Chứng minh rằng

\[\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{99}} - \frac{1}{{100}} > \frac{1}{2}.\]

Lời giải:

\[\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{99}} - \frac{1}{{100}}\]

\[ = \left( {\frac{1}{1} + \frac{1}{3} + ... + \frac{1}{{99}}} \right) - \left( {\frac{1}{2} + \frac{1}{4} + ... + \frac{1}{{100}}} \right)\]

\[ = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{99}} + \frac{1}{{100}}} \right) - 2 \cdot \left( {\frac{1}{2} + \frac{1}{4} + ... + \frac{1}{{100}}} \right)\]

\[ = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{99}} + \frac{1}{{100}}} \right) - \left( {\frac{1}{1} + \frac{1}{2} + ... + \frac{1}{{50}}} \right)\]

\[ = \frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{99}} + \frac{1}{{100}}\]

Ta có \(\frac{1}{{51}} > \frac{1}{{100}};\,\,\frac{1}{{52}} > \frac{1}{{100}};\,\,...;\,\,\frac{1}{{99}} > \frac{1}{{100}}.\)

Do đó \[\frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{99}} + \frac{1}{{100}} > \frac{1}{{100}} + \frac{1}{{100}} + ... + \frac{1}{{100}} + \frac{1}{{100}} = \frac{{50}}{{100}} = \frac{1}{2}.\]

Vậy \[\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{99}} - \frac{1}{{100}} > \frac{1}{2}.\]