Ta có:
$\dfrac{n}{n+2} = \dfrac{n+2-2}{n+2} = 1 - \dfrac{2}{n+2}$
$\dfrac{n+6}{n+8} = \dfrac{n+8 - 2}{n+8} = 1 - \dfrac{2}{n+8}$
Vì $\dfrac{2}{n+2} > \dfrac{2}{n+8}$
$⇒ \dfrac{n}{n+2} < \dfrac{n+6}{n+8}$
$c$) Ta có:
$\dfrac{1}{2^2} + \dfrac{1}{3^2} + .... + \dfrac{1}{2020^2} < \dfrac{1}{1.2} + \dfrac{1}{2.3} + .... + \dfrac{1}{2019.2020}$
$⇒ \dfrac{1}{2^2} + \dfrac{1}{3^2} + .... + \dfrac{1}{2020^2} < 1- \dfrac{1}{2019} < 1$
$⇒$ $\dfrac{1}{2^2} + \dfrac{1}{3^2} + .... + \dfrac{1}{2020^2} < \dfrac{1}{1.2} + \dfrac{1}{2.3} + .... + \dfrac{1}{2019.2020} < 1$
