Cho \[A = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{63}}\]. Chứng minh rằng A > 3.

Ta có: \[A = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{63}}\]

\[ = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{63}} + \frac{1}{{64}} - \frac{1}{{64}}\]

\[ = \left( {1 + \frac{1}{2}} \right) + \left( {\frac{1}{3} + \frac{1}{4}} \right) + \left( {\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}} \right) + ... + \left( {\frac{1}{{33}} + \frac{1}{{34}} + ... + \frac{1}{{64}}} \right) - \frac{1}{{64}}\]

\[ \Rightarrow A > 1 + \frac{1}{2} + 2 \times \frac{1}{4} + 4 \times \frac{1}{8} + ... + 32 \times \frac{1}{{64}} - \frac{1}{{64}}\]

\[ \Rightarrow A > 1 + \left( {\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}} \right) - \frac{1}{{64}}\]

\[ \Rightarrow A > 1 + 3 - \frac{1}{{64}}\]

\[ \Rightarrow A > 3 + \left( {1 - \frac{1}{{64}}} \right)\]

Mà \[1 - \frac{1}{{64}} > 0\] nên A > 3

Vậy A > 3