方案一： $1+2+3+ \cdots +n= \frac{n \times (n+1)}{2}$

$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\cdots+\frac{1}{(n-1)\times n}=1-\frac{1}{n}$

$\frac{1}{1 \times 2 \times \cdots \times m}+\frac{1}{2 \times 3\times \cdots \times (m+1)}+\cdots+\frac{1}{n\times (n+1) \times \cdots \times (n+m-1)} = S$

$$1+2+3+ \cdots +n= \frac{n \times (n+1)}{2}$$

$$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\cdots+\frac{1}{(n-1)\times n}=1-\frac{1}{n}$$

$$\frac{1}{1 \times 2 \times \cdots \times m}+\frac{1}{2 \times 3\times \cdots \times (m+1)}+\cdots+\frac{1}{n\times (n+1) \times \cdots \times (n+m-1)} = S$$

方案2： $\sum_{i=1}^n i= \frac{n \times (n+1)}{2}$

$\sum_{i=1}^{n-1} \frac{1}{i\times (i+1)}=1-\frac{1}{n}$

$\sum_{i=1}^n \frac{1}{\prod_{j=i}^{i+m-1}j}=S$

$$\sum_{i=1}^n i= \frac{n \times (n+1)}{2}$$

$$\sum_{i=1}^{n-1} \frac{1}{i\times (i+1)}=1-\frac{1}{n}$$

$$\sum_{i=1}^n \frac{1}{\prod_{j=i}^{i+m-1}j}=S$$