Example 2 Given that the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are both geometric sequences composed of positive numbers, with common ratios $p, q$ respectively, where $p>q$, and $p \neq 1, q \neq 1$. Let $c_{n}=a_{n}+b_{n}, S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{c_{n}\right\}$, find $\lim _{n \rightarrow \infty} \frac{S_{n}}{S_{n-1}}$
Solve: Because
$$
S_{n}=\frac{a_{1}\left(p^{n}-1\right)}{p-1}+\frac{b_{1}\left(q^{n}-1\right)}{q-1}
$$

Therefore,
$$
\frac{S_{n}}{S_{n-1}}=\frac{a_{1}(q-1)\left(p^{n}-1\right)+b_{1}(p-1)\left(q^{n}-1\right)}{a_{1}(q-1)\left(p^{n-1}-1\right)+b_{1}(p-1)\left(q^{n-1}-1\right)}
$$

Below, we discuss two cases:
(1) If $p>1$, then $\lim _{n \rightarrow \infty}\left(\frac{1}{p}\right)^{n}=0$
Since $p>q>0 \Rightarrow 0<\frac{q}{p}<1$, so $\lim _{n \rightarrow \infty}\left(\frac{q}{p}\right)^{n}=0$
Therefore, $\lim _{n \rightarrow \cdots} \frac{S_{n}}{S_{n-1}}=\lim _{n \rightarrow \infty} \frac{p^{n}\left[a_{1}(q-1)\left(1-\frac{1}{p^{n}}\right)+b_{1}(p-1)\left(\frac{q^{n}}{p^{n}}-\frac{1}{p^{n}}\right)\right]}{p^{n-1}\left[a_{1}(q-1)\left(1-\frac{1}{p^{n-1}}\right)+b_{1}(p-1)\left(\frac{q^{n-1}}{p^{n-1}}-\frac{1}{p^{n-1}}\right)\right]}$
$$
\begin{array}{l}
=p \lim _{n \rightarrow \infty} \frac{a_{1}(q-1)\left(1-\frac{1}{p^{n}}\right)+b_{1}(p-1)\left(\frac{q^{n}}{p^{n}}-\frac{1}{p^{n}}\right)}{a_{1}(q-1)\left(1-\frac{1}{p^{n-1}}\right)+b_{1}(p-1)\left(\frac{q^{n-1}}{p^{n-1}}-\frac{1}{p^{n-1}}\right)} \\
=p \times \frac{a_{1}(q-1)}{a_{1}(q-1)}=p
\end{array}
$$
(2) If $p<1$, then $0<q<p<1$, so $\lim _{n \rightarrow \infty} p^{n}=0$ and $\lim _{n \rightarrow \infty} q^{n}=0$.

Therefore, $\lim _{n \rightarrow \infty} \frac{S_{n}}{S_{n-1}}=\lim _{n \rightarrow \infty} \frac{a_{1}(q-1)\left(p^{n}-1\right)+b_{1}(p-1)\left(q^{n}-1\right)}{a_{1}(q-1)\left(p^{n-1}-1\right)+b_{1}(p-1)\left(q^{n-1}-1\right)}$
$$
=\frac{-a_{1}(q-1)-b_{1}(p-1)}{-a_{1}(q-1)-b_{1}(p-1)}=1
$$