II. (50 points) Given a sequence $\left\{a_{n}\right\}, n \in \mathbf{N}_{+}$, where $a_{1}=1, a_{n}=a_{1} a_{2} \cdots a_{n-1}+1, n \geqslant 2$. Find the smallest real number $M$, such that for any $m \in \mathbf{N}_{+}, \sum_{n=1}^{m} \frac{1}{a_{n}}<M$ holds.
II. Solution: From $a_{n}=a_{1} a_{2} \cdots a_{n-1}+1$, we get $\frac{1}{a_{1} a_{2} \cdots a_{n-1}}=\frac{1}{a_{n}}+\frac{1}{a_{1} a_{2} \cdots a_{n}}$, i.e., $\frac{1}{a_{n}}=\frac{1}{a_{1} a_{2} \cdots a_{n-1}}-\frac{1}{a_{1} a_{2} \cdots a_{n}}$.
For $m \geqslant 2$, substituting $n=2,3, \cdots, m$ into the above equation and summing up, we get
$\sum_{n=2}^{m} \frac{1}{a_{n}}=\frac{1}{a_{1}}-\frac{1}{a_{1} a_{2} \cdots a_{m}}$, i.e., $\sum_{n=1}^{m} \frac{1}{a_{n}}=\frac{2}{a_{1}}-\frac{1}{a_{1} a_{2} \cdots a_{m}}=2-\frac{1}{a_{1} a_{2} \cdots a_{m}}$.
Therefore, for every $m \in \mathbf{N}_{+}$, we have $\sum_{n=1}^{m} \frac{1}{a_{n}}<2$.
Next, we prove $M=2$, i.e., 2 is the smallest value satisfying $\sum_{n=1}^{m} \frac{1}{a_{n}}<M\left(m \in \mathbf{N}_{+}\right)$.
From the definition of the sequence, we get $a_{n} \geqslant a_{n-1}+1$. By mathematical induction, it is easy to see that for each positive integer $n, a_{n} \geqslant n$, thus $0<\frac{1}{a_{1} a_{2} \cdots a_{m}} \leqslant \frac{1}{a_{m}} \leqslant \frac{1}{m}, m \in \mathbf{N}_{+}$.
From this, we know that when $m$ is sufficiently large, $\frac{1}{a_{1} a_{2} \cdots a_{m}}$ becomes arbitrarily small, hence $M=2$.