<s>[INST] Determining if a recursively defined sequence converges and finding its limit Define $\lbrace x_n \rbrace$ by $$x_1=1, x_2=3; x_{n+2}=\frac{x_{n+1}+2x_{n}}{3} \text{if} \, n\ge 1$$ The instructions are to determine if this sequence exists and to find the limit if it exists. 
I prove that the sequence is eventually decreasing and that it is bounded below (see work below). Hence, if my work is correct, the sequence will have a limit. The problem is that if the limit is, say, $L$ then it should satisfy $L=\frac{L+2L}{3}$; however, every real number satisfies the latter equality. So, if this limit exists, how do I actually find it?
Proving the limit exists:
1) The sequence is bounded below by $1$.
Base Case: $x_1=1\ge 1$.
Induction: Suppose $x_n\ge 1$ for all $k=1,\ldots ,n+1$. Then $x_{n+2}=\frac{x_{n+1}+2x_{n}}{3}\ge \frac{3}{3}=1$.
2) The sequence is eventually decreasing: It actually starts decreasing after the 4th term. So I take as my base case $x_5<x_4$.
Base Case: Note that $$x_3=\frac{x_2+2x_1}{3}=\frac{3+2}{3}=5/3$$ and hence $$x_4=\frac{x_{3}+2x_{2}}{3}=\frac{5/3+2\times3}{3}=23/9$$ so $$x_5=\frac{x_{4}+2x_{3}}{3}=\frac{23/9+2\times5/3}{3}=\frac{53}{27}<x_4$$
Induction: Suppose $x_{n}<x_{n-1}.$ Then $$x_{n+1}-x_n=\frac{x_{n}+2x_{n-1}}{3}-x_n \\=2\frac{x_{n-1}-x_n}{3}<0$$
Therefore, the sequence is bounded below and is eventually decreasing. 
 [/INST] This is a linear recurrent sequence of order $2$. The characteristic 
polynomial $X^2-\frac{X+2}{3}$ has roots $1$ and $-\frac{2}{3}$. So there
are two constants $a$ and $b$ such that $x_n=a(1^n)+b(\frac{-2}{3})^n$. 
Using the initial conditions, we obtain $a=\frac{11}{5}$ and $b=\frac{9}{5}$.
The limit is therefore $\frac{11}{5}$.
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