13. $\left\{a_{n}\right\}$ is an arithmetic sequence, $\left\{b_{n}\right\}$ is a geometric sequence, and $b_{1}=a_{1}^{2}, b_{2}=a_{2}^{2}, b_{3}=a_{3}^{2}\left(a<a_{2}\right)$, also $\lim _{n \rightarrow+\infty}\left(b_{1}+b_{2}+\cdots+b_{n}\right)=\sqrt{2}+1$, try to find the first term and common difference of $\left\{a_{n}\right\}$.
$$
a_{1}=-\sqrt{2}, d=2 \sqrt{2}-2
$$
13. 【Analysis and Solution】This is a basic problem involving arithmetic and geometric sequences. The first three terms of the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $b_{i}=a_{i}^{2}(i=1,2,3)$, which can determine the relationship between the first term $a_{i}$ and the common difference $d$ of the sequence $\left\{a_{n}\right\}$. By $\lim _{x \rightarrow \infty}\left(b_{1}+b_{2}+\cdots+b_{n}\right)=\sqrt{2}+1$, we can find the values of $a_{1}$ and $d$.
Let the common difference of $\left\{a_{n}\right\}$ be $d$. Given $a_{1}0$,
From $b_{2}^{2}=b_{1} b_{3}$, we get $a_{2}^{4}=a_{1}^{2} a_{3}^{2}$.
$$
\therefore a_{2}^{2}=a_{1} a_{3} \text{ (discard, otherwise } a_{1}=a_{2}=a_{3} \text{), }
$$
or $a_{2}^{2}=-a_{1} a_{3}$.
$$
\therefore\left(a_{1}+d\right)^{2}=-a_{1}\left(a_{1}+2 d\right) \text {, }
$$

which simplifies to $2 a_{1}^{2}+4 a_{1} d+d^{2}=0$.
Solving this, we get $d=(-2 \pm \sqrt{2}) a_{1}$.
If $d=(-2-\sqrt{2}) a_{1}$, then $q=\frac{a_{2}^{2}}{a_{1}^{2}}=(\sqrt{2}+1)^{2}>1$, which does not meet the requirement.
If $d=(-2+\sqrt{2}) a_{1}$, then $q=\frac{a_{2}^{2}}{a_{1}^{2}}=(\sqrt{2}-1)^{2}$.
From $\lim _{x \rightarrow \infty}\left(b_{1}+b_{2}+\cdots+b_{n}\right)=\sqrt{2}+1$, we get
$$
\frac{b_{1}}{1-q}=\sqrt{2}+1 \text {, }
$$

which simplifies to $\frac{a_{1}^{2}}{[1-(\sqrt{2}-1)]^{2}}=\sqrt{2}+1$. Solving this, we get $a_{1}^{2}=2$.
From $a_{2}^{2}=-a_{1} a_{3}$ and $a_{1}<a_{2}$, we know $a_{1}<0$.
$$
\therefore a_{1}=-\sqrt{2}, d=(-2+\sqrt{2}) a_{1}=2 \sqrt{2}-2 \text {. }
$$