Example 6 Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms $S_{n}$ satisfies $S_{n}=2 a_{n}+(-1)^{n}, n \geqslant 1$.
(1) Write down the first 3 terms of the sequence $\left\{a_{n}\right\}$, $a_{1}, a_{2}, a_{3}$;
(2) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(3) Prove: For any integer $m>4$, $\frac{1}{a_{4}}+\frac{1}{a_{5}}+\cdots+\frac{1}{a_{m}}<\frac{7}{8}$.
Analyzing the relationship between the sum of the first $n$ terms $S_{n}$ and the general term $a_{n}$ of the sequence $\left\{a_{n}\right\}$ to find $a_{n}$, one should consider converting $a_{n}$ with $a_{n-1}$ or $a_{n+1}$ into a recursive relation, and then solve for $a_{n}$. For inequality proofs, consider using the bounding method; if bounding a single term is difficult to achieve the goal, you can try bounding multiple terms in combination.

Solution (1) When $n=1$, we have $S_{1}=a_{1}=2 a_{1}+(-1) \Rightarrow a_{1}=1$;
When $n=2$, we have $S_{2}=a_{1}+a_{2}=2 a_{2}+(-1)^{2} \Rightarrow a_{2}=0$;
When $n=3$, we have $S_{3}=a_{1}+a_{2}+a_{3}=2 a_{3}+(-1)^{3} \Rightarrow a_{3}=2$;
In summary, $a_{1}=1, a_{2}=0, a_{3}=2$.
(2) From the given, we have $a_{n}=S_{n}-S_{n-1}=2 a_{n}+(-1)^{n}-2 a_{n-1}-(-1)^{n-1}$, simplifying to $a_{n}=2 a_{n-1}+2(-1)^{n-1}$. The above can be rewritten as $a_{n}+\frac{2}{3}(-1)^{n}=2\left[a_{n-1}+\frac{2}{3}(-1)^{n-1}\right]$, hence the sequence $\left\{a_{n}+\frac{2}{3}(-1)^{n}\right\}$ is a geometric sequence with the first term $a_{1}+\frac{2}{3}(-1)^{1}$ and common ratio 2, so $a_{n}+\frac{2}{3}(-1)^{n}=\frac{1}{3} 2^{n-1}$. Therefore, $a_{n}=\frac{1}{3} \cdot 2^{n-1}-\frac{2}{3}(-1)^{n}=\frac{2}{3}\left[2^{n-2}-(-1)^{n}\right]$, the general term formula of the sequence $\left\{a_{n}\right\}$ is $a_{n}=\frac{2}{3}\left[2^{n-2}-(-1)^{n}\right]$.
(3) From the given, we have
$$
\begin{array}{l}
\frac{1}{a_{4}}+\frac{1}{a_{5}}+\cdots+\frac{1}{a_{m}}=\frac{3}{2}\left[\frac{1}{2^{2}-1}+\frac{1}{2^{3}+1}+\cdots+\frac{1}{2^{m-2}-(-1)^{m}}\right]=\frac{3}{2}\left[\frac{1}{3}+\right. \\
\left.\frac{1}{9}+\frac{1}{15}+\frac{1}{33}+\frac{1}{63}+\cdots+\frac{1}{2^{m-2}-(-1)^{m}}\right]=\frac{1}{2}\left[1+\frac{1}{3}+\frac{1}{5}+\frac{1}{11}+\frac{1}{21}+\cdots\right]4) .
\end{array}
$$