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Ta thấy rằng: \(2^2>1\times2\) , \(3^2>2\times3\),..., \(2011^2>2010\times2011\).
\(\Rightarrow A< \frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2010\times2011}=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+...+\frac{2011-2010}{2010\times2011}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)\(=1-\frac{1}{2011}< 1.\)
Vậy A < 1.
S=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2010.2011.2012}\)
=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2010.2011}-\frac{1}{2011.2012}\)
=\(\frac{1}{2}-\frac{1}{2011.2012}< \frac{1}{2}\)(Vì \(\frac{1}{2011.2012}>0\))
=> S <\(\frac{1}{2}\)
\(S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{2010.2011.2012}\)
\(S=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2012-2010}{2010.2011.2012}\)
\(S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2010.2011}-\frac{1}{2011.2012}\)
\(S=\frac{1}{1.2}-\frac{1}{2011.2012}=\frac{2023065}{4046132}\)
\(\text{Vì}\)\(\frac{2023065}{4046132}< \frac{1}{2}\Rightarrow S< P\)
\(A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}=\frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
=> \(A< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{2019}=\frac{3}{4}-\frac{1}{2019}=\frac{3}{4}\)
Vậy A<3/4
A< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2018.2019}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
=\(1-\frac{1}{2019}=\frac{2019-1}{2019}=\frac{2018}{2019}\)
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010.2011}\)\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}=1-\frac{1}{2011}=\frac{2010}{2011}>\frac{2010}{2680}=\frac{3}{4}\)
Hình như có gì đó sai sai :')
Ta có A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2011.2011}\)
\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2011.2012}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2012}=1-\frac{1}{2012}=\frac{2011}{2012}>\frac{3}{4}\)
Vậy A > 3/4