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Prove that the sequence converges and has a limit: $$x_{n+1}=\frac{1}{1+x_{n}},\quad\text{with}\; x_0 > - 1$$. I found the limit, it is equal to $$ \frac { 2 } { 1 + \sqrt { 5 } } $$ , it remains to prove the existence of the limit. I tried to prove that the limit exists according to Cauchy, considered the difference $x _ { n + 1 } - x _ { n } $, but nothing worked.

Yaroslav
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    I don't know how I would even use l'Hopital on this, to be honest. It's not a method commonly used on recursive sequences. Also, what is $x_0$? – Arthur Oct 09 '19 at 07:18
  • $x _ { 0 } > - 1$ – Yaroslav Oct 09 '19 at 07:28
  • Hint: find a closed subset of $\mathbb{R}$ containing $x_0$ on which $x \mapsto 1/(1+x)$ is a well-defined contraction mapping. – diracdeltafunk Oct 09 '19 at 07:29
  • Show that this is Cauchy? – xuq01 Oct 09 '19 at 07:29
  • One idea: Try to find an expression for $x_n$ of the form $$ x_n = \frac{Ax+B}{Bx+C} $$ – Matti P. Oct 09 '19 at 07:31
  • @MattiP. I tried to do this, I got A, B, C and D equal to the Fibonacci numbers, I found the limit, but I can not prove why it exists – Yaroslav Oct 09 '19 at 07:36
  • @xuq01 ,To prove that the difference of neighboring members of the sequence is infinitesimal? – Yaroslav Oct 09 '19 at 07:38
  • @diracdeltafunk I do not quite understand what you wanted to say – Yaroslav Oct 09 '19 at 07:41
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    @Yaroslav: Since comments are easily overlooked, you should edit your question to add the details of what you've tried (eg, how you got the Fibonacci $A$, $B$, $C$, $D$, and what value you found for the limit). This can save potential answerers from wasting time (theirs and yours) duplicating your effort and/or telling you things you already know. – Blue Oct 09 '19 at 07:41
  • @Yaroslav I've expanded my comment into a full solution. But no, to show a sequence is Cauchy it does not suffice to show that that difference of consecutive terms goes to $0$. For example, $\sum_{n=1}^\infty \frac{1}{n}$ does not converge, but the difference of two consecutive partial sums does go to $0$. – diracdeltafunk Oct 09 '19 at 07:57
  • @diracdeltafunk I unfortunately do not know what is the Banach Fixed Point Theorem – Yaroslav Oct 09 '19 at 08:03
  • @MattiP. Do not tell me what to do next? – Yaroslav Oct 09 '19 at 09:01
  • Well you should be using the answers and comments as hints only. You are the one who hasn't expanded the question: if you had added more context when you first asked the problem, we wouldn't need to ask you so much to find out what you need. – Toby Mak Oct 09 '19 at 09:51

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I'll expand upon my comment for clarity.

Let $f : [0,\infty) \to [0,\infty)$ be the map $f(x) = \frac{1}{1+x}$. $f$ is a contraction mapping on the complete metric space $[0,\infty)$ because $\lvert f'(x) \rvert = \frac{1}{(1+x)^2} \leq 1$ for all $x \in [0,\infty)$, with equality if and only if $x = 0$. By the Banach Fixed Point Theorem‚ $f$ has a unique fixed point, which is the limit of the sequence $z, f(z), f(f(z)), \dots$ for any $z \in [0,\infty)$. Since $f(x_0) \in [0,\infty)$ and $f(x_n) = x_{n+1}$ for all $n$, we see that $x_0, x_1, x_2, \dots$ converges to the unique fixed point of $f$. By inspection, this fixed point is $\frac{-1+\sqrt{5}}{2}$.

diracdeltafunk
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