ta có:
\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{64}\) = \(1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+..+\frac{1}{8}\right)+...+\left(\frac{1}{33}+...+\frac{1}{64}\right)\)
>\(\frac{3}{2}+\frac{1}{4}+\frac{1}{4}+\frac{4}{8}+\frac{4}{8}+..+\frac{32}{64}+\frac{32}{64}\)=\(\frac{3}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)>\(4\)
vạyA>4(đpcm)