Giải thích các bước giải:
Ta có:
$\dfrac{1}{n(n+1)}=\dfrac{n+1-n}{n(n+1)}=\dfrac{n+1}{n(n+1)}-\dfrac{n}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$
Mà:
$P=\dfrac{1}{2^2}+\dfrac{1}{3^2}+..+\dfrac{1}{2020^2}$
$\rightarrow P<\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+..+\dfrac{1}{2020^2}$
$\rightarrow P<\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+..+\dfrac{1}{2020.2020}$
$\rightarrow P<\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}..+\dfrac{1}{2019.2020}$
$\rightarrow P<\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}..+\dfrac{1}{2019}-\dfrac{1}{2020}$
$\rightarrow P<\dfrac{1}{2}-\dfrac{1}{2020}<\dfrac{1}{2}<\dfrac{3}{4}$
$\rightarrow đpcm$
