$\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 $x_1 &lt;x_2$

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
$0 &lt; a &lt; 2$
then

$\begin{array}\\
x(n+1)-x(n)
&amp;= ax(n)+(1-a)x(n-1)-x(n)\\
&amp;= (a-1)x(n)+(1-a)x(n-1)\\
&amp;= (a-1)(x(n)-x(n-1))\\
\text{so}\\
x(n+k)-x(n+k-1)
&amp;= (a-1)^k(x(n)-x(n-1))\\
\text{and}\\
x(n+k)-x(n)
&amp;=\sum_{j=1}^{k}(x(n+j)-x(n+j-1)\\
&amp;= \sum_{j=1}^{k}(a-1)^j(x(n)-x(n-1))\\
&amp;= (x(n)-x(n-1))\sum_{j=1}^{k}(a-1)^j\\
&amp;= (x(n)-x(n-1))\dfrac{(a-1)-(a-1)^{k+1}}{1-(a-1)}\\
&amp;= (x(n)-x(n-1))\dfrac{(a-1)-(a-1)^{k+1}}{2-a}\\
\text{so}
&amp;\text{putting } n = 1\\
x(k+1)-x(1)
&amp;= (x(1)-x(0))\dfrac{(a-1)-(a-1)^{k+1}}{2-a}\\
&amp;= (x(1)-x(0))(\dfrac{a-1}{2-a}-\dfrac{(a-1)^{k+1}}{2-a})\\
\end{array}
$

Since
$0 &lt; a &lt; 2$,
we have
$-1 &lt; a-1 &lt; 1$
so
$(a-1)^{k+1} \to 0$
and

$\begin{array}\\
x(k+1)
&amp;\to 
x(1)+(x(1)-x(0))\dfrac{a-1}{2-a}\\
&amp;=(1+\dfrac{a-1}{2-a})x(1)-\dfrac{a-1}{2-a}x(0)\\
&amp;=\dfrac{2-a+a-1}{2-a}x(1)-\dfrac{a-1}{2-a}x(0)\\
&amp;=\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)
&amp;\to
\dfrac{1}{2-1/2}x(1)+\dfrac{1-1/2}{2-1/2}x(0)\\
&amp;=\dfrac{1}{3/2}x(1)+\dfrac{1/2}{3/2}x(0)\\
&amp;=\dfrac23 x(1)+\dfrac13 x(0)\\
\end{array}
$
Suppose that $x_n &lt;= x_2$ for all n less than or equal to k then $x_{k+1}=\frac{{x_k}+x_{k-1}}{2} &lt;= \frac{2x_2}{2}=x_2$ so by induction sequence is bounded.Now suppose $x_{n-1} &lt;= x_{n}$ for all n less than or equal to k,then $x_{k+1}=\frac{{x_k}+x_{k-1}}{2} =&gt; \frac{2x_k}{2}=x_k$ so by induction sequence is increasing.As it is bounded and increasing it is convergent.