Lời giải:
Đặt \(P=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}\)
Ta có:
\(\frac{P}{2}=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{4}}+...+\frac{1}{2\sqrt{n}}\)
\(< \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{n-1}+\sqrt{n}}(1)\)
Mà:
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{n-1}+\sqrt{n}}=\frac{(\sqrt{2}-\sqrt{1})(\sqrt{2}+\sqrt{1})}{\sqrt{1}+\sqrt{2}}+\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}{\sqrt{2}+\sqrt{3}}+\frac{(\sqrt{4}-\sqrt{3})(\sqrt{4}+\sqrt{3})}{\sqrt{3}+\sqrt{4}}+....+\frac{(\sqrt{n}-\sqrt{n-1})(\sqrt{n}+\sqrt{n-1})}{\sqrt{n-1}+\sqrt{n}}\)
\(=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+...+(\sqrt{n}-\sqrt{n-1})\)
\(=\sqrt{n}-1(2)\)
Từ \((1);(2)\Rightarrow \frac{P}{2}< \sqrt{n}-1\Rightarrow P< 2\sqrt{n}-2\)
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Tương tự:
\(\frac{P}{2}>\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}+\frac{1}{2\sqrt{n}}=\sqrt{n}-\sqrt{2}+\frac{1}{2\sqrt{n}}\)
\(\Rightarrow P> 2\sqrt{n}-2\sqrt{2}+\frac{1}{\sqrt{n}}\)
Mà \(2\sqrt{n}-2\sqrt{2}+\frac{1}{\sqrt{n}}> 2\sqrt{n}-3\Rightarrow P>2\sqrt{n}-3\)
Ta có đpcm.