Example 5 Let the line $n x+(n+1) y=\sqrt{2}(n$ be a natural number) form a triangle with the two coordinate axes, with the area of the triangle being $S_{n}(n=1,2, \cdots, 2000)$. Then the value of $S_{1}+S_{2}+\cdots+S_{2000}$ is ( ).
(A) 1
(B) $\frac{1999}{2000}$
(C) $\frac{2000}{2001}$
(D) $\frac{2001}{2002}$
(2001, Hebei Province Junior High School Mathematics Innovation and Knowledge Application Competition)
Solution: The intersection points of the line $n x+(n+1) y=\sqrt{2}$ with the two coordinate axes are $\left(0, \frac{\sqrt{2}}{n+1}\right)$ and $\left(\frac{\sqrt{2}}{n}, 0\right)$, respectively.
Then $S_{n}=\frac{1}{n(n+1)}$. Therefore,
$$
\begin{array}{l}
S_{1}+S_{2}+\cdots+S_{2000} \\
=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\cdots+\frac{1}{2000 \times 2001} \\
=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{2000}-\frac{1}{2001} \\
=\frac{2000}{2001} .
\end{array}
$$

Therefore, the answer is (C).