题目

求级数sum _(n=1)^infty dfrac (1)(n(n+1)(n+2))的和。

求级数 $\sum_{n=1}^{\infty}\frac{1}{n\left(n+1\right)\left(n+2\right)}$ 的和。

题目解答

答案

令 $a_{n=}\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{\frac{1}{2}}{n}+\frac{-1}{n+1}+\frac{\frac{1}{2}}{n+2}$

注意到上式右端三项系数之和为零,所以

$S_n=\left(\frac{1}{2}\times\frac{1}{1}-\frac{1}{1+1}+\frac{1}{2}\times\frac{1}{1+2}\right)+\left(\frac{1}{2}\times\frac{1}{2}-\frac{1}{2+1}+\frac{1}{2}\times\frac{1}{2+2}\right)$

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

$=\frac{1}{4}-\frac{1}{2}\times\frac{1}{n+1}-\frac{1}{2}\times\frac{1}{n+2}$

所以 $s=\lim_{n\rightarrow\infty}\frac{1}{4}-\frac{1}{2}\times\frac{1}{n+1}-\frac{1}{2}\times\frac{1}{n+2}$

$=\frac{1}{4}$

所以本题答案为 $\frac{1}{4}$ 。

解析

步骤 1：部分分式分解
将级数的通项$\dfrac {1}{n(n+1)(n+2)}$分解为部分分式，即
$$\dfrac {1}{n(n+1)(n+2)}=\dfrac {A}{n}+\dfrac {B}{n+1}+\dfrac {C}{n+2}$$
其中$A$、$B$、$C$为待定系数。通过比较系数，可以得到$A=\dfrac {1}{2}$，$B=-1$，$C=\dfrac {1}{2}$。因此，
$$\dfrac {1}{n(n+1)(n+2)}=\dfrac {\dfrac {1}{2}}{n}+\dfrac {-1}{n+1}+\dfrac {\dfrac {1}{2}}{n+2}$$
步骤 2：级数求和
将级数的通项代入求和公式，得到
$$\sum _{n=1}^{\infty }\dfrac {1}{n(n+1)(n+2)}=\sum _{n=1}^{\infty }\left(\dfrac {\dfrac {1}{2}}{n}+\dfrac {-1}{n+1}+\dfrac {\dfrac {1}{2}}{n+2}\right)$$
步骤 3：求和计算
注意到上式右端三项系数之和为零，所以
$$\sum _{n=1}^{\infty }\dfrac {1}{n(n+1)(n+2)}=\lim _{n\rightarrow \infty }\left(\dfrac {1}{4}-\dfrac {1}{2}\times \dfrac {1}{n+1}-\dfrac {1}{2}\times \dfrac {1}{n+2}\right)$$