B=(1-1/2).(1-1/3).........(1-1/n+1) với n€N

$B=(1-\frac{1}{2}).(1-\frac{1}{3})...(1-\frac{1}{n+1})=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}=\frac{1}{n+1}^{}$ (Với $n∈N$)