- Đặt \(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{64}\)
- Ta có: \(A=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+...+\left(\frac{1}{33}+\frac{1}{34}+...+\frac{1}{64}\right)\)
\(\Rightarrow A>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+...+\left(\frac{1}{64}+\frac{1}{64}+...+\frac{1}{64}\right)\)
\(\Leftrightarrow A>1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(\Leftrightarrow A>4\)\(\left(ĐPCM\right)\)
