Ta có: n < 1/1.2 + 1/2.3 + 1/3.4 +...+ 1/2008.2009 + 1/2009.2010
n < 1/1-1/2 + 1/2-1/3 + 1/3-1/4 +...+ 1/2008-1/2009 + 1/2009-1/2010 (công thức)
n < 1/1- (1/2-1/2)- (1/3-1/3)-...- (1/2009-1/2009)-1/2010 (quy tắc dấu ngoặc)
n < 1/1 - 1/2010
n < 2009/2010
Vậy n<2009/2010<1
ta có \(N=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}.\)
ta lại có \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2010^2}< \frac{1}{2009.2010}\)
đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)
\(\Rightarrow N< A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...-\frac{1}{2009}+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}< 1\)
hay \(N< 1\left(đpcm\right)\)
