Lời giải

Ta có:

\(S = 1 + \frac{1}{2} + \frac{1}{3} + \cdot \cdot \cdot + \frac{1}{{63}} + \frac{1}{{64}}\)

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

\( > 1 + \frac{1}{2} + \left( {\frac{1}{4} + \frac{1}{4}} \right) + \left( {\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}} \right) + \cdot \cdot \cdot + \left( {\frac{1}{{64}} + \cdot \cdot \cdot + \frac{1}{{64}}} \right)\)

\( = 1 + \frac{1}{2} + 2 \cdot \frac{1}{4} + 4 \cdot \frac{1}{8} + 8 \cdot \frac{1}{{16}} + 16 \cdot \frac{1}{{32}} + 32 \cdot \frac{1}{{64}}\)

\( = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 4\)

Vậy \(S = 1 + \frac{1}{2} + \frac{1}{3} + \cdot \cdot \cdot + \frac{1}{{63}} + \frac{1}{{64}} > 4\).