6. Let $S_{n}$ be the sum of the elements of all 3-element subsets of the set $A=\left\{1, \frac{1}{2}, \cdots, \frac{1}{2^{n-1}}\right\}$. Then $\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2}}=$ $\qquad$ .
6.1 .

For any element in set $A$, it appears in a subset containing 3 elements $\mathrm{C}_{n-1}^{2}$ times, then
$$
\begin{array}{l}
S_{n}=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^{n-1}}\right) \mathrm{C}_{n-1}^{2} . \\
\text { Therefore, } \lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2}}=\lim _{n \rightarrow \infty} \frac{\frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}} \cdot \frac{(n-1)(n-2)}{2}}{n^{2}}=1 .
\end{array}
$$