This is a text book question.
Show that the sequence $\{u_n\}$ is convergent where $0<u_1<u_2$ and $u_{n+2}=\frac{2u_{n+1}+u_n}{3}$.
I am aware that it's possible to find the formula of this sequence using the roots of the characteristic equation $x^2=\frac{2x+1}{3}$ and following that apply limit $n\to\infty$. I found that $\lim\limits_{n\to \infty} u_n=\frac{u_1+3u_2}{4}$.
However, I want to show the existence of this limit analytically. Here's my attempt.
Given that $0<u_1<u_2$, these results follow.
$u_3=\frac{2u_2+u_1}{3}\implies u_1<u_3<u_2$
$ u_4=\frac{2u_3+u_2}{3}\implies u_3<u_4<u_2$
$u_5=\frac{2u_4+u_3}{3}\implies u_3<u_5<u_4$
The idea is that, for a given list of numbers (not all equal) $a_1, a_2, \cdots , a_n$, the arithmetic mean i.e., $\overline a=\frac{1}{n}\sum a_i$ lies somewhere between $\min(a_i)$ and $\max(a_i)$.
$\therefore 0<u_1<u_3<u_5<\cdots<u_{2n-1}<\cdots<u_2 \\ \text{ and } \\ u_2>u_4>u_6>\cdots>u_{2n}>\cdots>u_1\tag*{} $
Conclusion:
- The odd subsequence of $u$ is monotone increasing and it is bounded above by $u_2$.
- The even subsequence of $u$ is monotone decreasing and it is bounded below by $u_1$.
- By monotone convergence theorem, both these subsequences are convergent.
At this point, I think it's sufficient to show that these two subsequences converge to the same limit. I am not sure how to do this.
Initially, I tried to do this proof with the Cauchy criterion for convergence, but I couldn't work it out. I would be interested to see anyone accomplishing it.
Best Answer
THe sequence defined in the OP can be considered as a linear dynamical system on $\mathbb{R}^2$ as follows: $$\begin{pmatrix} x_{n+2}\\ x_{n+1}\end{pmatrix} =\begin{pmatrix} \frac23 & \frac13\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x_{n+1}\\ x_n\end{pmatrix}=\begin{pmatrix} \frac23 & \frac13\\ 1 & 0 \end{pmatrix}^{n+1}\begin{pmatrix} x_1\\ x_0\end{pmatrix} $$
The matrix $A=\begin{pmatrix} \frac23 & \frac13\\ 1 & 0 \end{pmatrix}$ has eigenvalues $\lambda_1=-\frac13$, $\lambda_2=1$. It follows that $$A=\begin{pmatrix} \frac23 & \frac13\\ 1 & 0 \end{pmatrix}= \begin{pmatrix} 1 & 1\\ -3 & 1 \end{pmatrix}\begin{pmatrix} -\frac13 & 0\\ 0 & 1 \end{pmatrix}\frac14\begin{pmatrix} 1 & -1\\ 3 & 1 \end{pmatrix} $$
Consequently \begin{align} A^n&= \begin{pmatrix} 1 & 1\\ -3 & 1 \end{pmatrix}\begin{pmatrix} \big(-\frac13\big)^n & 0\\ 0 & 1 \end{pmatrix}\frac14\begin{pmatrix} 1 & -1\\ 3 & 1 \end{pmatrix}\\&\xrightarrow{n\rightarrow\infty}\begin{pmatrix} 1 & 1\\ -3 & 1 \end{pmatrix}\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}\frac14\begin{pmatrix} 1 & -1\\ 3 & 1 \end{pmatrix}=\frac14\begin{pmatrix} 3 & 1\\ 3 & 1\end{pmatrix} \end{align}
Thus, the sequence $\mathbf{x}_n=[x_{n+1},x_n]^\intercal$ converges to $$\frac14\begin{pmatrix} 3 & 1\\ 3 & 1\end{pmatrix}\begin{pmatrix} x_1\\ x_0\end{pmatrix}=\begin{pmatrix} \frac{3x_1+x_0}{4} \\ \frac{3x_1+x_0}{4} \end{pmatrix} $$