n>n-1>...>2>1 \Rightarrow 1\frac{a}{b}\sqrt{x}n<1\frac{a}{b}\sqrt{x}n-1<...<1\frac{a}{b}\sqrt{x}1do đó tổng S > n\frac{a}{b}\sqrt{x}n=\sqrt{x}n
n>1 suy ra \sqrt{n}>\sqrt{1} suy ra \frac{1}{\sqrt{n}}<\frac{1}{\sqrt{1}}Do đó tổng S> \frac{n}{\sqrt{n}}=\sqrt{n}