Tính tổng

Tính tổng : $S=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\cdot \cdot \cdot+\dfrac{n-1}{n!}$


Đặt $S_n=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+ \cdot\cdot\cdot+\dfrac{n-1}{n!}$ với $n \ge 2$.

Ta có $S_n=S_{n-1}+\dfrac{n-1}{n!} =S_{n-1}+\dfrac{1}{(n-1)!}- \dfrac{1}{n!} \\ \Leftrightarrow S_n+\dfrac{1}{n!}=S_{n-1}+\dfrac{1}{(n-1)!}= \ldots=S_2+\dfrac{1}{2!}=1$

Vậy $S_n=\dfrac{n!-1}{n!}$.