How do I study monotony of $x_{n+1}=1+\frac 1{x_n}$ and how do I prove it's bounded

Question

The secuence ($x_n$) is defined by:

$x_{0}=1$

$x_{n+1}=1+\frac 1{x_n}$

First, I know $(x_n) > 0$ because there aren't negative numbers involved, so the second term is always positive. Then I see that the second term is at most $1$ and it can't be $0$ or less, so: $1<(x_n)\leq 2$. Then the sequence is bounded. The problem is I don't know how to rigorously show this in a more formal way.

Second, I need to know if the sequence is monotonic. I suspect it isn't based on some values that I've calculated. What's the approach to study monotony of a sequence that involves recurrence?

EDIT:

I just reasoned the following:

$(x_n)$ is increasing $\iff x_n <x_{n+1} \iff 1+\frac 1{x_{n-1}}<1+\frac 1{x_n} \iff x_{n-1} > x_n \iff (x_n)$ is decreasing

Which is a contradiction, then $(x_n)$ is not monotonic. Is this reasoning correct?

EDIT 2: I don't care about the convergence of the sequence. I just want to know how to see if this sequence is monotonic or is not, how to write it the answer rigorously, how to show it's bounded and additionally if the first edit reasoning is correct or is not. So the links to questions about convergence doesn't really help me.

You can show that the sequence is not monotonic by simply computing $x_0$, $x_1$ and $x_2$. — projectilemotion, Aug 12 '20 at 20:39
@rtybase The question is not about the sequence's convergence, but about whether or not the sequence is bounded and monotone. As the sequence is not monotone one cannot conclude convergence from boundedness. — projectilemotion, Aug 12 '20 at 21:12
@projectilemotion the question is not about the question, but about looking for answers, before asking. — rtybase, Aug 12 '20 at 21:22
@rtybase You don't actually know if someone searched for answers or not before asking here. So if you go around hinting that someone didn't, at least verify the whole question. None of these links answer what I want to know. — xtreyreader, Aug 12 '20 at 21:44
@xtreyreader I verified the entire question. You didn't search for answers, if you did you'd mention that as a courtesy. You'd also do the basic due diligence and check a few entries to see it's not monotone. Let's not complicate the matter, there is nothing wrong with duplicates/closing questions. — rtybase, Aug 12 '20 at 22:00
Also, @xtreyreader if a sequence converges then it is bounded. You can also use www.wolframalpha.com to check sequences. — rtybase, Aug 12 '20 at 22:14

score 1 · Answer 1 · answered Aug 12 '20 at 20:38

1

Clearly $x_n\geq 1$ for all $n$ so ${1\over x_n}\leq 1$ for all $n$ and thus $$x_{n+1} = 1+{1\over x_n}\leq 1+1=2$$ for all $n$

answered Aug 12 '20 at 20:38

nonuser

90,026

score 1 · Answer 2 · answered Aug 12 '20 at 20:45

1

the $x_i$ are quotients of consecutive Fibonacci numbers, and alternate above and below the limit,

answered Aug 12 '20 at 20:45

Will Jagy

139,541

How do I study monotony of $x_{n+1}=1+\frac 1{x_n}$ and how do I prove it's bounded

2 Answers2