Establishing monotonicity of the sequence of ratio of Fibonacci like sequence Let the sequence $(x_n)$ defined by $0 < x_1 < x_2$ and $$x_{n + 2} = x_{n + 1} + x_n\quad \text{for}~ n \geq 1.$$
I want to show that the sequence of ratio $(r_n)$ defined by $r_{n} = x_{n + 1}/x_n$ is convergent. I have shown that this sequence is bounded, that is $1 < r_n < 2$ except possibly for $r_1$.
I can show that the sequence of its reciprocal is contractive since $$\left |\frac{1}{r_{n + 2}} - \frac{1}{r_{n + 1}}\right| < \frac{1}{4}\left| \frac{1}{r_{n + 1}} - \frac{1}{r_n} \right|$$
and hence converges. 
But is there a way to establish the monotonicity(if possible) of its even and odd terms? My attempt was showing that $r_{2n} - r_{2n + 1}, r_{2n} - r_{2n - 1},$ and $r_{2n - 2} -r_{2n - 1}$ have the same sign, but this doesn't seem to work since the inequalities are reversed.
 A: So you have
$$
\left\{ \matrix{
  0 < x_{\,1}  < x_{\,2}  \hfill \cr 
  x_{\,n + 2}  = x_{\,n + 1}  + x_{\,n} \quad \left| {\;1 \le n} \right. \hfill \cr}  \right.
$$
and putting $r_{\,n}  = x_{\,n + 1} /x_{\,n} $, you get
$$
\left\{ \matrix{
  1 < r_{\,1}  \hfill \cr 
  {{x_{\,n + 2} } \over {x_{\,n + 1} }} = {{x_{\,n + 1} } \over {x_{\,n + 1} }} + {{x_{\,n} } \over {x_{\,n + 1} }}\quad  \Rightarrow \quad r_{\,n + 1}  = 1 + {1 \over {r_{\,n} }}\quad \left| {\;1 \le n} \right. \hfill \cr}  \right.
$$
implying that that it is always $1<r_n$.   
Then
$$
r_{\,2}  = 1 + {1 \over {r_{\,1} }}\quad r_{\,3}  = 1 + {1 \over {1 + {1 \over {r_{\,1} }}}}\quad r_{\,4}  = 1 + {1 \over {1 + {1 \over {1 + {1 \over {r_{\,1} }}}}}}
$$
and  it is clear that in the limit $r$ is approaching $\phi$, same as for Fibonacci numbers.
Now you have that
$$
\left\{ \matrix{
  \Delta \,r_{\,n}  = r_{\,n + 1}  - r_{\,n}  = 1 + {1 \over {r_{\,n} }} - r_{\,n}  =  - {{r_{\,n} ^2  - r_{\,n}  - 1} \over {r_{\,n} }} \hfill \cr 
  \Delta ^2 \,r_{\,n}  = r_{\,n + 2}  - 2r_{\,n + 1}  + r_{\,n}  = {{r_{\,n} ^2  + r_{\,n} ^2  - 3r_{\,n}  - 2} \over {r_{\,n} ^2  + r_{\,n} }} = {{\left( {r_{\,n}  + 2} \right)\left( {r_{\,n} ^2  - r_{\,n}  - 1} \right)} \over {\left( {r_{\,n}  + 1} \right)r_{\,n} }} \hfill \cr}  \right.
$$
which means that
$$
\eqalign{
  & {\rm sign}\left( {\Delta \,r_{\,n} } \right) =  - {\rm sign}\left( {r_{\,n}  - \varphi } \right)  \cr 
  & {\rm sign}\left( {\Delta ^2 \,r_{\,n} } \right) = \,{\rm sign}\left( {r_{\,n}  - \varphi } \right) \cr} 
$$
But
$$
r_{\,n}  < \varphi \quad  \Rightarrow \quad \varphi  < r_{\,n + 1}  = 1 + {1 \over {r_{\,n} }}\quad 
$$
and v.v.
Thus the sequence is oscillating.
Concerning instead the even and odd components, you have
$$
\left\{ \matrix{
  r_{\,n + 2}  - r_{\,n}  = 1 + {1 \over {1 + {1 \over {r_{\,n} }}}} - r_{\,n}  =  - {{r_{\,n} ^2  - r_{\,n}  - 1} \over {r_{\,n}  + 1}} \hfill \cr 
  r_{\,n + 4}  - 2r_{\,n + 2}  + r_{\,n}  = {{\left( {3r_{\,n}  + 1} \right)\left( {r_{\,n} ^2  - r_{\,n}  - 1} \right)} \over {\left( {3r_{\,n}  + 2} \right)\left( {r_{\,n}  + 1} \right)}} \hfill \cr}  \right.
$$
and
$$
r_{\,n}  < \varphi \quad  \Rightarrow \quad r_{\,n + 2}  = {{2\,r_{\,n}  + 1} \over {r_{\,n}  + 1}} < \varphi 
$$
Since the base sequence is oscillating, which gives all odd components
below $\phi$ and all even above, or v.v. depending on $r_1$, then the odd / even components are monotononically increasing / decreasing.
A: You can explicitly say that
$$
x_n=x_1F_{n-2}+x_2F_{n-1}
$$
where $F_k$ is the Fibonacci sequence with $F_0=0$, $F_1=1$ and thus $F_{-1}=1$.
Then 
$$
\frac{x_{n+1}}{x_n}=\frac{x_1F_{n-1}+x_2F_{n}}{x_1F_{n-2}+x_2F_{n-1}}
=\frac{x_1+x_2\frac{F_{n}}{F_{n-1}}}{x_1\frac{F_{n-2}}{F_{n-1}}+x_2}
$$
converges to
$$
\frac{x_1+x_2\Phi}{x_1\Phi^{-1}+x_2}=\Phi
$$
The difference to the limit is
$$
\frac{x_{n+1}}{x_{n}}-\Phi=\frac{x_1(F_{n-1}-\Phi F_{n-2})+x_2(F_{n}-\Phi F_{n-1})} {x_1F_{n-2}+x_2F_{n-1}}
$$
Now we know that
$$
F_{n}-\Phi F_{n-1}=-\frac1\Phi(F_{n-1}-\Phi F_{n-2})=\left(-\frac1\Phi\right)^{n-1}
$$
so that the above difference becomes
$$
...=\left(-\frac1\Phi\right)^{n-1}\frac{-x_1\Phi+x_2} {x_1F_{n-2}+x_2F_{n-1}}
$$
so that the sequence is alternatingly larger and smaller than the limit, thus there is no monotone convergence.
A: A different approach to the problem, given $x_{n+2}=x_{n+1}+x_n$ is a homogenous linear recurrence characteristic polynomial technique can be applied. The characteristic polynomials is
$$x^2-x-1=0$$
with the roots $r_1=\frac{1+\sqrt{5}}{2}$ and $r_2=\frac{1-\sqrt{5}}{2}$, so the general term of the sequence is
$$x_n=C_1r_1^n+C_2r_2^n$$
where $C_1,C_2$ are some constants which can be solved from
$$\left\{\begin{matrix}
x_1=C_1r_1+C_2r_2\\ 
x_2=C_1r_1^2+C_2r_2^2
\end{matrix}\right.$$
and from this it's clear that $\color{red}{C_1\ne0}$. Otherwise, if $C_1=0 \Rightarrow x_2=C_2r_2^2>0 \Rightarrow C_2>0$. But $r_2<0$, thus $x_1=C_2r_2<0$ contradicting the $x_1>0$ initial condition.
Now
$$\frac{x_{n+1}}{x_n}=\frac{C_1r_1^{n+1}+C_2r_2^{n+1}}{C_1r_1^n+C_2r_2^n}=\frac{r_1^{n+1}}{r_1^n}\cdot \frac{C_1+C_2\left(\frac{r_2}{r_1}\right)^{n+1}}{C_1+C_2\left(\frac{r_2}{r_1}\right)^n}=\\
r_1\cdot \frac{C_1+C_2\left(\frac{r_2}{r_1}\right)^{n+1}}{C_1+C_2\left(\frac{r_2}{r_1}\right)^n}$$
and 
$$\frac{r_2}{r_1}=-\frac{\sqrt{5}-1}{\sqrt{5}+1}=-\frac{2}{3+\sqrt{5}} \Rightarrow 0<\left|\frac{r_2}{r_1}\right|<1$$
thus (see point $(2)$ here) $\lim\limits_{n\rightarrow\infty}\left(\frac{r_2}{r_1}\right)^n=0$. As a result
$$\lim\limits_{n\rightarrow\infty}\frac{x_{n+1}}{x_n}=r_1$$
also known as golden ratio.
