Limit of $x_n = 0.5(x_{n-1} + x_{n-2})$ - help finishing proof... EDIT: Fixed geometric proof off-by-one error.
I am looking for the limit of $x_n = 0.5(x_{n-1} + x_{n-2})$ with $x_2 > x_1$ arbitrary. I can show $\forall n: |x_{n}-x_{n+1}| = \frac{c}{2^{n-1}}$ (where $c = x_2 - x_1$) and thus existence of the limit (cauchy sequence)(see appended). From there I wanted to proceed:
\begin{eqnarray*}
\lim(x_n) &=& x_1 + \lim( \sum_{k=1}^n (x_{k+1}-x_k)) \\
&=& x_1 + x_2 - x_1+ \lim( \sum_{k=2}^n \frac{(-1)^{k-1} c}{2^{k-1}}) \\
&=& x_2 + \lim(\sum_{k=1}^{n/2} \frac{c}{2^{2k}} -  \sum_{k=1}^{n/2} \frac{c}{2^{2k-1}} ) \\
&=& x_2  + c \lim(\sum_{k=1}^{n/2}  \frac{1}{4^k} -  \sum_{k=1}^{n/2} \frac{1}{2}\frac{1}{4^k}) \\
&=& x_2 + c (\frac{1}{4}\frac{1}{1-1/4}-\frac{1}{4}\frac{1}{2}\frac{1}{1-1/4}) \\
&=& x_2 + \frac{c}{8}\frac{1}{1-1/4} \\
&=& x_2  + \frac{1}{6}c.\end{eqnarray*}
I can tell this is wrong by the example in the book (Bartle Sherbert, 3.5), and I can tell it should be $x_1 + \frac{2}{3}c$ on the penultimate line the same way - but I cannot justify why that would be the case... 
Any ideas?
Thanks! Best,
Leon

Let $c = x_2 - x_1$. Then we prove by induction that $\forall n: |x_{n}-x_{n+1}| = \frac{c}{2^{n-1}}$. Base: firstly $|x_1 - x_2| = x_2 - x_1 = c/2^0 = c/2^{1-1}$, and secondly $|x_2 - x_3| = |x_2 - (1/2)(x_1+x_2)| = (1/2)|x_1-x_2| = c/2$. Inductive step:
\begin{eqnarray*}
|x_n - x_{n+1}| &=& | \frac{1}{2}(x_{n-1}+x_{n-2}) -  \frac{1}{2}(\frac{1}{2}(x_{n-1}+x_{n-2})+x_{n-1)} | \\
 &=& |\frac{1}{2}x_{n-2} - \frac{1}{4}x_{n-1}-\frac{1}{4}x_{n-1}  | \\
&=& \frac{1}{4}|x_{n-2}-x_{n-1}| \\
&=& \frac{c}{2^{n-1}}.
\end{eqnarray*}

$$\sum_{k=1}^{n/2} \frac1{2^{2k-1}}=  \sum_{k=1}^{n/2} \frac{1}{2}\frac{1}{2^{2k-2}}=  \sum_{k=1}^{n/2} \frac{1}{2}\frac{1}{4^{k-1}}=  \sum_{k=0}^{n/2-1} \frac{1}{2}\frac{1}{4^k}$$
