Prove that $x_1 = 0, x_2 = 0, x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded and monotonic. Then find its limit.
My attempt at boundedness:
(Using induction) For the base case we have $0 \leq x_1 = 0 \leq 2$. Assume that the sequence is bounded for $n = k$. Then,
\begin{align*}
0 \leq x_k &\leq 2 \\
\vdots \\
\text{lower bound } \leq x_{k + 1} &\leq \text{upper bound}
\end{align*}
I am thrown off by the term $x_{n + 2}$ in the recursive formula and I can't see the algebra to produce the above steps without getting $x_{n + 2}$ in the expression of the upper / lower bound.
Thank you.
Update:
I have added this to the prove:
We have $0 \leq x_1 = 0 \leq 2$ and $0 \leq x_2 = 0 \leq 2$. Assume that the sequence is bounded for $k+1$,
\begin{align*}
0 &\leq x_{k + 1} \leq 2 \\
0 &\leq x_k + x_{k+1} \leq 4 \\
0 &\leq x_k + \frac{1}{3} x_{k+1} \leq 4 \\
0 &\leq \frac{1}{6} x_{k} + \frac{1}{3} x_{k+1} \leq 4 \\
0 &\leq x_{k+2} \leq 4
\end{align*}
Therefore, by the principle of mathematical induction, the sequence is bounded.
Is this valid?
Best Answer
Observe that $x_1 = 0$, $x_2 = 0$, $x_3 = 1$, $x_4 = \frac{4}{3}$. We can prove by induction that $x_n <2$ for all $n$. Suppose that the inequality is true for $x_1, x_2,\ldots, x_{n+1}$. Then $$ x_{n + 2} = \frac{1}{3}x_{n + 1} + \frac{1}{6}x_n + 1 < \frac{2}{3} + \frac{2}{6} + 1 = 2. $$ Now we show that the sequence is monotonically increasing. Suppose that $x_1 \leq x_2 \leq x_3 \leq \ldots \leq x_{n+1}$ holds for some $n\geq 2$. Then $$ x_{n + 2} - x_{n + 1} = \frac{1}{3}(x_{n + 1} - x_n ) + \frac{1}{6}(x_n - x_{n - 1} ) \geq 0. $$ Thus $x_n$ is bounded from above and increasing, hence it is convergent. Its limit $x$ must satisfy $$ x = \frac{1}{3}x + \frac{1}{6}x + 1, $$ i.e., we must have $x=2$.