If $x_0$ and $x_1$ are positive numbers and $s_n = \frac12(x_n + x_{n-1})$, prove that the sequence converges. If $x_0$ and $x_1$ are positive numbers and $s_n = \frac12(x_n + x_{n-1})$, prove that the sequence converges.
(Hint : use the "Nested closed intervals Theorem")  
I know what the nested closed interval theorem is and how to prove it, but I have no idea how to prove that the sequence converges. I am also having trouble interpreting the sequence, for example if $n=1$ do you get $\frac12(1+0)$?
Can I prove this buy figuring out if it is an increasing or decreasing sequence and then find an upper or lower bound?
 A: Probably you mean ${\displaystyle x_{n+1} = {x_n + x_{n-1} \over 2}}$. If that's the case, the idea is that the interval $I_n = [x_n,x_{n+1}]$ (or $[x_{n+1},x_n]$ if $x_n > x_{n+1}$) is a subinterval of $I_{n-1}$ of half the length of $I_{n-1}$.
A: More generally,
if
$x_{n+1}
=ax_n+(1-a)x_{n-1}
$,
where
$0 < a < 1$,
then
$x_n$ converges.
This problem has 
$a = \frac12$.
Proof:
If
$x_{n+1}
=ax_n+(1-a)x_{n-1}
$,
where
$0 < a < 1$,
then
$\begin{array}\\
x_{n+1}-x_n
&=(a-1)x_n+(1-a)x_{n-1}\\
&=(a-1)(x_n-x_{n-1})\\
\text{by induction}\\
x_{n+k}-x_{n+k-1}
&=(a-1)^k(x_n-x_{n-1})\\
\text{so that, setting }n=1,\\
x_{k+1}-x_{k}
&=(a-1)^k(x_1-x_{0})\\
\end{array}
$
Since
$-1 < a-1 < 0$,
$x_n$
converges.
Also,
summing
$x_{k+1}-x_{k}
=(a-1)^k(x_1-x_{0})
$
from
$0$ to $n-1$,
$\begin{array}\\
x_n-x_0
&=\sum_{k=0}^{n-1} (x_{k+1}-x_{k})\\
&=\sum_{k=0}^{n-1} ((a-1)^k(x_1-x_{0}))\\
&=(x_1-x_{0})\sum_{k=0}^{n-1} (a-1)^k\\
&=(x_1-x_{0})\frac{1-(a-1)^n}{1-(a-1)} \\
&=\frac{(x_1-x_{0})(1-(a-1)^n)}{2-a} \\
&=\frac{x_1-x_{0}}{2-a} 
-\frac{(x_1-x_{0})(a-1)^n)}{2-a} \\
\text{so that}\\
x_n
&=x_0+\frac{x_1-x_{0}}{2-a} 
-\frac{(x_1-x_{0})(a-1)^n)}{2-a} \\
&=\frac{x_0(2-a)+x_1-x_{0}}{2-a} 
-\frac{(x_1-x_{0})(a-1)^n)}{2-a} \\
&=\frac{x_0(1-a)+x_1}{2-a} 
-\frac{(x_1-x_{0})(a-1)^n)}{2-a} \\
\text{so that}\\
\lim_{n \to \infty} x_n
&=\frac{x_0(1-a)+x_1}{2-a} \\
\end{array}
$
A: Although @martycohen's answer is quite complete, I'd like to share another method of solving the problem in this particular case. Working in base $2$ helps in figuring out the limit, for here is what the sequence $\{ x_n \}$ looks like in base $2$:
$$
\begin{align}
x_0 &= 0\\
x_1 &= 1\\
x_2 &= 0.1\\
x_3 &= 0.11\\
x_4 &= 0.101\\
x_5 &= 0.1011\\
x_6 &= 0.10101\\
x_7 &= 0.101011\\
x_8 &= 0.1010101\\
x_9 &= 0.10101011\\
x_{10}&= 0.101010101\\
&\ \, \vdots
\end{align}
$$
The pattern should be obvious, and the limit is exactly what you'd expect it to be: $$\lim_{n \to \infty} x_n = 0.\overline{10} = 0.10101010\dots.$$
Thus, the limit equals
$$
\sum_{i=1}^\infty \frac{1}{2^{2k-1}} = \frac{1/2}{1-(1/4)} = \frac{2}{3}.
$$
