Suppose $x_0$, $x_1$ are arbitrary and a sequence $(x_n)$ be defined by $x_{n+1}=\frac{x_n+x_{n-1}}{2}, n\geq 1.$ I want show, by induction, the following:
- $x_0\leq x_n\leq x_1,\forall n\geq 1$;
- $(x_n)$ is not monotone;
- $\lvert x_{n+1}-x_{n+2}\rvert=\frac{k}{2^{n}}, k>0,\forall n\geq 1.$
For (1); Let $x_0=1$ and $x_1=2$ so that $1\leq x_1\leq 2$ is true for $n=1$, and suppose $1\leq x_k\leq2,$ is true for $n=k,$ I tried to show that its true for $n=k+1,$ but stuck there!
For your second assertion, suppose $x_0 > x_1$ ($x_0 < x_1$ is similar, $x_0 = x_1$ is trivial), I think you can prove by induction that $x_{2k} > x_{2k-1}$ and $x_{2k+1} < x_{2k}$. For $k = 1$ it's trivial. Assume the statement holds for some $k\in\mathbb{N}$. Then for $k+1$ you get $x_{2(k+1)} = \frac{x_{2k+1}+x_{2k}}2 > x_{2k+1}$ by our assumption. The second case is similar. This clearly shows your sequence is not monotone.
As for (3), you can prove, it seems, that $|x_{n-1} - x_n| = \frac{|x_0 - x_1|}{2^{n-1}}$. You again can prove this by induction on $n$ (that is a kind of obvious, since each time you calculate $x_{n+1}$, you take the middle point of the interval between $x_n$ and $x_{n-1}$, hence if the distance between $x_n$ and $x_{n-1}$ was $\frac{|x_0 - x_1|}{2^{n-1}}$, the distance between $x_{n+1}$ and $x_n$ will be twice shorter.).