Lời giải
Với mọi $n\in\mathbb{N}$ ta có:
\(\frac{1}{\sqrt{1}}> \frac{1}{\sqrt{n}}\)
\(\frac{1}{\sqrt{2}}> \frac{1}{\sqrt{n}}\)
.....
Do đó:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}> \underbrace{\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+...+\frac{1}{\sqrt{n}}}_{\text{n số}}=\frac{n}{\sqrt{n}}=\sqrt{n}\)
(chứng minh xong vế 1)
Vế 2:
\(\frac{1}{2\sqrt{1}}+\frac{1}{2\sqrt{2}}+...+\frac{1}{2\sqrt{n}}< \frac{1}{\sqrt{0}+\sqrt{1}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(=\frac{\sqrt{1}-\sqrt{0}}{1-0}+\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{n}-\sqrt{n-1}}{n-(n-1)}\)
\(=\sqrt{1}-\sqrt{0}+\sqrt{2}-\sqrt{1}+...+\sqrt{n}-\sqrt{n-1}=\sqrt{n}\)
\(\Rightarrow \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n}\) (đpcm)
Vậy....
