A \(=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{100}}\)
A > \(\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+....+\frac{1}{\sqrt{100}}\) (có 100 số hạng \(\frac{1}{\sqrt{100}}\))
= \(\frac{1}{\sqrt{100}}.100=\frac{100}{\sqrt{100}}=\frac{100}{10}=10\)
Vậy A > 10