Finding the limit of $a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2})$ for arbitrary $a_{0}$ and $a_{1}$. Here is the problem statement:

Let $a$, $b$ $\in$ $\Bbb{R}$. A sequence $(a_{n})_{n \in \Bbb{N}}$ is defined recursively by
$$a_{0}:=a, \qquad a_{1}:=b, \qquad a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2}) \quad \text{for} \quad n \geq 2.$$
Prove that $\lim_{n\to\infty}a_{n}$ exists and compute its value.

Now I have proved that the limit exists. My idea was to use the Monotone Convergence Theorem as follows:
Proof: Without loss of generality, assume $a<b$. By inspection of the first two terms of $a_{n}$, we see that
$$a_{3} = \frac{1}{2}(a_{2}+a_{1})=\frac{1}{2}\left[\frac{1}{2}(a_{1}+a_{0}) + a_{1}\right] = \frac{1}{2}(a_{1}+a_{0})+\frac{1}{4}(a_{1}-a_{0})=a_{2}+\frac{1}{4}(a_{1}-a_{0})>a_{2} \quad (\because a_{1}>a_{0}).$$
Therefore we need to show that the sequence is increasing, so our inductive hypothesis is $a_{n}<a_{n+1}$.
$$a_{n}<a_{n+1}\implies a_{n+1}=\frac{1}{2}(a_{n}+a_{n-1})<\frac{1}{2}(a_{n+1}+a_{n})=a_{n+2}.$$
Thus $a_{n}$ is increasing for $n\geq 2$. Now we need to show that it is bounded. By inspection, $a_{3}=\frac{1}{2}(a_{2}+a_{1})<\frac{1}{2}(a_{1}+a_{1}) = a_{1}$, and $a_{4}<a_{1}$ and so on. So again assume by induction $a_{n}<a_{1}$ and we have
$$a_{n+1}=\frac{1}{2}(a_{n}+a_{n-1})<\frac{1}{2}(a_{1}+a_{1}) = a_{1}.$$
Thus $a_{n}$ is bounded above. Hence by the Monotone Convergence Theorem, sequence is convergent and $\lim a_{n}$ exists.$\quad\square$
However, I'm stuck in finding the limit. Any hints would be appreciated.
 A: The characteristic equation is
$$\lambda^2=\frac{1}{2}(\lambda+1)$$
Its solutions are
$$\lambda=1;\;\lambda=-1/2$$
The general solution of the recurrence is
$$a_n=m+n\left(-\frac{1}{2}\right)^n$$
for $n=0$ we get $a_0=m+n=a$
for $n=1$ we have $a_1=m-\frac12n=b$
so we get
$$m= \frac{1}{3} (a+2 b),n= \frac{2 (a-b)}{3}$$
and the recurrence can be written as
$$a_n=\frac{1}{3} (a+2 b)+ \frac{2 (a-b)}{3}\left(-\frac{1}{2}\right)^n$$
as $n\to\infty$ we have $a_n\to \frac{1}{3} (a+2 b)$.
Hope this is useful
A: $$
\begin{bmatrix}
a_{n}\\
a_{n-1}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
a_{n-1}\\
a_{n-2}
\end{bmatrix}
$$
$$
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix}
= 
\begin{bmatrix}
1 & 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & \frac{1}{2}
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
1 & -1
\end{bmatrix}
$$
$$
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix}^{n}
= 
\begin{bmatrix}
1 & 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & \frac{1}{2}^{n}
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
1 & -1
\end{bmatrix}
$$
$$
\begin{bmatrix}
a_{n}\\
a_{n-1}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix} ^ {n}
\begin{bmatrix}
a_{1}\\
a_{0}
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & \frac{1}{2}^{n}
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
1 & -1
\end{bmatrix}
\begin{bmatrix}
a_{1}\\
a_{0}
\end{bmatrix}
$$
A: Hint:
$$a_n=\frac12(a_{n-1}+a_{n-2})\implies a_n-a_{n-1}=-\frac{1}{2}(a_{n-1}-a_{n-2}).$$
Can you use this to find a closed form for $a_n-a_{n-1}$, and then for $a_n$?
