Compare the following values: $\frac{3^{2000}+2}{3^{2001}+2}, \frac{3^{2001}+2}{3^{2002}+2}, \frac{3^{2002}+2}{3^{2003}+2}$.
Analyze the values of the function $f(x)=\frac{3^{x-1}+2}{3^{x}+2}$ when $x$ takes the values $2001$, $2002$, and $2003$.

Solution: Construct the function $f(x)=\frac{3^{x-1}+2}{3^{x}+2}(x \in \mathbf{R})$, then $f(x)=\frac{3^{x}+6}{3\left(3^{x}+2\right)}=$ $\frac{3^{x}+2+4}{3\left(3^{x}+2\right)}=\frac{1}{3}+\frac{4}{3\left(3^{x}+2\right)}$. Clearly, $f(x)$ is a decreasing function on $(-\infty,+\infty)$, so $f(2001)>f(2002)>f(2003)$. Therefore, $\frac{3^{2000}+2}{3^{2001}+2}>\frac{3^{2001}+2}{3^{2002}+2}>\frac{3^{2002}+2}{3^{2003}+2}$.