\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}.50=\frac{1}{2}\)

Vậy \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{1}{2}\)