\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2012^2}+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}+\dfrac{1}{2012.2013}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2012}-\dfrac{1}{2013}\)
\(=1-\dfrac{1}{2013}\)
\(\Rightarrow A< 1-\dfrac{1}{2013}\)
\(\Rightarrow A< 1\) ( đpcm )
mình gợi ý nè :
Chứng minh A <\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2012^2}\) + \(\dfrac{1}{2013^2}\)
Vì \(\dfrac{1}{2^2}\) < \(\dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\)
...
\(\dfrac{1}{2012^2}\) < \(\dfrac{1}{2011.2012}\)
\(\dfrac{1}{2013^2}\) < \(\dfrac{1}{2012.2013}\)
\(\Rightarrow\) \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2012^2}\) + \(\dfrac{1}{2013^2}\) < \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ... + \(\dfrac{1}{2011.2012}\) +
+ \(\dfrac{1}{2012.2013}\)
Hay A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2012}-\dfrac{1}{2013}\)
A < \(1-\dfrac{1}{2013}\)
A < \(\dfrac{2012}{2013}\)
Mà \(\dfrac{2012}{2013}\) < 1
\(\Rightarrow\) A < \(\dfrac{2012}{2013}\) < 1
Hay A < 1
Vậy A < 1
1/2^2<1/1*2=1-1/2
1/3^2<1/2*3=1/2-1/3
...
1/2012^2<1/2011*2012=1/2011-1/2012
1/2013^2<1/2012*2013=1/2012-1/2013
suy ra 1/2^2+1/3^2+..+1/2012^2+1/2013^2<1-1/2+1/2-1/3+..+1/2011-1/2012+1/2012-1/2013=1-1/2013<1
suy ra A<1
Vậy A<1
