Đáp án:+Giải thích các bước giải:
Đặt `A=1-1/2^2- 1/3^3- ...-1/2004^2`
Ta có:
`1=1`
`1/2.3<1/2^2`
`1/3.4<1/3^3`
`.....`
`1/(2003.2004)>1/2004^2`
`=>A>1-1/2.3-1/3.4-...-1/(2004.2005)`
`=>A>1-(1/2*3+1/3*4+...+1/(2004.2005))`
`=>A>1-((3-2)/1.2+(4-3)/2.3+...+(2005-2004)/(2004.2005)`
`=>A>1-(2/2*3- 3/2*3+ 3/3*4- 4/3*4+...+2005/(2004.2005)- 2004/(2004*2005)`
`=>A>1-(1/2-1/3+1/3-1/4+...+1/2004-1/2005)`
`=>A>1-(1/2-1/2005)`
`=>A>1-2003/4010`
`=>A>2007/4010`
`**2007/4010=2007/4010`
`**1/2004=(1*2007)/(2004*2007)=2007/4022028`
Vì `2007/4010>2007/4022028` nên `2007/4010>1/2004`
Mà `A>2007/4010`
`=>A>1/2004`
`=>1-1/2^2- 1/3^3- ...-1/2004^2>1/2004`
