Lê Quốc Trần Anh
05/06/2017 at 10:16-
We have :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{2017^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{2016.2017}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{2016}-\dfrac{1}{2017}=1-\dfrac{1}{2017}< 1\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{2017^2}< 1\)
Lê Quốc Trần Anh selected this answer. -
FA KAKALOTS 28/01/2018 at 22:15e have :
122+132+.....+120172<11.2+12.3+.....+12016.2017
=11−12+12−13+....+12016−12017=1−12017<1
⇒122+132+......+120172<1
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Denote \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2017^2}\)
We have :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{2016^2}< \dfrac{1}{2015.2016};\dfrac{1}{2017^2}< \dfrac{1}{2016.2017}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}+\dfrac{1}{2016.2017}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2016}-\dfrac{1}{2017}\)
\(=1-\dfrac{1}{2017}< 1\)
So A < 1