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Ta co:
\(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n+1}.\sqrt{n}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Ap vào bài toan được
\(S_n=\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(< \frac{1}{2}\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{n+1}}\right)< \frac{1}{2}\)
Lời giải:
Liên hợp ta thấy:
\(2(\sqrt{n+1}-\sqrt{n})=2.\frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}=\frac{2}{\sqrt{n+1}+\sqrt{n}}<\frac{2}{\sqrt{n}+\sqrt{n}}=\frac{1}{\sqrt{n}}(1)\)
\(2(\sqrt{n}-\sqrt{n-1})=2.\frac{n-(n-1)}{\sqrt{n}+\sqrt{n-1}}=\frac{2}{\sqrt{n}+\sqrt{n-1}}>\frac{2}{\sqrt{n}+\sqrt{n}}=\frac{1}{\sqrt{n}}(2)\)
Từ \((1);(2)\Rightarrow 2(\sqrt{n+1}-\sqrt{n})< \frac{1}{\sqrt{n}}< 2(\sqrt{n}-\sqrt{n-1})\)
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Áp dụng vào bài toán:
\(S=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>1+2(\sqrt{3}-\sqrt{2})+2(\sqrt{4}-\sqrt{3})+...+2(\sqrt{101}-\sqrt{100})\)
\(\Leftrightarrow S>1+2(\sqrt{101}-\sqrt{2})>18(*)\)
Và:
\(S< 1+2(\sqrt{2}-\sqrt{1})+2(\sqrt{3}-\sqrt{2})+....+2(\sqrt{100}-\sqrt{99})\)
\(\Leftrightarrow S< 1+2(\sqrt{100}-\sqrt{1})=19(**)\)
Từ $(*); (**)$ suy ra $18< S< 19$ (đpcm)
Ta có \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
=> biểu thức trên sẽ \(< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+......+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n+1}}< 2\)\(\left(dpcm\right)\)
Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}}\)
\(=\frac{\sqrt{n}}{n\left(n+1\right)}\)
\(=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
=> Biểu thức trên sẽ bé hơn \(2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n+1}}< 2\Rightarrowđpcm\)
Xét \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{\left(n+1\right)n}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\) = \(\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)=\left(\frac{\sqrt{n}}{\sqrt{n+1}}+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\) < \(2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Vậy \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.....+\frac{1}{\left(n+1\right)\sqrt{n}}<2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\) = \(2\left(1-\frac{1}{\sqrt{n+1}}\right)<2\) (đpcm)