Gọi biểu thức trên là A.
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}.\)
Ta thấy:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}.\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}.\)
..................
\(\dfrac{1}{2013^2}=\dfrac{1}{2012.2013}.\)
\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}\)
\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}.\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2012}-\dfrac{1}{2013}.\)
\(=1-\dfrac{1}{2013}.\)
\(< 1\left(đpcm\right).\)
Vậy \(A< 1.\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2012\cdot2013}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2012}-\dfrac{1}{2013}\\ =1-\dfrac{1}{2013}\\ < 1\)
Vậy ...
