15. Let $A=\{1,2, \cdots, n\}$, and let $S_{n}$ denote the sum of the elements in all non-empty proper subsets of $A$, and $B_{n}$ denote the number of subsets of $A$. Find the value of $\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2} B_{n}}$.
15. It is known that $B_{n}=2^{n}$.

First, consider the number of times the digit 1 appears in all non-empty proper subsets of $A$. It is known that it appears 1 time in the one-element sets, i.e., $C_{n-1}^{0}$ times, in the two-element sets $C_{n-1}^{1}$ times, in the three-element sets $C_{n-1}^{2}$ times, ..., and in the $(n-1)$-element sets $C_{n-1}^{n-2}$ times. Therefore, the total number of times the digit 1 appears in all non-empty proper subsets of $A$ is
$$
\begin{array}{l}
C_{n-1}^{0}+C_{n-1}^{1}+\cdots+C_{n-1}^{n-2}. \\
\text{Then } S_{n}=(1+2+\cdots+n)\left(C_{n-1}^{0}+C_{n-1}^{1}+\cdots+C_{n-1}^{n-2}\right) \\
=\frac{n(n+1)}{2}\left(2^{n-1}-1\right).
\end{array}
$$

Therefore, $\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2} B_{n}}=\frac{n(n+1)\left(2^{n-1}-1\right)}{2 n^{2} \times 2^{n}}=\frac{1}{4}$.