CMR : \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2015}<1\)
K
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30 tháng 11 2015
Với mọi số nguyên dương n ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)
Ta có: \(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}}\)
\(\Rightarrow\frac{1}{\left(n+1\right)\sqrt{n}}<\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\frac{2}{\sqrt{n}}=\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\). Do đó ta có:
\(A<\frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}+\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}+...+\frac{2}{\sqrt{2015}}-\frac{2}{\sqrt{2016}}=2-\frac{2}{\sqrt{2016}}<2\)
Vậy A < 2.
9 tháng 5 2015
\(\frac{1}{2^2}+\)\(\frac{1}{3^2}+\)\(\frac{1}{4^2}+\)...+\(\frac{1}{2015^2}+\)\(\frac{1}{2015}\)
<\(\frac{1}{1.2}+\)\(\frac{1}{3.4}+\)\(\frac{1}{4.5}+\)...+\(\frac{1}{2014.2015}\)+\(\frac{1}{2015}\)
Ta có:\(\frac{1}{1.2}+\)\(\frac{1}{3.4}+\)\(\frac{1}{4.5}+\)...+\(\frac{1}{2014.2015}\)+\(\frac{1}{2015}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}\)
=1
=>\(\frac{1}{2^2}+\)\(\frac{1}{3^2}+\)\(\frac{1}{4^2}+\)...+\(\frac{1}{2015^2}+\)\(\frac{1}{2015}\) \(<1\)
K
9 tháng 5 2015
Ta có : \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};\frac{1}{4^2}<\frac{1}{3.4};...;\frac{1}{2015^2}<\frac{1}{2014.2015}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}<\frac{1}{1}-\frac{1}{2015}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2015}<1\)(đpcm)
9 tháng 8 2016
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2016^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
\(A< 1-\frac{1}{2016}\)
\(A< \frac{2015}{2016}\left(đpcm\right)\)
9 tháng 8 2016
\(A=\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{2016.2016}< \frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{2015.2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}\)
\(=\frac{2015}{2016}\)
\(\Rightarrow A< \frac{2015}{2016}\)
1/22+1/32+1/42+......+1/20152+1/20162 < 1/1.2+1/2.3+1/3.4+.....+1/2014.2015+1/2015.2016
Mà: 1/1.2+1/2.3+1/3.4+.....+1/2014.2015+1/2015.2016
=1-1/2+1/2-1/3+1/3-1/4+.......+1/2014-1/2015+1/2015-1/2016
=1-1/2016
=2016/2016-1/2016
=2015/2016 <1
Nên 1/22+1/32+1/42+......+1/20152+1/20162 < 1