Determine the convergence of a sequence given by $a_n= \frac{a_{n-1} + a_{n-2}}2$ let $a_0$ and $a_1$ be any two real numbers, and define 
$a_n= \dfrac{a_{n-1} + a_{n-2}}{2}$
Determine the convergence of a sequence. 
Alright, what I have so far. I have two cases $a_0$ > $a_1$ or $a_0$ < $a_1$
for $a_0 > a_1$
The sequence converges to $\frac{1}{3}$($a_0 - a_1$) + $a_0$
for $a_0 < a_1$
The sequence converges to $\frac{2}{3}$($a_1 - a_0$) +  $a_0$
Any additional input? 
 A: Hint 1:
$$
a_n-a_{n-1}=-\tfrac12(a_{n-1}-a_{n-2})
$$
Show by induction that
$$
a_n-a_{n-1}=\left(-\tfrac12\right)^{n-1}(a_1-a_0)
$$
Hint 2:
$$
a_n=a_0+\sum_{k=1}^n\left(-\tfrac12\right)^{k-1}(a_1-a_0)
$$
A: $a_n = \dfrac{a_{n-1}+a_{n-2}}{2}$. One overkill way of finding the limit, is getting the $a_n$ in terms of $n$ via a generating function. 
Write $F(x) = \sum_{i=0}^{\infty}a_i x^i$. Thus, 
$$
\frac{F(x)-a_0}{x}=\sum_{i=0}^{\infty}a_{i+1}x^{i}
$$
$$
\frac{F(x)-a_0-a_1}{x^2} = \sum_{i=0}^{\infty}a_{i+2}x^{i}
$$
So, 
$$
\frac{F(x)-a_0-a_1}{x^2} = \frac{1}{2}\cdot \frac{F(x)-a_0}{x} + \frac{1}{2}F(x)
$$
Solving for $F(x)$, one obtains:
$$
\frac{(x-2)a_0-2a_1}{x^2+x-2}=F(x)
$$
Using partial fractions, one then obtains:
$$
F(x) = \frac{-(a_0+2a_1)}{3(x-1)}+\frac{2(2a_0+a_1)}{3(x+2)}
$$
Now write each term as a power series, obtaining:
$$
F(x) =\frac{(a_0+2a_1)}{3}\sum_{k=0}^{\infty} x^k +\frac{(2a_0+a_1)}{3}\sum_{k=0}^{\infty} \left(-\frac{1}{2}\right)^k(x)^k
$$
Now, $a_{n+1}$ is the $(n+1)^{st}$ coefficient and therefore the coefficient of $x^n$ given above. Explicitly,
$$
a_{n+1} = \frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^n
$$
e.g. if $a_0=0, a_1=1$, we have: 
$$
a_2=\frac{1}{2},a_3=\frac{\frac{1}{2}+1}{2}=\frac{3}{4},a_4=\frac{\frac{3}{4}+\frac{1}{2}}{2}=\frac{5}{8}
$$
Now we expect that $$a_4 = \frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^3 = \frac{5}{8}$$
And 
$$
 \frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^3= \\\frac{2}{3}+\frac{1}{3}\left(\frac{-1}{2}\right)^3=\frac{5}{8}
$$
Now, it is easy to see that
$$
\lim_{n\to \infty}a_{n+1} = \lim_{n\to \infty} \frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^n = \frac{(a_0+2a_1)}{3}
$$
