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Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{n}};\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{n}}....;\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n}}\)
=>\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+...+\frac{1}{\sqrt{n}}\)
\(=n.\frac{1}{\sqrt{n}}=\sqrt{n}\left(dpcm\right)\)
Ta có \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>...\)\(>\frac{1}{\sqrt{n}}\)
Suy ra \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\)\(\frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\)\(+...+\frac{1}{\sqrt{n}}=n.\frac{1}{\sqrt{n}}=\sqrt{n}\)
Đặt A =\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}\)
=> A > \(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+.....+\frac{1}{\sqrt{n}}\)
=> A > \(\frac{1}{\sqrt{n}}.n\)
=> A > \(\sqrt{n}\)
=> \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}>\sqrt{n}\)(Đpcm)
Ta có : \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{n}}\)(vì 1 < n) (1)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{n}}\)(vì 2 < n) (2)
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\(\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n}}\)(n)
Cộng các vế trái với nhau,các vế phải với nhau,ta có :
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n}}.n\left(=\sqrt{n}\right)\)(từ 1 đến n có n số tự nhiên).Vậy ta có đpcm.
a) \(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)
\(=\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}\)
\(=\frac{n^2\left(n^2+2n+1+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
\(=\frac{n^4+2n^2\left(n+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
\(=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
=>đpcm
b) Từ công thức trên ta có:
\(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
=> \(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\frac{n^2+n+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)
Ta có:
\(S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(=2010+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(2010+\left(1-\frac{1}{2011}\right)=2010+\frac{2010}{2011}=2010\frac{2010}{2011}\)
Ta có : \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{n}};\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{n}}...;\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n}}\)
\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+...+\frac{1}{\sqrt{n}}\)
\(=n.\frac{1}{\sqrt{n}}=\sqrt{n}\left(dpcm\right)\)