Example 1 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=a, a_{2}=b, 2 a_{n+2}=a_{n+1}+a_{n} \text {. }
$$
(1) If $b_{n}=a_{n+1}-a_{n}$, prove: for $a \neq b$, $\left\{b_{n}\right\}$ is a geometric sequence;
(2) If $\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\cdots+a_{n}\right)=4$, find the values of $a$ and $b$.
(2011, Joint Autonomous Admission Examination of Excellent Talent Cultivation Cooperative Universities)
(1) Proof From the problem, we have
$$
\begin{array}{l}
2\left(a_{n+2}-a_{n+1}\right)=-\left(a_{n+1}-a_{n}\right) \\
\Rightarrow b_{n+1}=-\frac{1}{2} b_{n}, \text { and } b_{1}=a_{2}-a_{1}=b-a \neq 0 .
\end{array}
$$

Therefore, $\left\{b_{n}\right\}$ is a geometric sequence with the first term $b-a$ and the common ratio $-\frac{1}{2}$.
(2) Solution From (1), we know $b_{n}=\left(-\frac{1}{2}\right)^{n-1} b_{1}$, that is,
$$
a_{n+1}-a_{n}=\left(-\frac{1}{2}\right)^{n-1}(b-a) \text {. }
$$

Then $a_{n}=a_{1}+\sum_{k=1}^{n}\left(a_{k+1}-a_{k}\right)$
$$
=a+\frac{2}{3}(b-a)\left[1-\left(-\frac{1}{2}\right)^{n-1}\right] \text {. }
$$

Thus, $a_{1}+a_{2}+\cdots+a_{n}$
$$
=n a+\frac{2}{3}(b-a) n-\frac{4}{9}(b-a)+\frac{4}{9}(b-a)\left(-\frac{1}{2}\right)^{n} \text {. }
$$

Since $\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\cdots+a_{n}\right)=4$, we have,
$$
\left\{\begin{array}{l}
a+\frac{2}{3}(b-a)=0 \\
-\frac{4}{9}(b-a)=4 .
\end{array}\right.
$$

Solving these equations, we get $(a, b)=(6,-3)$.