$\langle x_n\rangle $ be a sequence such that $x_{(n+2)}=\frac{(x_{n+1}+ x_{n})}{2} $for all $n \in \mathbb{N}$, prove that $x_n$ is convergent If $\langle x_n\rangle $ is a sequence of positive real numbers such that $$x_{(n+2)}=\frac{(x_{n+1}+ x_{n})}{2}$$ for all $n \in \mathbb{N},\ $ let $x_1 <x_2$
then subsequence of odd terms is increasing and subsequence of even terms is  decreasing .But how to prove it mathematically? We have 
$$x_{(n+2)}-x_n=\frac{(x_{n+1}- x_{n})}{2}= \frac{(x_{n}- x_{n-2})}{4}$$ How to proceed from here? Any hint please.

(I'm pretty sure that
I and many others 
have done this before
but I'll work it out again.)
If
$x(n+1) = ax(n)+(1-a)x(n-1)
$
where
$0 < a < 2$
then
$\begin{array}\\
x(n+1)-x(n)
&= ax(n)+(1-a)x(n-1)-x(n)\\
&= (a-1)x(n)+(1-a)x(n-1)\\
&= (a-1)(x(n)-x(n-1))\\
\text{so}\\
x(n+k)-x(n+k-1)
&= (a-1)^k(x(n)-x(n-1))\\
\text{and}\\
x(n+k)-x(n)
&=\sum_{j=1}^{k}(x(n+j)-x(n+j-1)\\
&= \sum_{j=1}^{k}(a-1)^j(x(n)-x(n-1))\\
&= (x(n)-x(n-1))\sum_{j=1}^{k}(a-1)^j\\
&= (x(n)-x(n-1))\dfrac{(a-1)-(a-1)^{k+1}}{1-(a-1)}\\
&= (x(n)-x(n-1))\dfrac{(a-1)-(a-1)^{k+1}}{2-a}\\
\text{so}
&\text{putting } n = 1\\
x(k+1)-x(1)
&= (x(1)-x(0))\dfrac{(a-1)-(a-1)^{k+1}}{2-a}\\
&= (x(1)-x(0))(\dfrac{a-1}{2-a}-\dfrac{(a-1)^{k+1}}{2-a})\\
\end{array}
$
Since
$0 < a < 2$,
we have
$-1 < a-1 < 1$
so
$(a-1)^{k+1} \to 0$
and
$\begin{array}\\
x(k+1)
&\to 
x(1)+(x(1)-x(0))\dfrac{a-1}{2-a}\\
&=(1+\dfrac{a-1}{2-a})x(1)-\dfrac{a-1}{2-a}x(0)\\
&=\dfrac{2-a+a-1}{2-a}x(1)-\dfrac{a-1}{2-a}x(0)\\
&=\dfrac{1}{2-a}x(1)+\dfrac{1-a}{2-a}x(0)\\
\end{array}
$
This is the case
$a = \frac12$
so
$\begin{array}\\
x(k)
&\to
\dfrac{1}{2-1/2}x(1)+\dfrac{1-1/2}{2-1/2}x(0)\\
&=\dfrac{1}{3/2}x(1)+\dfrac{1/2}{3/2}x(0)\\
&=\dfrac23 x(1)+\dfrac13 x(0)\\
\end{array}
$
