Cho hàm số $f\left(x \right)=\ln \dfrac{2020x}{x+1}.$ Tính tổng $S={{f}^{'}}\left(1 \right)+{{f}^{'}}\left(2 \right)+...+{{f}^{'}}\left(2020 \right)?$

$f\left(x \right)=\ln \dfrac{2020x}{x+1}\Rightarrow f'\left(x \right)=\dfrac{1}{x\left(x+1 \right)}=\dfrac{1}{x}-\dfrac{1}{x+1}$
Khi đó: $S=f'\left(1 \right)+f'\left(2 \right)+...+f'\left(2020 \right)=\sum\limits_{k=1}^{2020}{\left(\dfrac{1}{k}-\dfrac{1}{k+1} \right)=1-\dfrac{1}{2021}=\dfrac{2020}{2021}.}$