`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 :
$\dfrac{1}{2^2}< \dfrac{1}{1.2}$
$\dfrac{1}{3^2}< \dfrac{1}{2.3}$
$\dfrac{1}{4^2}< \dfrac{1}{3.4}$
$.........$
$\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right).n}$
$⇒ M < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}$
$⇒ M< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}$
$⇒ M< 1-\dfrac{1}{n}< 1 < 2$ `(Đpcm)`
Xin hay nhất !