1/(1+1)+2/(1+2)+ 3/(1+3)+…+2019/(1+2019)=

Ta có

$A=\dfrac{1}{2} + \dfrac{2}{3}  + \cdots + \dfrac{2019}{2020} = \left( 1 – \dfrac{1}{2} \right) + \left( 1 – \dfrac{1}{3} \right) + \cdots + \left( 1 – \dfrac{1}{2020} \right)$

$= \underbrace{1 + \cdots + 1}_{2019 \, \text{số 1}} – \left( \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{2020} \right)$

$= 2019 – \left( \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{2020} \right)$

Ta sẽ tính

$S = \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{2020}$