Finding a Sequences Limit Let $x_1$>$x_2$ be arbitrary real numbers and $x_n$=$x_{n-1}/3 + 2(x_{n-2})/3$ for all n>2.  Find the formula for $x_n$ and its limit.
I used induction to show that the sequence is strictly increasing, but I couldn't really get beyond that, I'm not sure what to do.  Honestly I'm not even sure what finding the formula for $x_n$ is.  Any help?  Hints are appreciated!  Thanks
 A: Hint
Use the classical way : the characteristic equation being given by $$r^2=\frac13 r+\frac23$$ its roots are $r_1=-\frac23$ , $r_2=1$ so the general solution is simply $$x_n=c_1 \left(-\frac{2}{3}\right)^n+c_2$$
I am sure that you can take from here.
A: Hint: Rewrite as 
$$x_n-x_{n-1}=-\frac{2}{3}(x_{n-1}-x_{n-2}).\tag{1}$$
Let $y_n=x_n-x_{n-1}$. We can find a simple expression for $y_n$ in terms of $y_2=x_2-x_1$.
Then we can express $x_n$ as the sum of a finite geometric series (hint, telescoping) and the limit as the sum of an infinite geometric series.
The expression (1) shows that $(x_n)$ is neither an increasing nor a decreasing sequence.
A: $$\begin{align}
x_n&=\frac{x_{n-1}}3+\frac{2x_{n-2}}3\\
3x_n&=x_{n-1}+2x_{n-2}\\
3x_n-3x_{n-1}&=2x_{n-1}+2x_{n-2}\\
x_n-x_{n-1}&=-\frac23(x_{n-1}-x_{n-2})\\
\Delta x_n&=-\frac23 \Delta x_{n-1}\\
&=\left(-\frac23\right)^2 \Delta x_{n-2}\\
&=\left(-\frac23\right)^3 \Delta x_{n-3}\\
&\vdots\\
&=\left(-\frac23\right)^{n-1} \Delta x_{1}\\
\Rightarrow\Delta x_r&=\left(-\frac23\right)^{r-1} \Delta x_{1}\\
\end{align}$$
Summing from $r=n$ down to $1$ by telescoping, and put $x_0=a,x_1=b$:
$$\begin{align}
x_n-x_1&=-\frac23\left[\frac{1-(-\frac23)^{n-1}}{1-(-\frac23)}\right](x_1-x_0)\\
&=-\frac25\left[1-\left(-\frac23\right)^{n-1}\right](x_1-x_0)\\
x_n&=b-\frac25\left[1-\left(-\frac23\right)^{n-1}\right](b-a)\\
&=\frac15(3b+2a)+\frac25\left(-\frac23\right)^{n-1}(b-a)\qquad \blacksquare
\end{align}$$
As $n\to\infty$,
$$x_n\to\frac15(3b+2a)\qquad \blacksquare$$
