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

2/ Ta có : \(A=\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+....+\dfrac{1}{99\times100}\)

\(=\left(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}\right)+\left(\dfrac{1}{5\times6}+....+\dfrac{1}{99\times100}\right)\)

\(=\dfrac{7}{12}+\left(\dfrac{1}{5\times6}+....+\dfrac{1}{99\times100}\right)>\dfrac{7}{12}\) (1)

\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+.....+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+.....+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{100}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{100}\right)\times2\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)\)

\(=\dfrac{1}{51}+\dfrac{1}{52}+....+\dfrac{1}{100}\)

Dãy số trên có : \(100-51+1=50\) số hạng.

mà 50 chia hết cho 10 nên ta nhóm 10 số vào một nhóm.

\(A=\left(\dfrac{1}{51}+...+\dfrac{1}{60}\right)+\left(\dfrac{1}{61}+...+\dfrac{1}{70}\right)+\left(\dfrac{1}{71}+...+\dfrac{1}{80}\right)+\left(\dfrac{1}{81}+...+\dfrac{1}{90}\right)+\left(\dfrac{1}{91}+...+\dfrac{1}{100}\right)\)

\(< \dfrac{1}{50.10}+\dfrac{1}{60.10}+\dfrac{1}{70.10}+\dfrac{1}{80.10}+\dfrac{1}{90.10}=\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7.3}=\dfrac{167}{210}< \dfrac{175}{210}=\dfrac{5}{6}\)

\(\Rightarrow A< \dfrac{5}{6}\) (2)

Từ (1) và (2) \(\Rightarrow\) đpcm