Với b>1và \(b\in N\) có \(\frac{1}{b^2}>\frac{1}{b\left(b+1\right)}=\frac{1}{b}-\frac{1}{b+1}\)(1)
\(\frac{1}{b^2}< \frac{1}{\left(b-1\right)b}=\frac{1}{b-1}-\frac{1}{b}\) (2).Từ (1),(2) => \(\frac{1}{b}-\frac{1}{b+1}< \frac{1}{b^2}< \frac{1}{b-1}-\frac{1}{b}\)