`3/(1^2 . 2^2)+5/(2^2 . 3^2)+7/(3^2 . 4^2)+...+19/(9^2 . 10^2)`
`=(2^2 - 1^2 )/(1^2 . 2^2)+(3^2 -2^2)/(2^2 . 3^2)+(4^2 -3^2)/(3^2 . 4^2)+...+(10^2-9^2)`
`=(1/(1^2)-1/(2^2))+(1/(2^2)-1/(3^2))+(1/(3^2)+1/(4^2))+...+(1/(9^2)-1/(10^2))`
`=1/(1^2) - 1/(2^2) + 1/(2^2) + 1/(3^2) - 1/(3^2) + ... + 1/(9^2) - 1/(10^2)`
`=1/(1^2)-1/(10^2)=1-1/(10^2)=1-1/100`
Ta có : `1-1/100<1` nên `3/(1^2 . 2^2)+5/(2^2 . 3^2)+7/(3^2 . 4^2)+...+19/(9^2 . 10^2)<1`
`->đpcm`