Prove that $x_{n+2} := \frac{1}{2}(x_n + x_{n+1})$ converges, if $x_0 = 1$ and $x_1 = 2$? This question is related to my other question, where I had just to find the limit (which is $\frac{5}{3}$) of the following defined sequence:
$$x_0 = 1 \\ \\
x_1 = 2 \\ \\
x_{n + 2} = \frac{1}{2} (x_n + x_{n + 1})$$
Now, I need to prove that $\frac{5}{3}$ is really the limit, so I started my saying:
For any $\epsilon > 0$, there's an $N > 0$, such that if $n > N$, then $|x_n - \frac{5}{3}| < \epsilon$. 
This means that we want to show that for any $n > N$, the distance between our limit $\frac{5}{3}$ and $x_n$ is less than $\epsilon$. Or, again, I need to show: $$0 <|x_n - \frac{5}{3}| < \epsilon$$ for all $n$ greater than a certain $N$.
Now, how do I proceed proving this limit?
 A: Geometrically, $x_{n+2}$ is the average of $x_{n+1}$ and $x_n$. Therefore it lies smack in the middle of $x_{n+1}$ and $x_n$:
$$
x_n < x_{n+2} < x_{n+1}.
$$
Intuitively, one can imagine that we keep shrinking the interval where the all future $x_N$ can live, and the interval's size gets cut in half each time. More precisely,
$$
|x_{n+1}-x_n| = \left|\frac{1}{2}(x_n + x_{n-1}) - x_n\right| = \frac{1}{2}|x_{n-1}-x_n|.
$$
By induction,
$$
|x_{n+1}-x_n| = 2^{-(n+1)}|x_1 - x_0|.
$$
And all future $x_N$ with $N \geq n+1$ lie in $(x_n,x_{n+1})$, so
$$
|x_N - x_n| \leq |x_{n+1}-x_n| = 2^{-(n+1)}|x_1 - x_0|.
$$
With this it's fairly transparent that the sequence is Cauchy.
A: Setting
$$
u_n=x_{n+1}-x_n,\quad \forall n\ge 0,
$$
we get:
$$
u_{n+1}:=x_{n+2}-x_{n+1}=-\frac12(x_{n+1}-x_n)=-\frac12u_n \quad \forall n\ge 0;
$$
and so
$$
x_{n+1}-x_n=u_n=\left(-\frac12\right)^nu_0 \quad \forall n\ge 0.
$$
We deduce that
\begin{eqnarray}
x_n&=&x_0+(x_1-x_0)+(x_2-x_1)+\ldots+(x_{n-1}-x_{n-2})+(x_n-x_{n-1})\\
&=&x_0+u_0+u_1+\ldots+u_{n-2}+u_{n-1}\\
&=&x_0+u_0\sum_{k=0}^{n-1}\left(-\frac12\right)^k=x_0+u_0\frac{1-(-1/2)^n}{1+1/2}\\
&=&x_0+\frac{2u_0}{3}\left[1-\left(-\frac{1}{2}\right)^n\right]\\
&=&1+\frac{2}{3}\left[1-\left(-\frac{1}{2}\right)^n\right].
\end{eqnarray}
Taking the limit we get
$$
\lim x_n=1+\frac23=\frac53.
$$
