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\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2.n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1}\)
\(\Rightarrow S_n=\frac{1}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1}=1-\frac{\sqrt{n+1}}{n+1}\)
\(\Rightarrow S\left(n\right)\) hữu tỉ khi và chỉ khi \(\frac{\sqrt{n+1}}{n+1}=\frac{1}{\sqrt{n+1}}\) hữu tỉ
\(\Leftrightarrow\sqrt{n+1}\) hữu tỉ
\(\Leftrightarrow n+1=k^2\) với \(k\in Z\) ; \(k>1\)
\(\Rightarrow n=k^2-1\) với \(k\in Z;k>1\)
Vậy với mọi n có dạng \(n=k^2-1\) sao cho k là số nguyên lớn hơn 1 thì \(S\left(n\right)\) hữu tỉ
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\(\hept{\begin{cases}\frac{2}{2\sqrt{n}}< \frac{2}{\sqrt{n-1}+\sqrt{n}}=2\left(\sqrt{n}-\sqrt{n-1}\right)\\\frac{2}{2\sqrt{n}}>\frac{2}{\sqrt{n+1}+\sqrt{n}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\end{cases}}\)
Từ đây ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}< 2\left(\sqrt{1}-\sqrt{0}+\sqrt{2}-\sqrt{1}+...+\sqrt{n}-\sqrt{n-1}\right)\)
\(=2\left(\sqrt{n}-0\right)=2\sqrt{n}\)
Tương tự ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n+1}-\sqrt{n}\right)\)
\(=2\left(\sqrt{n+1}-1\right)>\sqrt{n}\)
Gọi \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}=A\)là A
Có \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>...>\frac{1}{\sqrt{n}}\)
=> \(A>n.\frac{1}{\sqrt{n}}=\sqrt{n}\)(1)
Ta có: \(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}< \frac{2}{\sqrt{n}+\sqrt{n-1}}=2\left(\sqrt{n}+\sqrt{n-1}\right)\)
=> \(\frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
Khi đó: \(\frac{1}{\sqrt{1}}< 2\left(\sqrt{1}-\sqrt{0}\right)\)
\(\frac{1}{\sqrt{2}}< 2\left(\sqrt{2}-\sqrt{1}\right)\)
...
\(\frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
=> \(A< 2\left(\sqrt{n}-\sqrt{n-1}+...+\sqrt{1}\right)\)
=> \(A< 2\sqrt{n}\)(2)
Từ (1) và (2) => \(\sqrt{n}< A< 2\sqrt{n}\)
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\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2}=1+\frac{1}{n}-\frac{1}{n+1}\)
\(S=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+....+1+\frac{1}{n}-\frac{1}{n+1}\)
\(=n+1-\frac{1}{n+1}=\frac{\left(n+1\right)^2-1}{n+1}=\frac{2009^2-1}{2009}\Rightarrow n+1=2009\Rightarrow n=2008\)
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Bài 1:
Có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Có: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)
xong bn áp dụng lên trên lm tiếp
Bài 3:
theo bđt cô si ta có:
\(\sqrt{\frac{b+c}{a}\cdot1}\le\left(\frac{b+c}{a}+1\right):2=\frac{b+c+a}{2a}\)
=> \(\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\) (1)
Tương tự ta có :
\(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\) (2)
\(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\) (3)
Cộng vế vs vế (1)(2)(3) ta có:
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2a+2b+2c}{a+b+c}=2\)
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Xét số hạng tổng quát ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{\left(n+1\right)n}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)< \sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\sqrt{n}\cdot\frac{2}{\sqrt{n}}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)
Áp dụng vào bài tập, ta có:
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}+...+\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)
\(=2-\frac{2}{\sqrt{n+1}}< 2\left(đpcm\right)\)