4. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{7}=\frac{16}{3}$, and $a_{n}=\frac{3 a_{n-1}+4}{7-a_{n-1}}(n=2,3, \cdots)$, then
A. There does not exist $m \in \mathbf{N}^{*}$, such that when $n \geqslant m$, $a_{n}-\frac{4}{3}$
$\mathbf{C}$. There exists $m \in \mathbf{N}^{*}$, such that when $n \geqslant m$, $a_{n}>\frac{a_{n-1}+a_{n+1}}{2}$
D. There exists $m \in \mathbf{N}^{*}$, such that when $n \geqslant m$, $a_{n}<\frac{a_{n-1}+a_{n+1}}{2}$
The fixed point $x$ of the sequence $\left\{a_{n}\right\}$ satisfies $x=\frac{3 x+4}{7-x} \Rightarrow x^{2}-4 x+4=0 \Rightarrow x=2$, then $a_{n}-2=\frac{3 a_{n-1}+4}{7-a_{n-1}}-2=\frac{5 a_{n-1}-10}{7-a_{n-1}}$ $\Rightarrow \frac{1}{a_{n}-2}=\frac{1}{5} \cdot \frac{7-a_{n-1}}{a_{n-1}-2}=\frac{1}{5}\left(-1+\frac{5}{a_{n-1}-2}\right)=\frac{1}{a_{n-1}-2}-\frac{1}{5}$. Thus, $\frac{1}{a_{n}-2}=\frac{1}{a_{7}-2}-\frac{1}{5}(n-7)=\frac{17-2 n}{10} \Rightarrow a_{n}=2-\frac{10}{2 n-17}\left(n \in \mathbf{N}^{*}\right)$. Therefore, $2 a_{n}-\left(a_{n-1}+a_{n+1}\right)=\frac{10}{2 n-19}+\frac{10}{2 n-15}-\frac{20}{2 n-17}$ $=20\left[\frac{2 n-17}{(2 n-15)(2 n-19)}-\frac{1}{2 n-17}\right]=\frac{80}{(2 n-15)(2 n-17)(2 n-19)}$, so the answer is $C$.