13. Let $\left\{a_{n}\right\}$ be an arithmetic sequence, $\left\{b_{n}\right\}$ 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 $\left\{a_{n}\right\}$.
$b_{2}^{2}=b_{1} \cdot b_{3} \Rightarrow a_{2}^{4}=a_{1}^{2} \cdot a_{3}^{2} \Rightarrow a_{2}^{2}=-a_{1} \cdot a_{3}$ (if $a_{2}^{2}=a_{1} \cdot a_{3}$ then it contradicts the condition $a_{1}1$.
Thus $\frac{a_{2}}{a_{1}}=-1+\sqrt{2} \Rightarrow q=\left(\frac{a_{2}}{a_{1}}\right)^{2}=3-2 \sqrt{2} \Rightarrow \frac{b_{1}}{1-q}=\sqrt{2}+1$
$\Rightarrow \frac{b_{1}}{2 \sqrt{2}-2}=\sqrt{2}+1 \Rightarrow b_{1}=2$.
Noting that $a_{1}<0 \Rightarrow a_{1}=-\sqrt{2} \Rightarrow d=-\sqrt{2}(-2+\sqrt{2})=2 \sqrt{2}-2$.
Therefore, the first term of the sequence $\left\{a_{n}\right\}$ is $a_{1}=-\sqrt{2}$, and the common difference is $d=2 \sqrt{2}-2$.