Example 2 Let $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ be an arithmetic sequence and a geometric sequence, respectively, and $a_{1}=b_{1}>0, a_{2}=b_{2}>0$. Try to compare the size of $a_{n}$ and $b_{n}$.
Let $\{a_{n}\}$ have a common difference of $d$, and $\{b_{n}\}$ have a common ratio of $q$. Clearly, $q>0$.
$$
\begin{aligned}
& \because a_{2}=b_{2}, \\
& \therefore a_{1}+d=a_{1} q, \text { i.e., } d=a_{1}(q-1) . \\
& \therefore a_{n}-b_{n}=a_{1}+(n-1) d-a_{1} q^{n-1} \\
= & a_{1}+(n-1) a_{1}(q-1)-a_{1} q^{n-1} \\
= & a_{1}(1-q)\left[\frac{1-q^{n-1}}{1-q}-(n-1)\right] \\
= & a_{1}(1-q)\left(1+q+q^{2}+\cdots+q^{n-2}-(n-1)\right] .
\end{aligned}
$$

If $q>1$, then
$$
\begin{array}{l}
1+q+q^{2}+\cdots+q^{n-2}>n-1 . \\
\because a_{1}>0, \therefore a_{n}<b_{n} .
\end{array}
$$

If $q=1$, then $a_{n}=b_{n}$.
If $0<q<1$, then
$$
1+q+q^{2}+\cdots+q^{n-2}<n-1 \text {. }
$$

Thus, $a_{n}<b_{n}$.
In summary, we have $a_{n} \leqslant b_{n}$.