[tex]\dfrac{1}{3^n}-\dfrac{1}{3^{n+1}} = \dfrac{1}{3^n}-\dfrac{1}{3^n\cdot 3}=\\ \\ =\dfrac{1}{3^n}\cdot \left(1-\dfrac{1}{3}\right) = \dfrac{1}{3^n}\cdot \dfrac{3-1}{3}=\\ \\ = \dfrac{1}{3^n}\cdot \dfrac{2}{3} = 2\cdot \dfrac{1}{3^{n+1}} = 2\cdot \left(\dfrac{1}{3}\right)^{n+1}\\ \\ \Rightarrow n+1 = 2019 \Rightarrow \boxed{n = 2018}[/tex]
