II. (50 points) Given the sequence $\left\{a_{n}\right\}$ defined by
$$
a_{1}=\frac{2}{3}, a_{n+1}=a_{n}^{2}+a_{n-1}^{2}+\cdots+a_{1}^{2}\left(n \in \mathbf{N}_{+}\right)
$$

Determine. If for any $n\left(n \in \mathbf{N}_{+}\right)$,
$$
\frac{1}{a_{1}+1}+\frac{1}{a_{2}+1}+\cdots+\frac{1}{a_{n}+1}<M
$$

holds true. Find the minimum value of $M$.
Given that when $n \geqslant 2$, $a_{n+1}=a_{n}^{2}+a_{n}$. Also, $a_{2}=a_{1}^{2}=\frac{4}{9}=\left(\frac{1}{3}\right)^{2}+\frac{1}{3}$. Let's set $b_{1}=\frac{1}{3}$ and $b_{n}=a_{n}$ for $n \geqslant 2$.
Then, $b_{n+1}=b_{n}^{2}+b_{n}\left(n \in \mathbf{N}_{+}\right)$.
We will prove by mathematical induction that:
$$
\frac{1}{b_{1}+1}+\frac{1}{b_{2}+1}+\cdots+\frac{1}{b_{n}+1}=\frac{3 b_{n+1}-1}{b_{n+1}} \text {. }
$$

When $n=1$,
$$
b_{2}=\frac{4}{9}, \frac{1}{b_{1}+1}=\frac{3 b_{2}-1}{b_{2}}=\frac{3}{4} \text {. }
$$

Equation (2) holds.
Assume that for $n=k$, equation (2) holds.
Then for $n=k+1$,
$$
\begin{array}{l}
\frac{1}{b_{1}+1}+\frac{1}{b_{2}+1}+\cdots+\frac{1}{b_{k}+1}+\frac{1}{b_{k+1}+1} \\
=\frac{3 b_{k+1}-1}{b_{k+1}}+\frac{1}{b_{k+1}+1} \\
=\frac{3 b_{k+1}\left(b_{k+1}+1\right)-1}{b_{k+1}\left(b_{k+1}+1\right)} .
\end{array}
$$

From equation (1), we know
$$
\frac{3 b_{k+1}\left(b_{k+1}+1\right)-1}{b_{k+1}\left(b_{k+1}+1\right)}=\frac{3 b_{k+2}-1}{b_{k+2}} \text {. }
$$

Therefore, the conclusion holds for $n=k+1$.
Thus, $\frac{1}{a_{1}+1}+\frac{1}{a_{2}+1}+\cdots+\frac{1}{a_{n}+1}$
$$
\begin{array}{l}
=\frac{1}{b_{1}+1}+\frac{1}{b_{2}+1}+\cdots+\frac{1}{b_{n}+1}+\frac{1}{1+a_{1}}-\frac{1}{1+b_{1}} \\
=\frac{3 b_{n+1}-1}{b_{n+1}}-\frac{3}{20}=\frac{57}{20}-\frac{1}{b_{n+1}} .
\end{array}
$$

We know that $b_{n+1}$ is a positive number, and
$$
b_{n+1}-b_{n}=b_{n}^{2} \geqslant b_{1}^{2} \geqslant \frac{1}{9} \text {, }
$$

Therefore, as $n$ approaches infinity, $b_{n+1}$ approaches infinity.
Thus, the minimum value of $M$ is $\frac{57}{20}$.