Need help finding the limit of this sequence. If $x_1< x_2$ are real numbers, and $x_n = \frac{1}{2}(x_{n-2} + x_{n-1})$ for every $n > 2$, how do I show $\lim x_n = \frac{1}{3}x_1 + \frac{2}{3}x_2$?
 A: Due to the form of the recursion, you can always write 
$$
x_n = a_k x_{n-k} + b_k x_{n-k-1},
$$
where $a_k$ and $b_k$ are some appropriate coefficients. Now, assume $n$ is very large and solve $a_k$, $b_k$. Using the recursion we get
$$
x_n = (a_k/2+b_k) x_{n-k-1} + a_k/2 \cdot x_{n-k-2},
$$
so $a_{k+1} = a_k/2+b_k$ and $b_{k+1} = a_k/2$. Using these we hence obtain the recursion $a_{k+1} = \frac{1}{2}(a_k+a_{k-1})$. With standard techniques (assuming $a_k = u^k$ and solving a quadratic equation that follows), we get the full solution as $a_k = C_1 + C_2(-1/2)^k$. From the initial values $a_1 = 1/2$ and $a_2 = 3/4$ we get the solution that
$$
a_k = 2/3 + (-1/2)^k \cdot 1/3
$$
and
$$
b_k = a_{k-1}/2 = 1/3 + (-1/2)^{k-1} \cdot 1/6.
$$
Now, we assumed $n \rightarrow \infty$, so we can take $k \rightarrow \infty$ as well and get that $a_k \rightarrow 2/3$ and $b_k \rightarrow 1/3$. At the limit the coefficient $a_k$ is the coefficient of $x_2$ and $b_k$ the coefficient of $x_1$, so the claim follows. 
