If $\lim\limits_{n\to +\infty}x_n=+\infty$ then $\left(\frac{x_n}{x_{n+1}}\right)_{n\in\mathbb{N}}$ converges

Question

As in the title, in an exercise (Elementary Real Analysis by Thomson and Bruckner p.38), we have to prove that if $\lim\limits_{n\to +\infty}x_n=+\infty$ then $\left(\frac{x_n}{x_{n+1}}\right)_{n\in\mathbb{N}}$ converges using only the definition of convergence and divergence and without using subsequences (since in the book the exercise is put before studying other things).

Here's what I tried so far:

Since $\lim\limits_{n\to +\infty}x_n=+\infty$ then it's easy to prove that the sequence $(x_n)_{n\in\mathbb{N}}$ is bounded below. Furthermore, $\exists n_0\in\mathbb{N},\,\forall n\ge n_0,\,x_n>0$

Let $\varepsilon>0$.

$\exists N\in\mathbb{N},\,\forall n\ge N,\,x_n>\varepsilon$

Let $N_0=\max (N,n_0)$ and let $n\ge N_0$. Thus $x_n$ and $x_{n+1}$ are positive.

$x_n>\varepsilon\Rightarrow \dfrac{x_n}{x_{n+1}}>\dfrac{\varepsilon}{x_{n+1}}$

$x_{n+1}>\varepsilon\Rightarrow \dfrac{x_n}{\varepsilon}>\dfrac{x_n}{x_{n+1}}$

Thus $\forall n\ge N_0,\,\dfrac{\varepsilon}{x_{n+1}}<\dfrac{x_n}{x_{n+1}}<\dfrac{x_n}{\varepsilon}$

Note: After having spent some time with the exercise, I have a doubt that the statement we have to prove isn't true, but I couldn't find any counter example. Anyway it's just a doubt.

Answer 1

This is not true. Take $x_n=\begin{cases} n & \text{for } n \text{ even} \\ 2n & \text{for } n \text{ odd} \end{cases}$

Answer 2

Not true, for instance $x_n = 2^{[\frac{n}{2}]}$