\(\begin{array}{l}f\left( x \right) = \ln \left( {\frac{{x + 1}}{x}} \right) = \ln \left( {x + 1} \right) - \ln x\\ \Rightarrow f'\left( x \right) = \frac{1}{{x + 1}} - \frac{1}{x}\end{array}\)
Khi đó ta có:
\(\begin{array}{l}S = f'\left( 1 \right) + f'\left( 2 \right) + ... + f'\left( {2019} \right)\\S = \frac{1}{2} - \frac{1}{1} + \frac{1}{3} - \frac{1}{2} + \frac{1}{4} - \frac{1}{3} + ... + \frac{1}{{2020}} - \frac{1}{{2019}}\\S = \frac{1}{{2020}} - 1 = - \frac{{2019}}{{2020}}\end{array}\)
