ta có:\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>...>\frac{1}{\sqrt{n}}\)
→\(1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+...+\frac{1}{\sqrt{n}}\)(n hạng tử \(\frac{1}{\sqrt{n}}\))
→\(1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}>\frac{n}{\sqrt{n}}=\sqrt{n}\)