# $a_{n+1}=1+\frac{1}{a_n}.$ Find the limit.

Let $a_1=1$ and $a_{n+1}=1+\frac{1}{a_n}.$

Is $a_n$ convergent?

How could i find its limit?

I found even terms of the sequence decrease and odd terms are increase. But i cant find upper and lower bounds to use monotone convergence theorem.

• Here is one hint: If the limit exists (call it $A$), then $A=1+1/A$. Can you see why? Oh, and one more question: Was a value for $a_1$ (or $a_0$) given? – Harald Hanche-Olsen Dec 6 '14 at 11:50
• $a_1=1$ Ok but how could i find the sequences $(a_{2n})$ and $(a_{2n+1})$ approaches to same limit? – user35 Dec 6 '14 at 11:52
• You could show the sequence is contractive: $|a_{n+2}-a_{n+1}|\le C|a_{n+1}-a_n|$ with $0<C<1$. (Note this, by itself, would imply the sequence converges.) – David Mitra Dec 6 '14 at 11:58
• As for the bounds: there is an obvious lower bound. Once you've found that, there is an almost as obvious upper bound. – David Mitra Dec 6 '14 at 12:04
• first prove $a_n\in(1,2)$ then we can use high-low limit to prove it's convergent. – Idele Dec 7 '14 at 8:27

If $f:I \rightarrow I$ is differentiable and $p$ is a fixed point such that $f'(p) <1$ then the sequence given by $x_{n+1}=f(x_n)$ este converges to $p$.
For $$f:(0,\infty) \rightarrow (0,\infty),f(x)=1+\frac{1}{x},f'(x)=-\frac{1}{x^2},p=\frac{1+\sqrt{5}}{2}, |f'(p)|=\frac{2}{3+\sqrt{5}}<1.$$
You can prove by induction that $$a_n = \frac{F_{n+1}}{F_n}$$ where $\{F_n\}_{n\geq 1}=\{1,1,2,3,5,8,\ldots\}$ is the Fibonacci sequence.
Binet formula hence gives: $$\lim_{n\to +\infty} a_n = \phi = \frac{1+\sqrt{5}}{2}.$$