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 $\lim _{n \rightarrow+\infty}\left(b_{1}+b_{2}+\cdots+b_{n}\right)=\sqrt{2}+1$. Find the first term and common difference of $\left\{a_{n}\right\}$.
13. Let the required common difference be $d$.

Since $a_{1}0$.
This leads to $a_{1}^{2}\left(a_{1}+2 d\right)^{2}=\left(a_{1}+d\right)^{4}$.
Simplifying, we get $2 a_{1}^{2}+4 a_{1} d+d^{2}=0$.
Solving, 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
$$
q=\frac{a_{2}^{2}}{a_{1}^{2}}=(\sqrt{2}-1)^{2} .
$$

But $\lim _{n \rightarrow+\infty}\left(b_{1}+b_{2}+\cdots+b_{n}\right)=\sqrt{2}+1$ exists, so $|q|<1$. Therefore, $q=(\sqrt{2}+1)^{2}$ is impossible.
Thus, $\frac{a_{1}^{2}}{1-(\sqrt{2}-1)^{2}}=\sqrt{2}+1 \Rightarrow a_{1}^{2}=(2 \sqrt{2}-2)(\sqrt{2}+1)=2$.
Therefore, $a_{1}=-\sqrt{2}$,
$$
d=(-2+\sqrt{2}) a_{1}=(-2+\sqrt{2})(-\sqrt{2})=2 \sqrt{2}-2 .
$$