7. Let $A=\{1,2, \cdots, n\}$, and let $S_{n}$ denote the sum of the elements in all non-empty proper subsets of $A$; $B_{n}$ denote the number of subsets of $A$, then $\lim _{n \rightarrow+\infty} \frac{S_{n}}{n^{2} \cdot B_{n}}$ is $\qquad$ .
7. $\frac{1}{4}$ Detailed Explanation: It is easy to know that $B_{n}=2^{n}$. First, we examine the number of times the element 1 appears in all non-empty proper subsets of $A$. It is easy to see that it appears 1 time in the 1-element set, i.e., $C_{n-1}^{0}$ times, $C_{n-1}^{1}$ times in the 2-element set, $C_{n-1}^{2}$ times in the 3-element set, $\cdots$, and $C_{n-1}^{n-2}$ times in the $n-1$ element set. Therefore, the total number of times the number 1 appears in all non-empty proper subsets of $A$ is $C_{n-1}^{0}+C_{n-1}^{1}+\cdots+C_{n-1}^{n-2}$. Thus, $S_{n}=(1+2+\cdots+n)\left(C_{n-1}^{0}+C_{n-1}^{1}+\cdots+C_{n-1}^{n-2}\right)=\frac{n(n+1)}{2}\left(2^{n}-1\right)$,

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