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

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

Thus, $a_{n}=3 S_{n}-4(n \geqslant 2)$.
Given $a_{1}=1$, we have $a_{2}=\frac{1}{2}$.
When $n \geqslant 2$, $3 S_{n}=a_{n}+4,3 S_{n+1}=a_{n+1}+4$.
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, $a_{n}=\left\{\begin{array}{ll}1, & n=1, \\ -\left(-\frac{1}{2}\right)^{n-1}, & n \geqslant 2 \text{. }\end{array}\right.$
Thus, when $n \geqslant 2$, we have
$$
a_{n}+S_{n}=\frac{4}{3} a_{n}+\frac{4}{3} \text{. }
$$

Therefore, $\lim _{n \rightarrow \infty}\left(a_{n}+S_{n}\right)=\frac{4}{3} \lim _{n \rightarrow \infty} a_{n}+\frac{4}{3}=\frac{4}{3}$.