4. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, the sum of the first $n$ terms is $S_{n}$, for any integer $n \geqslant 2, S_{n-1}-8, a_{n}, 8 S_{n}-4$ are always in arithmetic progression. Then $\lim _{n \rightarrow \infty}\left(a_{n}+S_{n}\right)=$ $\qquad$ .
4. $\frac{4}{3}$.

From the problem, we get
$$
2 a_{n}=S_{n-1}-8+8 S_{n}-4=9 S_{n}-a_{n}-12,
$$

Thus, $a_{n}=3 S_{n}-4(n \geqslant 2)$.
Given $a_{1}=1$, we have $a_{2}=\frac{1}{2}$. When $n \geqslant 2$,
$$
3 S_{n}=a_{n}+4, \quad 3 S_{n+1}=a_{n+1}+4 .
$$

Subtracting the two equations, we get
$$
3 a_{n+1}=a_{n+1}-a_{n} .
$$

Therefore, $\frac{a_{n+1}}{a_{n}}=-\frac{1}{2}(n \geqslant 2)$. Hence,
$$
a_{n}=\left\{\begin{array}{ll}
1, & n=1, \\
-\left(-\frac{1}{2}\right)^{n-1}, & n \geqslant 2 .
\end{array}\right.
$$

Thus, when $n \geqslant 2$, we have $a_{n}+S_{n}=\frac{4}{3} a_{n}+\frac{4}{3}$. Therefore,
$$
\lim _{n \rightarrow \infty}\left(a_{n}+S_{n}\right)=\frac{4}{3} \lim _{n \rightarrow \infty} a_{n}+\frac{4}{3}=\frac{4}{3} .
$$