Ta có :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+............+\dfrac{1}{2012^2}+\dfrac{1}{2013^2}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...............+\dfrac{1}{2011.2012}+\dfrac{1}{2012.2013}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2012}-\dfrac{1}{2013}\)
\(\Rightarrow A< 1-\dfrac{1}{2013}< 1\)
\(\Rightarrow A< 1\rightarrowđpcm\)
Lời giải:
Dễ thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(....\)
\(\dfrac{1}{2012^2}=\dfrac{1}{2012.2012}< \dfrac{1}{2011.2012}\)
\(\dfrac{1}{2013^2}=\dfrac{1}{2013.2013}< \dfrac{1}{2012.2013}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}+\dfrac{1}{2012.2013}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2012}-\dfrac{1}{2013}\)
\(\Rightarrow A< 1-\dfrac{1}{2013}\)
\(\Rightarrow A< 1\left(dpcm\right)\)
