`a,` Đặt `A=1/[1.2]+1/[2.3]+1/[3.4]+...+1/[n(n+1)]`
`⇒ A= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/n - 1/[n+1]`
`⇒ A= 1- 1/[n+1]`
`⇒ A < 1` `(Đpcm)`
`b,` Đặt `M=1/[2^2] + 1/[3^2] + 1/[4^2] +..+ 1/[n^2]`
Ta có :
221<1.21
321<2.31
421<3.41
.........
n21<(n−1).n1
⇒M<1.21+2.31+3.41+...+(n−1).n1
⇒M<1−21+21−31+31−41+...+n−11−n1
⇒ M<1−n1<1<2 `(Đpcm)`
Xin hay nhất !
