[数分提高]2014-2015-2第2教学周第2次课

已知 $$\bex x_n=\sum_{i=1}^n \frac{1}{i(i+1)(i+2)(i+3)}. \eex$$ 试证: $\sed{x_n}$ 收敛, 并求其极限.

 

证明: $$\beex \bea x_n&=\frac{1}{3}\sum_{i=1}^n \frac{(i+3)-i}{i(i+1)(i+2)(i+3)}\\ &=\frac{1}{3}\sez{ \sum_{i=1}^n \frac{1}{i(i+1)(i+2)} -\sum_{i=1}^n \frac{1}{(i+1)(i+2)(i+3)} }\\ &=\frac{1}{3}\sez{\frac{1}{1\cdot 2\cdot 3} -\frac{1}{(n+1)(n+2)(n+3)}}\\ &\to \frac{1}{18}\quad\sex{n\to\infty}. \eea \eeex$$