limit of the sequence $a_n=1+\frac{1}{a_{n-1}}$ and $a_1=1$ Problem: Find with proof limit of the sequence $a_n=1+\frac{1}{a_{n-1}}$ with  $a_1=1$  or show that the limit does not exist.
My attempt:
I have failed to determine the existence. However if the limit exists then it is easy to find it.
The sequence is not  monotonic and I have failed to find any monotonic subsequence subsequence. The sequence is clearly bounded below by $1$. I have observed that the sequence is a  continued fraction so it alternatively increases and decreases. 
So, please help me.
 A: Hint:
Note that $a_{n+2}=2-\frac1{a_n+1}$.
Function $x\mapsto 2-\frac1{x+1}$ is monotonic on $[1,\infty)$.
That can be used to prove that $(a_{2n-1})$ is increasing and $(a_{2n})$ is decreasing.
A: As you already stated, the sequence alternately increases and decreases. Therefore you have  found a monotone subsequence, namely $a_1,a_3,a_5,\dots$. And you have another monotone subsequence, namely $a_2,a_4,a_6,\dots$. You can easily find the recursion formula $a_n = f(a_{n-2})$.
Now prove that (a) both of these subsequences are indeed monotone and bounded and (b) that their limits are the same.
Alternative: Prove that $|a_j - a_{j-1}|$ is bounded by a geometric sequence, which will prove that the sequence in question is Cauchy.    
A: $$a_n=1+\frac{1}{a_{n-1}}$$
$$a_1=1,a_2=2, a_3=\frac{3}{2},a_4=\frac{5}{3},a_5=\frac{8}{5},a_6=\frac{13}{8},a_7=\frac{21}{13}...$$
Cauchy sequence definition:
($\forall\epsilon >0),(\exists q\in\mathbb{N}),(\forall n,m\in\mathbb{N})(n,m>q)\Rightarrow |a_n-a_m|<\epsilon$
Let's assume:
$$a_{n-1}-a_{n}=0$$
And show for every member of sequence.
$$a_{n+1}-a_n=(1+\frac{1}{a_{n}})-(1+\frac{1}{a_{n-1}})=\frac{1}{a_n}-\frac{1}{a_{n-1}}=\frac{a_{n-1}-a_n}{a_n*a_{n-1}}=0$$ 
So $a_n$ has a limit.
And we can write this:
$$\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} a_{n-1}=L $$
$$L=1+\frac{1}{L}$$
$$L^2-L-1=0$$
$$L_1=\frac{1+\sqrt5}{2}$$
$$L_2=\frac{1-\sqrt5}{2}$$
But all members of sequence are positive numbers so our limit is $L_1$.
