$(a_n)$ is a positive sequence such that:
$$\forall \epsilon > 0 ,\quad \exists N\in \mathbb{N}, \quad \forall n,m >N ,\quad |\frac{a_{n}}{a_{m}} - 1|< \epsilon $$
how to prove this sequence is bounded? and after, how to prove this sequence have a limit?
thanks!!
Suppose the sequence is unbounded. Then we can obtain a subsequence $a_{n_k}$ such that $\forall k \in \mathbb{N}$ we have $a_{n_k} \geq k$. But then for any $k \in \mathbb{N}$
$$ \frac{a_{n_k}}{a_{n_r}} \leq \frac{a_{n_k}}{r} \rightarrow 0 \quad \textrm{as} \quad r \rightarrow \infty.$$
This contradicts the property in the question.
Let the bound be $K$. Then, given $\epsilon >0$ use the property in the question to find $N$ such that
$$n,m > N \Longrightarrow |\frac{a_n}{a_m} - 1| < \frac{\epsilon}{K}.$$
Multiply through by $a_m$ and use the bound $K$ on the right-hand side of the resulting inequality to see that the sequence is Cauchy, and hence convergent.