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 $\lim_{n\to\infty}\frac{a_{3n}}{a_n}=\frac{1}{3}$

The limit of $\frac{a_{3n}}{a_n}$ need not be $\frac13$, because it need not exist.

Each $n>0$ can be written uniquely as $2^k + r$, where $2^k \le n < 2^{k+1}$. Define $a_n$ (in terms of these $k$ and $r$) piecewise as follows: $$a_n = \begin{cases} \frac{2^{1 - r/2^{k-1}}}{2^k} & \mbox{if }r < 2^{k-1} \\ \frac{1}{2^k} & \mbox{if }r \ge 2^{k-1}. \end{cases}$$

Essentially, for the first half of the range from $2^k$ to $2^{k+1}$, $a_n$ decreases from $\frac{1}{2^{k-1}}$ to $\frac1{2^k}$ geometrically, by factors of $2^{-1/2^{k-1}}$. For the second half of that range, $a_n$ stays at $\frac1{2^k}$.

It is identically true that $\frac{a_{2n}}{a_n} = \frac12$. Moreover, if $n \ge 2^k$, then $2^{-1/2^{k-1}} \le \frac{a_{n+1}}{a_n} \le 1$, so $\frac{a_{n+1}}{a_n} \to 1$.

However, when $n = 2^k$, $\frac{a_{3n}}{a_n} = \frac14$, while $\frac{a_{9n}}{a_{3n}} = \frac1{2^{5/4}}$, so $\frac{a_{3n}}{a_n}$ does not converge.

You might complain that the sequence $a_n$ is not strictly decreasing. If this is a problem, just replace $a_n$ by $a_n' = a_n\left(1 + \frac1n\right)$. We have:

• $\frac{a'_{n+1}}{a'_n} = \frac{a_{n+1}}{a_n} \cdot \frac{1+\frac1{n+1}}{1+\frac1n} = \frac{a_{n+1}}{a_n} \left(1 - \frac1{(n+1)^2}\right)$, which still converges to 1.
• $\frac{a'_{2n}}{a'_n} = \frac{a_{2n}}{a_n} \cdot \frac{1 + \frac1{2n}}{1 + \frac1n} = \frac{a_{2n}}{a_n} \left(1 - \frac1{2n+2}\right)$, which still converges to $\frac12$.
• $\frac{a'_{3n}}{a'_n} = \frac{a_{3n}}{a_n} \cdot \frac{1 + \frac1{3n}}{1 + \frac1n} = \frac{a_{3n}}{a_n} \left(1 - \frac2{3n+3}\right)$, which still does not converge to anything.
• Since we already had $a_{n+1} \le a_n$, we now have $a'_{n+1} = a_{n+1} \left(1 + \frac1{n+1}\right) \le a_n \left(1 + \frac1{n+1}\right) < a_n \left(1 + \frac1n\right) = a'_n$.

If the limit $\frac{a_{3n}}{a_n}$ does exist, then we may proceed as in Andras's answer to show that it must equal $\frac13$:

If $\frac{a_{3n}}{a_n} \to c$, then $\frac{a_{3^kn}}{a_n} \to c^k$ for any $k$, but we can bound $\frac{a_{3^kn}}{a_n}$ between $\frac{a_{2^\ell n}}{a_n}$ and $\frac{a_{2^{\ell+1} n}}{a_n}$ for $\ell$ such that $2^\ell < 3^k < 2^{\ell+1}$ (i.e., $\ell = \lfloor k \log_2 3\rfloor$). These two ratios converge to $\frac{1}{2^{\ell}}$ and $\frac1{2^{\ell+1}}$, so we get that $2^{-\ell/k} < c < 2^{-(\ell+1)/k}$. Taking $k$ arbitrarily large, $\frac{\ell}{k} \to \log_2 3$, so $c$ must be $2^{-\log_2 3} = \frac13$.

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.