Ta có:
\(\dfrac{1}{n}.\dfrac{1}{n+1}=\dfrac{1}{n\left(n+1\right)}\) (1)
\(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{1}{n}.\dfrac{1}{n+1}=\dfrac{1}{n}-\dfrac{1}{n+1}\) ( \(n\in Z,n\ne0\) )