Find the Limit.? What is the limit of
$Y_{n} := \frac{1}{3}Y_{n-1} + \frac{2}{3}Y_{n-2} $
for $n > 2$ and $Y_{1}<Y_{2}$.
I have tried solving, but the limit tends to exist between $y_{1}$ and $y_{2}$ and hence I am not able to evaluate what would be the ratio between which they would lie?

$Y_{n} := \frac{1}{3}Y_{n-1} + \frac{2}{3}Y_{n-2}$ is homogenous linear diference equation which has characteristic equation $x^2=\frac{1}{3}x+\frac{2}{3}$ or $3x^2-x-2=0$, $(3x+2)(x-1)=0$. Solution is $$Y_{n}=C_1(-\frac{2}{3})^n+C_2(1)^n=C_1(-\frac{2}{3})^n+C_2.$$
Constants $C_1$, $C_2$ can be found from $Y_1, Y_2$:
$$-\frac{2}{3}C_1+C_2=Y_1$$
$$\frac{4}{9}C_1+C_2=Y_2$$
from where $C_1=\frac{9}{10}(Y_2-Y_1)$ and $C_2=\frac{2}{5}Y_1+\frac{3}{5}Y_2$ and
$$Y_n=\frac{9}{10}(-\frac{2}{3})^n(Y_2-Y_1)+\frac{2}{5}Y_1+\frac{3}{5}Y_2.$$
Therefore, $$\lim_{n\to\infty} Y_n=\frac{2}{5}Y_1+\frac{3}{5}Y_2.$$
