Example 2. Consider the sequence $\left\{a_{n}\right\}$ defined by $a_{1}=2, a_{n+1}=\frac{a_{n}}{a_{n}+3}$ $(n \geqslant 1)$. Solve the following problems:
(1) If $b_{n}=\frac{1}{a_{n}}$, find the relationship between $b_{n+1}$ and $b_{n}$;
(2) Find the general term of the sequence $\left\{a_{n}\right\}$.
Solution: (1) From the given, we have
$$
a_{n}=-\frac{1}{b_{n}}, \quad a_{n+1}=\frac{1}{b_{n+1}} .
$$

Substituting the above into the known equation
$$
a_{n+1}=\frac{a_{n}}{a_{n}+3}
$$

we get
$$
3_{n} \frac{1}{b_{n}+3}=\frac{\frac{1}{b_{n}}}{b_{n}} .
$$

From this, we obtain $b_{n+1}=3 b_{n}+1$.
(2) From the result in (1), we have $b_{n+1}-b_{n}$ $=3\left(b_{n}-b_{n-1}\right)$.
Let $c_{n}=b_{n+1}-\dot{b}_{n}$, then $c_{n+1}=3 c_{n}$.
Since $\quad b_{1}=\frac{1}{2}, \quad a_{2}=\frac{2}{2+3}=\frac{2}{5}$,
$\therefore b_{2}=\frac{5}{2}$.
$\therefore \quad c_{1}=b_{2}-b_{1}=\frac{5}{2}-\frac{1}{2}=2$.
When $n \geqslant 2$,
$b_{n}=\frac{1}{2}+\sum_{k=1}^{n} 2 \cdot 3^{k-1}=2 \cdot 3^{n-1}-1$
The above formula still holds when $n=1$.
Thus, $b_{n}=\frac{2 \cdot 3^{n-1}-1}{2}$,

Therefore, $a_{n}=\frac{2}{2 \cdot 3^{n-1}-1}$.