Question

N=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)

tính N<1

Answer 1

Ta có 

1/2^2<1/1.2

1/3^2<1/2.3

......

1/2009^2<1/2008.2009

1/2010^2<1/2009.2010

=>1/2^2+1/3^2+...+1/2010^2<1/1.2+1/2.3+....+1/2009.2010

=>N<1/1.2+1/2.3+....+1/2009.2010

=>N<1-1/2010

=>N<2009/2010<1

Vậy N<1

Answer 2

\(N=\) \(\frac{1}{2^2}\) \(+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2008.2009}+\frac{1}{2009.2010}\)

\(N< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}+\frac{1}{2009}-\frac{1}{2010}\)

\(N< 1-\frac{1}{2010}\)       

\(N< \frac{2009}{2010}< 1\)

\(\Rightarrow N< 1\)