Proving convergence of a sequence $a_{n+1} = \frac{1}{1+a_{n}}$ and finding the limit. I am trying to prove the sequence (for n = 1,2,3...)
$$a_{1} = 1,\,a_{n+1} = \frac{1}{1+a_{n}}$$
converges and then find its limit. I can find its limit easily but I cannot find a way to prove that it converges. I know there exists a theorem stating that a monotone increasing and bounded sequence has a limit. I do not think this sequence is increasing. Also, how can I apply the theorem
$$\lim_{x\to\infty} q^{n}=0 \,{if} |q| < 1$$ to this question? I think it is irrelevant but it is good to know different approaches to prove convergence of a sequence. Thank you.
 A: One way to proceed is $a_{2n} - a_{2(n-1)} = \dfrac{1}{1+a_{2n-1}} - \dfrac{1}{1+a_{2n-3}}= - \dfrac{a_{2n-1} - a_{2n-3}}{(1+a_{2n-1})(1+a_{2n-3})}= \dfrac{a_{2n-2} - a_{2n-4}}{K}, K > 0$. Thus using induction you can prove $\{a_{2n}\}$ is a convergent sub-sequence ( either bounded above or below you can check initial values of $a_2, a_4$. Similarly you can find a similar expression for $a_{2n-1}- a_{2n-3}$ and by induction and Bolzanos theorem you can show $\{a_{2n-1}\}$ is a convergent sub-sequence. There is a theorem that you would have to prove that this lead to $\{a_n\}$ is also convergent, say to $L$,and you can solve for $L = \dfrac{1}{1+L}, 0 < L < 1$ and you can get the answer you are looking for....
Note: Both sub-sequences mentioned above converge to the same limit $L$ indeed as you can show that too. And the theorem I mentioned can be applied and $L = \dfrac{\sqrt{5} - 1}{2}$ . 
A: Write $f(x)={1\over{1+x}}$, compute it's derivative, it is negative, let $l>0$ such that $f(l)=l$, $a_0\geq l$ implies that $f(a_0)\geq l$ and recursively $f(a_n)\geq l$.
MVT implies that $|f(a_n)-f(a_{n-1})|=|a_{n+1}-a_n|\leq |f'(l)||a_n-a_{n-1}|$, since $f'(l)|<1$, we deduce that $(a_n)$ is a Cauchy sequence and $lim_na_n=l$.
A: $$a_{n+1}-\frac{\sqrt5-1}{2}=\frac{1}{1+a_n}-\frac{2}{\sqrt5+1}=\frac{2\left(\frac{\sqrt5-1}{2}-a_n\right)}{\sqrt5+1)(1+a_n)}.$$
Thus, for all $n\geq2$ we obtain $$\left|a_n-\frac{\sqrt5-1}{2}\right|=\frac{2\left|\frac{\sqrt5-1}{2}-a_{n-1}\right|}{\sqrt5+1)(1+a_{n-1})}<$$
$$<\frac{2}{\sqrt5+1}\cdot\left|\frac{\sqrt5-1}{2}-a_{n-1}\right|<...<\left(\frac{2}{\sqrt5+1}\right)^{n-1}\left|\frac{\sqrt5-1}{2}-a_1\right|\rightarrow0$$ because $\frac{2}{\sqrt5+1}<1$.
Id est, $$\lim_{n\rightarrow+\infty}a_n=\frac{\sqrt5-1}{2}$$ and we are done!
A: As I saw the task
$a_{n+1}$=$\frac{1}{1+{a_n}}$=$\frac{1}{1+\frac{1}{a_{n-1}}}$=$\frac{1}{1+\frac{1}{1+\frac{1}{a_{n-2}}}}$ ...... making common denominators:
$a_{n+1}$=$\frac{1}{1+{a_n}}$=$\frac{1+a_{n-1}}{2+{a_{n-1}}}$=$\frac{2+a_{n-2}}{3+{a_{n-2}}}$=$\frac{3+a_{n-3}}{5+{a_{n-3}}}$=.....=$\frac{F_{n}+a_{n-n+1}}{F_{n+1}+{a_{n-n+1}}}$ (=$\frac{F_{n}+1}{F_{n+1}+1}$), 
where $F_n$ the n. Fibonacci number.
$\lim_{n\rightarrow+\infty}{(a_{n+1})}$= $\lim_{n\rightarrow+\infty}$($\frac{1+{1\over{F_{n}}}}{1+{1\over{F_{n-1}}}}$)$\frac{F_{n}}{F_{n+1}}$=$\frac{\sqrt5-1}{2}$
