13. (2001 National High School Mathematics Competition) 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 $\lim \left(b_{1}+b_{2}+\cdots+b_{n}\right)=\sqrt{2}+1$, find the first term and common difference of $\left\{a_{n}\right\}$.
13. Solution: Let the required common difference be $d$. Since $a_{1}0$.

From this, we get $a_{1}^{2}\left(a_{1}+2 d\right)^{2}=\left(a_{1}+d\right)^{4}$, which simplifies to $2 a_{1}^{2}+4 a_{1} d+d^{2}=0$. Solving this, we get $d=(-2 \pm \sqrt{2}) a_{1}$. Since $-2 \pm \sqrt{2}<0$, it follows that $a_{1}<0$.
If $d=(-2-\sqrt{2}) a_{1}$, then $q=\frac{a_{2}^{2}}{a_{1}^{2}}=(\sqrt{2}+1)^{2}$;
If $d=(-2+\sqrt{2}) a_{1}$, then $q=\frac{a_{2}^{2}}{a_{1}^{2}}=(\sqrt{2}-1)^{2}$.
However, $\lim _{n \rightarrow \infty}\left(b_{1}+b_{2}+\cdots+b_{n}\right)=\sqrt{2}+1$, so $|q|<1$. Therefore, $q=(\sqrt{2}+1)^{2}$ is impossible. Thus, $\frac{a_{1}^{2}}{1-(\sqrt{2}-1)^{2}}=\sqrt{2}+1$, solving this gives $a_{1}^{2}=(2 \sqrt{2}-2)(\sqrt{2}+1)=2$.
Therefore, $a_{1}=-\sqrt{2}, d=(-2+\sqrt{2}) a_{1}=2 \sqrt{2}-2$, meaning $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term $-\sqrt{2}$ and common difference $2 \sqrt{2} - 2$.