Cho A = 1/1.2 + 1/3.4 + 1/5.6 + + 1/99.100. Chứng minh rằng: 7/12 < A < 5/6
Cho \[A = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + .... + \frac{1}{{99.100}}\]. Chứng minh rằng: \(\frac{7}{{12}}\) < A < \(\frac{5}{6}\).
Ta có \[A = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + .... + \frac{1}{{99.100}}\]
\[\begin{array}{l}A = \left( {\frac{1}{{1.2}} + \frac{1}{{3.4}}} \right) + \left( {\frac{1}{{5.6}} + ... + \frac{1}{{99.100}}} \right)\\A = \frac{7}{{12}} + \left( {\frac{1}{{5.6}} + ... + \frac{1}{{99.100}}} \right) > \frac{7}{{12}}\end{array}\]
(vì \[\frac{1}{{5.6}} + ... + \frac{1}{{99.100}} > 0\])
Ta có:
\[\begin{array}{l}A = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + ... + \frac{1}{{99.100}}\\ \Rightarrow A = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + ... + \frac{1}{{99}} - \frac{1}{{100}}\end{array}\]
\[ \Rightarrow A = \left( {\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{{99}}} \right) - \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{{100}}} \right)\]
\[ \Rightarrow A = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + ... + \frac{1}{{99}} + \frac{1}{{100}}} \right) - 2\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{{100}}} \right)\]
\[\begin{array}{l} \Rightarrow A = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + \frac{1}{{99}} + \frac{1}{{100}}} \right) - \left( {1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{50}}} \right)\\ \Rightarrow A = \frac{1}{{51}} + \frac{1}{{52}} + \frac{1}{{53}} + ... + \frac{1}{{100}}\end{array}\]
Tổng A có (100 – 51) : 1 + 1 = 50 (số hạng)
Như vậy, ta nhóm 10 số vào 1 nhóm được:
\[\begin{array}{l}A = \left( {\frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{60}}} \right) + \left( {\frac{1}{{61}} + \frac{1}{{62}} + ... + \frac{1}{{70}}} \right)\\ + \left( {\frac{1}{{71}} + \frac{1}{{72}} + ... + \frac{1}{{80}}} \right) + \left( {\frac{1}{{81}} + \frac{1}{{82}} + ... + \frac{1}{{90}}} \right)\\ + \left( {\frac{1}{{91}} + \frac{1}{{92}} + ... + \frac{1}{{100}}} \right)\end{array}\]
Ta thấy:
\[\begin{array}{l}\left( {\frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{60}}} \right) < 10\cdot \frac{1}{{50}} = \frac{1}{5}\\\left( {\frac{1}{{61}} + \frac{1}{{62}} + ... + \frac{1}{{70}}} \right) < 10\cdot \frac{1}{{60}} = \frac{1}{6}\\\left( {\frac{1}{{71}} + \frac{1}{{72}} + ... + \frac{1}{{80}}} \right) < 10\cdot \frac{1}{{80}} = \frac{1}{7}\\\left( {\frac{1}{{81}} + \frac{1}{{82}} + ... + \frac{1}{{90}}} \right) < 10\cdot \frac{1}{{90}} = \frac{1}{8}\end{array}\]
\[\begin{array}{l}\left( {\frac{1}{{91}} + \frac{1}{{92}} + ... + \frac{1}{{100}}} \right) < \frac{1}{9}\\ \Rightarrow \left( {\frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{60}}} \right) + \left( {\frac{1}{{61}} + \frac{1}{{62}} + ... + \frac{1}{{70}}} \right) + \left( {\frac{1}{{71}} + \frac{1}{{72}} + ... + \frac{1}{{80}}} \right)\\ + \left( {\frac{1}{{81}} + \frac{1}{{82}} + ... + \frac{1}{{90}}} \right) + \left( {\frac{1}{{91}} + \frac{1}{{92}} + ... + \frac{1}{{100}}} \right) < \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} < \frac{5}{6}\\ \Rightarrow A < \frac{5}{6}\end{array}\]
Vậy \(\frac{7}{{12}} < A < \frac{5}{6}\).