Cho biểu thức A = 1/1.2 + 1/ 3.4 + 1/ 5.6 + .... + 1/99.100. Chứng minh 7/12 <A< 5/6

Ta có: A\( = \frac{1}{{1 \cdot 2}} + \frac{1}{{3 \cdot 4}} + \frac{1}{{5 \cdot 6}} + \cdot \cdot \cdot + \frac{1}{{99 \cdot 100}} > \frac{1}{{1 \cdot 2}} + \frac{1}{{3 \cdot 4}} = \frac{7}{{12}}\).

Lại có:

A\( = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdot \cdot \cdot + \frac{1}{{99}} - \frac{1}{{100}} = \frac{5}{6} - \left( {\frac{1}{4} - \frac{1}{5}} \right) - \left( {\frac{1}{6} - \frac{1}{7}} \right) - \cdot \cdot \cdot - \left( {\frac{1}{{99}} - \frac{1}{{100}}} \right) < \frac{5}{6}\).

Vậy \(\frac{7}{{12}} < {\rm{A}} < \frac{5}{6}\).