# The accumulation points of $\{\frac{a_n}{a_m}:n,m\in \mathbb N\}$

I'm trying to solve the following problem:

Suppose that $$(a_n)_{n\in\mathbb N}$$ is a sequence such that $$a_n>0$$ for any $$n\in\mathbb N$$ and that $$\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\to\infty$$. Prove that $$0$$ is the only accumulation point of $$\left\{\dfrac{a_n}{a_m}:n,m\in \mathbb N\right\}$$.

Here it is my attempt: The only accumulation point is $$0$$ because of $$\displaystyle\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}\to\infty$$. If we fix denominator then limit goes infinity, conversely, if we fix numerator then limit goes to $$0$$.

However, in these kind of questions I cannot prove that why other points cannot be accumulation points.

Is my reasoning logical and how to solve this kind of question giving reason that other points cannot be accumulation point.

The accumulation points of $$S=\left\{\dfrac{a_n}{a_m}: n, m \in\mathbb N\right\}$$ are precisely the points $$b\in\mathbb R$$ that are limit of sequences $$(b_n)_{n\in\mathbb N}\subset S$$ such that $$b_n\neq b_m$$ for any distinct $$n,m\in\mathbb N$$. Your reasoning fails because it approaches only one kind of sequences that could be made with the elements of $$S$$: the ones that fix the numerator and the ones that fix the denominator. However, it is really true that $$0$$ is the onliest accumulation point of $$S$$, as we prove below.

First of all, notice that

$$a>0$$ is an accumulation point of $$S$$ if, and only if, $$a^{-1}$$ is an accumulation point of $$S$$.

To see that, just pick $$\dfrac{a_{n_k}}{a_{m_k}}\to a$$ then, by the continuity of the function $$x\mapsto \dfrac{1}{x}$$, we get that $$\dfrac{a_{m_k}}{a_{n_k}}\to a^{-1}$$.

Therefore, it is enough to check if $$S$$ has accumulation points at the set $$(0,1]$$, so we may consider only sequences of the kind $$\left(\dfrac{a_{n_k}}{a_{m_k}}\right)_{k\in\mathbb N}$$ such that $$n_k< m_k$$. Actually, we will see below that any sequence of this kind converges to $$0$$, so $$S$$ can't have accumulation points at $$(0,1]$$ and, by the equivalence we stated, it will follow that $$S$$ can't have accumulation points at $$(0,+\infty)$$.

Pick a sequence $$\left(\dfrac{a_{n_k}}{a_{m_k}}\right)_{k\in\mathbb N}$$ such that $$n_k< m_k$$ for any $$k\in\mathbb N$$ and $$\dfrac{a_{n_k}}{a_{m_k}}\neq \dfrac{a_{n_l}}{a_{m_l}}$$ for any distinct $$k,l\in\mathbb N$$.

Fix $$\varepsilon>0$$ and put $$M=\varepsilon^{-1}$$. Since $$\displaystyle\lim_{n\to+\infty}\dfrac{a_{n+1}}{a_n}=+\infty$$, we may fix $$n_0\in\mathbb N$$ such that $$\dfrac{a_{n+1}}{a_n}>M,\ \forall n> n_0.$$

Consider the set $$\left\{\dfrac{a_n}{a_m}:n< m\mbox{ and }m\leq n_0\right\}$$. Since this set is finite and the elements of the sequence are all distinct from each other, there exists $$k_0\in\mathbb N$$, such that $$\mbox{m_k>n_0, \forall k>k_0.}$$ Finally, notice that, for any $$k>k_0+1$$, we have $$m_{k-1}>n_0$$, so $$\dfrac{a_{m_k}}{a_{m_{k-1}}}>M$$, and consequently, $$\frac{a_{n_k}}{a_{m_k}} \leq \frac{a_{m_{k-1}}}{a_{m_k}}<\frac{1}{M}=\varepsilon.$$ Since this happens for any $$\varepsilon>0$$, it follows that $$\dfrac{a_{n_k}}{a_{m_k}}\to 0$$.