}=\frac{1}{2}$ then $\lim_{n\to\infty}\frac{a_{3n}}{a_n}=\frac{1}{3}$ Let $\{a_n\}$ be a decreasing sequence and $a_n>0$ for all $n$. 
If $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ and $\displaystyle\lim_{n\to\infty}\frac{a_{2n}}{a_n}=\frac{1}{2}$, 
how to prove or disprove that
$\displaystyle\lim_{n\to\infty}\frac{a_{3n}}{a_n}=\frac{1}{3}$ ?
Thank you.

This is not an answer, but a longer train of thoughts / conjectures that might lead to a full answer.
It might be an idea to use $\displaystyle\lim_{n\to\infty}\frac{a_{2n}}{a_n}=\frac{1}{2}$ and apply it $r$ times. This gives
$$
\displaystyle\lim_{n\to\infty}\frac{a_{2^rn}}{a_n}=\frac{1}{2^r}
$$ 
which holds for all $r$. Now take some $m$ and identify $r$ such that $2^r \leq m < 2^{r+1}$. Then one has 
$$
\frac{1}{2^r} \geq \displaystyle\lim_{n\to\infty}\frac{a_{m n}}{a_n} > \frac{1}{2^{r+1}}
$$
In particular, this holds when $m$ is chosen to be any power $3^p$. This gives a countable infinite set of inequalities of the type above, all of which must hold. They do hold if 
$$
\displaystyle\lim_{n\to\infty}\frac{a_{3 n}}{a_n}=\frac{1}{3}
$$ 
implying
$$
\displaystyle\lim_{n\to\infty}\frac{a_{3^p n}}{a_n}=\frac{1}{3^p}
$$ 
The conjecture is that there is no other way that they can hold.
