Example 6 (2001 National High School Mathematics League Question) Let $\left\{a_{n}\right\}$ be an arithmetic sequence, and $\left\{b_{n}\right\}$ be a geometric sequence, and $b_{1}=a_{1}^{2}, b_{2}=a_{2}^{2}, b_{3}=a_{3}^{2}\left(a_{1}<a_{2}\right)$, and $\lim \left(b_{1}+b_{2}+\cdots+b_{n}\right)=1+\sqrt{2}$. Try to find the first term and common difference of the sequence $\left\{a_{n}\right\}$.
Let $\{a_n\}$ be an arithmetic sequence with common difference $d$, and $\{b_n\}$ be a geometric sequence with common ratio $q$. Then, according to the given conditions:
$a_{1}0$.
Since $\lim _{n \rightarrow 0}(b_{1}+b_{2}+\cdots+b_{n})=1+\sqrt{2}$, and $q=\frac{b_{2}}{b_{1}}=\frac{a_{2}^{2}}{a_{1}^{2}}>0$, we know $q \in(0,1)$. According to the properties of the geometric sequence $\{b_n\}$ and the given conditions, we have $b_{2}^{2}=b_{1} b_{3}$, which implies
$a_{2}^{1}=a_{1}^{2} a_{3}^{2}$, thus $a_{2}^{2}= \pm a_{1} a_{3}$.
If $a_{2}^{2}=a_{1} a_{3} \Leftrightarrow (a_{1}+d)^{2}=a_{1}(a_{1}+2 d)$, then $d=0$, which contradicts equation (1).
Therefore, it must be that $a_{2}^{2}=-a_{1} a_{3}$,

Thus, $a_{2}^{2}=-a_{1}(2 a_{2}-a_{1})$,
which simplifies to $a_{2}^{2}+2 a_{1} a_{2}-a_{1}^{2}=0$,
then $\frac{a_{2}}{a_{1}}=-1 \pm \sqrt{2}$.
Since $q=\frac{b_{2}}{b_{1}}=\frac{a_{2}^{2}}{a_{1}^{2}}<1$ and considering equation (1), we have $a_{1}+a_{2}<0$, thus $a_{1}<0$.
Given $0<q<1$, we have
$$
q=\frac{b_{2}}{b_{1}}=\frac{a_{2}^{2}}{a_{1}^{2}}=(-1+\sqrt{2})^{2},
$$

which simplifies to $\frac{a_{2}}{a_{1}}=-1+\sqrt{2}$.
Therefore, $q=\frac{b_{2}}{b_{1}}=\frac{a_{2}^{2}}{a_{1}^{2}}=3-2 \sqrt{2}$.
Thus, $1+\sqrt{2}=\lim _{n \rightarrow \infty}(b_{1}+b_{2}+\cdots+b_{n})=\frac{b_{1}}{1-q}$.
Combining equations (5) and (6) and considering equation (4), we get
$$
a_{1}=-\sqrt{2} \text {, }
$$

and from equation (5), we have $a_{2}=-2+\sqrt{2}$,
so $d=a_{2}-a_{1}=2 \sqrt{2}-2$.