$\begin{array}{l}\quad\dfrac1{2.9}+\dfrac1{3.12}+\dfrac1{4.15}+\ldots+\dfrac1{2020.6063}\\=\dfrac3{3.2.9}+\dfrac3{3.3.12}+\dfrac3{3.4.15}+\ldots+\dfrac3{3.2020.6063}\\=\dfrac3{6.9}+\dfrac3{9.12}+\dfrac3{12.15}+\ldots+\dfrac3{6060.6063}\\=\dfrac16-\dfrac19+\dfrac19-\dfrac1{12}+\dfrac1{12}-\dfrac1{15}+\ldots+\dfrac1{6060}-\dfrac1{6063}\\=\dfrac16-\dfrac1{6063}<\dfrac16\\\text{- Vậy $\dfrac1{2.9}+\dfrac1{3.12}+\dfrac1{4.15}+\ldots+\dfrac1{2020.6063}<\dfrac16$} \end{array}$