7. In the right triangle $\triangle ABC$, $\angle C=90^{\circ}, AB=c$. Along the direction of vector $\overrightarrow{AB}$, points $M_{1}, M_{2}, \cdots, M_{n-1}$ divide the segment $AB$ into $n$ equal parts. Let $A=M_{0}, B=M_{n}$. Then
$$
\begin{array}{l}
\lim _{n \rightarrow+\infty} \frac{1}{n}\left(\overrightarrow{C A} \cdot \overrightarrow{C M_{1}}+\overrightarrow{C M_{1}} \cdot \overrightarrow{C M_{2}}+\cdots+\overrightarrow{C M_{n-1}} \cdot \overrightarrow{C B}\right) \\
=
\end{array}
$$
7. $\frac{c^{2}}{3}$.

Let $C B=a, C A=b$. Then $a^{2}+b^{2}=c^{2}$. Hence
$$
\overrightarrow{C M}_{i}=\frac{n-i}{n} \overrightarrow{C A}+\frac{i}{n} \overrightarrow{C B} \text {. }
$$

By $\overrightarrow{C A} \cdot \overrightarrow{C B}=0$, we get
$$
\begin{array}{l}
\lim _{n \rightarrow \infty} \frac{1}{n}\left(\sum_{i=1}^{n} \overrightarrow{C M_{i-1}} \cdot \overrightarrow{C M_{i}}\right) \\
=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n}\left[\frac{i(i-1)}{n^{2}} a^{2}+\right. \\
\left.\frac{(n-i)(n-i+1)}{n^{2}} b^{2}\right] \\
=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} \frac{i(i-1)}{n^{2}}\left(b^{2}+a^{2}\right) \\
=\lim _{n \rightarrow \infty} \frac{c^{2}}{n^{3}} \sum_{i=1}^{n}\left(i^{2}-i\right)=\lim _{n \rightarrow \infty} \frac{n^{2}-1}{3 n^{2}} c^{2}=\frac{c^{2}}{3} .
\end{array}
$$