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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.

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    $\begingroup$ This is not true. Exercise: find $(x_n)$ such that $\lim x_n=+\infty$, $\limsup x_n/x_{n+1}=+\infty$ and $\liminf x_n/x_{n+1}=0$. $\endgroup$ – Did Jan 18 '15 at 10:43
  • $\begingroup$ Thank you Did but I don't know yet what $\lim\sup$ and $\lim\inf$ are. I'll go back to your exercise once I study this. $\endgroup$ – Scientifica Jan 18 '15 at 10:46
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    $\begingroup$ Exercise: find $(x_n)$ such that $\lim x_n=+\infty$ and there exists two subsequences $(x_{A(n)})$ and $(x_{B(n)})$ such that $\lim x_{A(n)}/x_{A(n)+1}=+\infty$ and $\lim x_{B(n)}/x_{B(n)+1}=0$. $\endgroup$ – Did Jan 18 '15 at 10:48
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    $\begingroup$ "Suppose there exist a subsequence (so a sequence)" ?? Of course if you change the exercise, I cannot guarantee the result remains true. Please read what is written, not what you imagine should be written. $\endgroup$ – Did Jan 18 '15 at 11:42
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    $\begingroup$ @Did The sequence $x_n=n$ if $n$ is odd and $x_n=n^2$ if $n$ is even. It's clear that $(x_n)$ diverges to $+\infty$. $A(n)=2n$ and $B(n)=2n+1$. $\frac{x_{A(n)}}{x_{A(n)+1}}=\frac{(2n)^2}{2n+1}$ diverges to $+\infty$ and $\frac{x_{B(n)}}{x_{B(n)+1}}=\frac{2n+1}{(2n)^2}$ converges to $0$. That's the answer to your exercise and in fact is a counter example for the exercise of my question. Thank you! $\endgroup$ – Scientifica Jan 18 '15 at 11:54
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This is not true. Take $x_n=\begin{cases} n & \text{for } n \text{ even} \\ 2n & \text{for } n \text{ odd} \end{cases}$

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    $\begingroup$ $(x_n/x_{n+1})$ converge $\endgroup$ – idm Jan 18 '15 at 10:39
  • $\begingroup$ @idm the sequence diverges. He's right! $\endgroup$ – Scientifica Jan 18 '15 at 10:41
  • $\begingroup$ @idm ?? $ $ $ $ $\endgroup$ – Did Jan 18 '15 at 10:41
  • $\begingroup$ Yes, sorry :-) I cancelled my downvote. $\endgroup$ – idm Jan 18 '15 at 10:42
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Not true, for instance $x_n = 2^{[\frac{n}{2}]}$

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  • $\begingroup$ You're right! Thanks for your answer! $\endgroup$ – Scientifica Jan 18 '15 at 17:29

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