ও
$$=\frac{2\cdot{2019}}{1+2({\frac{1}{2\cdot{3}}}+{\frac{1}{3\cdot{4}}}+..+{\frac{1}{2019\cdot{2020}}})}=$$
$$=\frac{2\cdot{2019}}{1+2\cdot({\frac{1}{2}}-{\frac{1}{2020}})}=$$
$$=\frac{2\cdot{2019}}{1+{\frac{2018}{2020}}=\frac{2\cdot{2019}\cdot{2020}}{2\cdot{2020}}=2020$$
\begin{flushright}
@mathematikaa
\end{flushright}


