Solving for $x$ in $2019$

This is an infinite geometric series with initial value $1$ and common ratio $x$ .

$\frac{1}{1-x} = 2019 \\ 1 = 2019 - 2019 x \\ x = \frac{2018}{2019}$

Given that:

$\begin{aligned} 1 + x + x^2 + x^3 + \cdots & = 2019 & \small \color{#3D99F6} \text{Multiply both sides by }x. \\ x + x^2 + x^3 + x^4 + \cdots & = 2019x & \small \color{#3D99F6} \text{Add 1 on both sides.} \\ \color{#3D99F6}1 + x + x^2 + x^3 + \cdots & = 1 + 2019x & \small \color{#3D99F6} \text{Note that } 1 + x + x^2 + x^3 + \cdots = 2019 \\ \color{#3D99F6}2019 & = 1 + 2019x & \small \color{#3D99F6} \text{Rearranging} \\ \implies x & = \boxed {\dfrac {2018}{2019}} \end{aligned}$

$1 + x + x^2 + x^3 +.... = 2019$

$1 + x ( 1 + x + x^2 + ...... ) = 2019$

$1 + 2019x = 2019$ . (From 1st line)

$2019x = 2018$

$\boxed{x = \frac{2018}{2019}}$