Five. (20 points) As shown in Figure 1, in the square $ABCD$ with side length 1, points $M_{1}$, $M_{2}, \cdots, M_{n} (n \geqslant 1)$ sequentially divide the broken line $BC + CD$ into $n+1$ equal parts. Try to find:
(1) $\sum_{k=1}^{n} \mathcal{\Lambda} M_{k}^{2}$;
(2) $\lim _{n \rightarrow \infty}\left(\frac{1}{n} \sum_{k=1}^{n} A M_{k}^{2}\right)$.
When $n$ is odd, point $M_{\frac{n+1}{2}}$ coincides with $C$, and the other division points are symmetric about the line $A C$.

When $k \leqslant \frac{n+1}{2}$, point $M_{k}\left(1, \frac{2 k}{n+1}\right)$ lies on side $B C$, and $A M_{4}^{2}=1+\frac{4 k^{2}}{(n+1)^{2}}$. At this time,
$$
\begin{array}{l}
\sum_{i=1}^{\frac{n+1}{2}} A M_{k}^{2}=\frac{n+1}{2}+\frac{4}{(n+1)^{2}}\left[1^{2}+2^{2}+\cdots+\left(\frac{n+1}{2}\right)^{2}\right] \\
=\frac{n+1}{2}+\frac{4}{(n+1)^{2}} \cdot \frac{\frac{n+1}{2} \cdot \frac{n+3}{2}(n+2)}{6} \\
=\frac{n+1}{2}+\frac{(n+2)(n+3)}{6(n+1)} .
\end{array}
$$

Then $\sum_{k=1}^{n} A M_{k}^{2}=2 \sum_{k=1}^{\frac{n+1}{2}} A M_{k}^{2}-A C^{2}$
$$
=n+1+\frac{(n+2)(n+3)}{3(n+1)}-2=\frac{4 n^{2}+5 n+3}{3(n+1)} \text {. }
$$

Thus, $\lim _{n \rightarrow \infty}\left(\frac{1}{n} \sum_{k=1}^{n} A M_{k}^{2}\right)=\lim _{n \rightarrow \infty} \frac{4 n^{2}+5 n+3}{3(n+1) n}=\frac{4}{3}$.
When $n$ is even, no division point coincides with point $C$, and the division points on side $B C$ are symmetric to the division points on side $D C$ about the line $A C$.

When $k \leqslant \frac{n}{2}$, point $M_{k}\left(1, \frac{2 k}{n+1}\right)$ lies on side $B C$, and $A M_{k}^{2}=1+\frac{4 k^{2}}{(n+1)^{2}}$, at this time,
$$
\begin{array}{l}
\sum_{k=1}^{\frac{n}{2}} A M_{k}^{2}=\frac{n}{2}+\frac{4}{(n+1)^{2}}\left[1^{2}+2^{2}+\cdots+\left(\frac{n}{2}\right)^{2}\right] \\
=\frac{n}{2}+\frac{4}{(n+1)^{2}} \cdot \frac{\frac{n}{2}\left(\frac{n}{2}+1\right)(n+1)}{6} \\
=\frac{n}{2}+\frac{n(n+2)}{6(n+1)} . \\
\text { Then } \sum_{k=1}^{n} A M_{k}^{2}=2 \sum_{k=1}^{\frac{n}{2}} A M_{k}^{2}=n+\frac{n(n+2)}{3(n+1)} \\
=\frac{4 n^{2}+5 n}{3(n+1)} .
\end{array}
$$

Then $\sum_{k=1}^{n} A M_{k}^{2}=2 \sum_{k=1}^{\frac{n}{2}} A M_{k}^{2}=n+\frac{n(n+2)}{3(n+1)}$

Thus, $\lim _{n \rightarrow \infty}\left(\frac{1}{n} \sum_{i=1}^{n} A M_{i}^{2}\right)=\lim _{n \rightarrow \infty} \frac{4 n^{2}+5 n}{3(n+1) n}=\frac{4}{3}$.