Is the following scheme for generating $p_n=(1/3)^n$ stable or not. $p_n=(5/6)p_{n-1}-(1/6)p_{n-2}$. Is the scheme for generating $p_n=(1/3)^n$ stable or not?
$$p_n= \frac{5}{6} p_{n-1} - \frac{1}{6}p_{n-2}.$$
 A: From your recursion, we have
$$
\left[\begin{array}{c}p_n\\p_{n-1}\end{array}\right]
=\left[\begin{array}{c}5/6&-1/6\\1&0\end{array}\right]
\left[\begin{array}{c}p_{n-1}\\p_{n-2}\end{array}\right]
$$
The eigenvalues of the matrix are
$$
\frac12\quad\text{for }\left[\begin{array}{c}1/2\\1\end{array}\right]
$$
and
$$
\frac13\quad\text{for }\left[\begin{array}{c}1/3\\1\end{array}\right]
$$
Since
$$
\left[\begin{array}{c}1/2&1/3\\1&1\end{array}\right]^{-1}
\left[\begin{array}{c}p_1\\p_0\end{array}\right]
=\left[\begin{array}{c}6p_1-2p_0\\-6p_1+3p_0\end{array}\right]
$$
we get
$$
\left[\begin{array}{c}p_n\\p_{n-1}\end{array}\right]
=\dfrac1{2^{n-1}}\left[\begin{array}{c}1/2\\1\end{array}\right](6p_1-2p_0)
+\dfrac1{3^{n-1}}\left[\begin{array}{c}1/3\\1\end{array}\right](-6p_1+3p_0)
$$
Thus, unless you start with a multiple of $\left[\begin{array}{c}1/3\\1\end{array}\right]$, the system will tend to
$$
\left[\begin{array}{c}p_n\\p_{n-1}\end{array}\right]
\to\dfrac1{2^{n-1}}\left[\begin{array}{c}1/2\\1\end{array}\right](6p_1-2p_0)
$$

Another approach:
Your recursion has the characteristic equation
$$
0=x^2-\frac56x+\frac16=\left(x-\frac12\right)\left(x-\frac13\right)
$$
Thus, the solutions are of the form
$$
\begin{align}
p_n
&=\frac{a}{2^n}+\frac{b}{3^n}\\
&=\frac{-2p_0+6p_1}{2^n}+\frac{3p_0-6p_1}{3^n}
\end{align}
$$
Note that if $p_0=3p_1$, then $p_n$ grows like $3^{-n}$; otherwise, it grows like $2^{-n}$.
