设 $\sed{a_n}$ 为等差数列, $a_{n+1}-a_n=d>0\ (n=1,2,\cdots)$, $m$ 为一正整数. 计算 $$\bex S=\vsm{n}\frac{1}{a_n\cdot a_{n+1}\cdots a_{n+m}}. \eex$$

解答: 由 $$\beex \bea \frac{1}{a_na_{n+1}\cdots a_{n+m}} &=\frac{a_{n+m}-a_n}{a_na_{n+1}\cdots a_{n+m}}\cdot \frac{1}{md} =\sex{\frac{1}{a_n\cdots a_{n+m-1}}-\frac{1}{a_{n+1}\cdots a_{n+m}}}\frac{1}{md},\\ S_N&=\sum_{n=1}^N \frac{1}{a_na_{n+1}\cdots a_{n+m}}\\ &=\frac{1}{md}\sex{\frac{1}{a_1\cdots a_m}-\frac{1}{a_{N+1}\cdots a_{N+1+m}}},\\ S&=\frac{1}{a_1\cdots a_m md}. \eea \eeex$$