Let $\{a_n\}$ be a positive sequence which satisfies $a_{n+2}=a_{n+1}+a_n$ for $n=1,2,\ldots$. Let $z_n=a_{n+1}/a_n$. How can I prove that $\lim_{n\rightarrow\infty}z_n$ exists?
I looked at $z_{n+1}=1+1/z_n$, but I still don't know how to go from here.
Corrected version.
By hypothesis the $z_n$ are positive, so the recurrence
$$z_{n+1}=1+\frac1{z_n}$$
tells you that $z_n>1$ for $n\ge 2$. It follows that $1<z_n<2$ for $n\ge3$. If the sequence converges, it must converge to some $L\in[1,2]$. Moreover, that $L$ must be a fixed point of the function
$$f(x)=1+\frac1x\;:$$
$f$ is continuous, so it must satisfy
$$L=\lim_{n\to\infty}z_{n+1}=\lim_{n\to\infty}f(z_n)=f(L)\;.$$
Solving
$$L=1+\frac1L$$
for the unique solution in the interval $[1,2]$, we see that the only possible limit is
$$\varphi=\frac{1+\sqrt5}2\approx1.618\;.$$
Note that if $z>\varphi$, then $$f(z)=1+\frac1z<1+\frac1\varphi=\varphi\;,$$ while if $z<\varphi$, then $$f(z)=1+\frac1z>1+\frac1\varphi=\varphi\;.$$ That is, the sequence is alternately above and below $\varphi$.
Suppose that $1<a\le x<y$. The mean value theorem says that there is a $u\in[x,y]$ such that
$$\left|\frac{f(x)-f(y)}{x-y}\right|=|f\,'(u)|=\frac1{u^2}\le\frac1{a^2}\;.\tag{1}$$
You can easily check that if $1<z<2$, then $\frac32<f(z)<2$, so $z_n>\frac32$ for $n\ge 4$. We may therefore take $a=\frac32$ in $(1)$ to see that for each $n\ge 4$,
$$\left|\frac{z_{n+2}-z_{n+1}}{z_{n+1}-z_n}\right|\le\frac1{(3/2)^2}=\frac49$$
and hence
$$|z_{n+2}-z_{n+1}|\le\frac49|z_{n+1}-z_n|\;.$$
To finish, just combine this with the fact that the terms are alternatively larger and smaller than $\varphi$ to see that if $z_n<\varphi$, then
$$z_n<z_{n+2}<z_{n+4}<\ldots<\varphi<\ldots<z_{n+5}<z_{n+3}<z_{n+1}$$
with $\lim\limits_{k\to\infty}(z_{n+2k+1}-z_{n+2k})=0$: this shows that the sequence converges to $\varphi$.