Example 5 Toss a fair coin $n$ times, and let $p_{n}$ denote the probability of not getting three consecutive heads.
(1) Find $p_{1}, p_{2}, p_{3}, p_{4}$;
(2) Investigate the recursive formula for the sequence $\left\{p_{n}\right\}$, and provide a proof;
(3) Discuss the monotonicity and limit of the sequence $\left\{p_{n}\right\}$, and explain the probabilistic meaning of this limit.
(1) Obviously, $p_{1}=p_{2}=1$,
$$
p_{3}=1-\frac{1}{8}=\frac{7}{8}, p_{4}=1-\frac{3}{16}=\frac{13}{16} \text {. }
$$
(2) There are three cases in total.
(i) If the $n$-th toss is a tail, then the probability of not having three consecutive heads in the first $n$ tosses is the same as not having three consecutive heads in the first $n-1$ tosses, which is $\frac{1}{2} p_{n-1}$;
(ii) If the $n$-th toss is a head and the $(n-1)$-th toss is a tail, then the probability of not having three consecutive heads in the first $n$ tosses is the same as not having three consecutive heads in the first $n-2$ tosses, which is $\frac{1}{4} p_{n-2}$;
(iii) If the $n$-th toss is a head, the $(n-1)$-th toss is a head, and the $(n-2)$-th toss is a tail, then the probability of not having three consecutive heads in the first $n$ tosses is the same as not having three consecutive heads in the first $n-3$ tosses, which is $\frac{1}{8} p_{n-3}$.
In summary, $p_{n}=\frac{1}{2} p_{n-1}+\frac{1}{4} p_{n-2}+\frac{1}{8} p_{n-3}(n \geqslant 4)$.(1)
Thus, $p_{n-1}=\frac{1}{2} p_{n-2}+\frac{1}{4} p_{n-3}+\frac{1}{8} p_{n-4}(n \geqslant 5)$.
Subtracting $\frac{1}{2} \times$ (2) from (1) gives
$p_{n}=p_{n-1}-\frac{1}{16} p_{n-4}(n \geqslant 5)$.
(3) From (2), we know that when $n \geqslant 3$, $P_{n}$ is monotonically decreasing and has a lower bound of 0, so the limit of $p_{n}$ exists. Taking the limit on both sides of $p_{n}=p_{n-1}-\frac{1}{16} p_{n-4}$ gives $\lim _{n \rightarrow \infty} p_{n}=0$.