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1/ Trước hết ta chứng minh \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Áp dụng :
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n+1}}< 2\) (đpcm)
Với mọi \(n\ge2\)
\(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}>\frac{2}{\sqrt{n}+\sqrt{n+1}}=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=2\left(\sqrt{n+1}-\sqrt{n}\right)\) (1)
Lại có : \(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}< \frac{2}{\sqrt{n}+\sqrt{n-1}}=\frac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n}+\sqrt{n-1}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\)
\(=2\left(\sqrt{n}-\sqrt{n-1}\right)\) (2)
Từ (1) và (2) suy ra \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
Áp dụng với n = 2,3,4,...,100 được đpcm.
\(a,A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+...+\frac{\sqrt{99}-\sqrt{100}}{99-100}\)
\(=\frac{1-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{99}-\sqrt{100}}{-1}\)
\(=\frac{1-\sqrt{100}}{-1}=9\)
\(b,B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+..+\frac{1}{\sqrt{99}}\)
\(=\frac{2}{\sqrt{1}+\sqrt{1}}+\frac{2}{\sqrt{2}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{99}}>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)\(\Rightarrow B>2\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{99}+\sqrt{100}}\right)\)
\(\Rightarrow B>2\left(\frac{\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{99}-\sqrt{100}}{-1}\right)\)
\(\Rightarrow B>2\left(\frac{1-\sqrt{100}}{-1}\right)\)
\(\Rightarrow B>2.9=18\left(ĐPCM\right)\)
\(S=\frac{2}{2}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{\sqrt{100}}\)
\(S>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{4}}+...+\frac{2}{\sqrt{100}+\sqrt{101}}\)
\(S>2\left(\sqrt{2}-\sqrt{1}\right)+2\left(\sqrt{3}-\sqrt{2}\right)+...+2\left(\sqrt{101}-\sqrt{100}\right)\)
\(S>2\left(\sqrt{101}-1\right)>2\left(\sqrt{100}-1\right)=18\) (1)
\(S< 1+\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)
\(S< 1+2\left(\sqrt{2}-1\right)+2\left(\sqrt{3}-\sqrt{2}\right)+...+2\left(\sqrt{100}-\sqrt{99}\right)\)
\(S< 1+2\left(\sqrt{100}-1\right)=19\) (2)
(1); (2) \(\Rightarrow18< S< 19\)
\(\frac{2}{\sqrt{k+1}+\sqrt{k}}<\frac{2}{2\sqrt{k}}<\frac{2}{\sqrt{k}-\sqrt{k-1}}\)
\(2\left(\sqrt{k+1}-\sqrt{k}\right)<\frac{1}{\sqrt{k}}<2\left(\sqrt{k}-\sqrt{k-1}\right)\)
\(2\sqrt{3}-2\sqrt{2}<\frac{1}{\sqrt{2}}<2\sqrt{2}-2\sqrt{1}\)
\(2\sqrt{4}-2\sqrt{3}<\frac{1}{\sqrt{3}}<2\sqrt{3}-2\sqrt{2}\)
\(2\sqrt{5}-2\sqrt{4}<\frac{1}{\sqrt{4}}<2\sqrt{4}-2\sqrt{3}\)
.......................................................................
\(2\sqrt{101}-2\sqrt{100}<\frac{1}{\sqrt{100}}<2\sqrt{100}-2\sqrt{99}\)
Cộng từng vế ta dc
\(2\sqrt{101}-2\sqrt{2}<\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{100}}<2\sqrt{100}-2\sqrt{1}\)
\(17<\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{100}}<18\)