Example. The infinite sequence
$$
a_{1}, a_{2}, a_{3}, \ldots
$$

has the relation
$$
a_{n}=4+\frac{1}{3} u_{n-1}(n=2,3,1, \cdots)
$$

Find $\lim _{n \rightarrow \infty} a_{n}$.
$$
\begin{array}{l}
a_{2}=4+\frac{1}{3} a_{1}, \\
a_{3}=4+\frac{1}{3} a_{2}=4+\frac{1}{3}\left(4+\frac{1}{3} a_{1}\right) \\
=4+4 \cdot \frac{1}{3}+\left(\frac{1}{3}\right)^{2} a_{1} \\
a_{4}=4+\frac{1}{3} a_{3} \\
=4+4 \cdot \frac{1}{3}+4\left(\frac{1}{3}\right)^{2}+\left(\frac{1}{3}\right)^{2} a_{1} \\
a_{n}=4+4 \cdot \frac{1}{3}+4\left(\frac{1}{3}\right)^{2}+\cdots \\
+4\left(\frac{1}{3}\right)^{n-2}+\left(\frac{1}{3}\right)^{n-1} a_{1} \\
\therefore \lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty}\left\{4+4 \cdot \frac{1}{3}\right. \\
+4\left(\frac{1}{3}\right)^{2}+\cdots \\
\left.+4\left(\frac{1}{3}\right)^{n-2}\right\} \\
+\lim _{n \rightarrow}\left(\frac{1}{3}\right)^{n-1} a_{1}=6. \\
\end{array}
$$

Above, by simplifying $a_{2}, a_{3}, a_{4}, \cdots$ successively, we find $a_{n}$.