Define $x_{2n}=\frac{x_{2n-1}+2x_{2x-2}}{3}\,\,\,\,\,,x_{2n+1}=\frac{2x_{2n}+x_{2n-1}}{3}$ for $n \in \mathbb{N^+}$. Let $x_0=a$ and $x_1=b$ with $b>a$. Define $$x_{2n}=\frac{x_{2n-1}+2x_{2n-2}}{3}\,\,\,\,\,,x_{2n+1}=\frac{2x_{2n}+x_{2n-1}}{3}$$ for $n \in \mathbb{N^+}$. Show that the sequence $(x_n)$ converges. my attempt is to prove that $(x_{2n})$ is increasing and $(x_{2n+1})$ is decreasing. But I don know how to prove it. Anyone can help ?

Closed forms for linear recurrence relations can be found using matrices and diagonalization.
Let $v_i = \begin{bmatrix}x_i & x_{i - 1}\end{bmatrix}^\intercal$. We have $v_1 = \begin{bmatrix}b & a\end{bmatrix}^\intercal$ and for any positive integer $n$,
$$\begin{align}
v_{2n} &= \begin{bmatrix}1/3 & 2/3\\1 & 0\end{bmatrix}v_{2n - 1}\\
v_{2n + 1} &= \begin{bmatrix}2/3 & 1/3\\1 & 0\end{bmatrix}v_{2n}\\
\end{align}$$
It is interesting to note that both matrices are right-stochastic (rows sum to 1).
Let $M = \begin{bmatrix}2/3 & 1/3\\1 & 0\end{bmatrix}\begin{bmatrix}1/3 & 2/3\\1 & 0\end{bmatrix} = \begin{bmatrix}5/9 & 4/9\\1/3 & 2/3\end{bmatrix}$;
$$\begin{align}
v_{2n} &= \begin{bmatrix}1/3 & 2/3\\1 & 0\end{bmatrix}M^{n - 1}\begin{bmatrix}b\\a\end{bmatrix}\\
v_{2n + 1} &= M^n\begin{bmatrix}b\\a\end{bmatrix}\\
\end{align}$$
We may diagonalize $M = \begin{bmatrix}-4/3 & 1\\1 & 1\end{bmatrix}\begin{bmatrix}2/9 & 0\\0 & 1\end{bmatrix}\begin{bmatrix}-4/3 & 1\\1 & 1\end{bmatrix}^{-1}$
Substitute $u_i = \begin{bmatrix}-4/3 & 1\\1 & 1\end{bmatrix}^{-1}v_i$ to get
$$\begin{align}
u_{2n + 1} &= \begin{bmatrix}(2/9)^n & 0\\0 & 1\end{bmatrix}u_1 = \frac{1}{7}\begin{bmatrix}3(2/9)^n(a - b)\\4a - 3b\end{bmatrix}\\
\implies \begin{bmatrix}x_{2n + 1}\\x_{2n}\end{bmatrix} = v_{2n + 1} &= \begin{bmatrix}-4/3 & 1\\1 & 1\end{bmatrix}\begin{bmatrix}3(2/9)^n(a - b)\\4a - 3b\end{bmatrix}
\end{align}$$
This should answer your questions.
