Applying the conclusion of # 4282 can give $$\frac{1}{\binom{n}{2019}}=\frac{2019}{2018}\left(\frac{1}{\binom{n-1}{2018}}-\frac{1}{\binom{n}{2018}}\right)$$

It follows that $$\begin{align*} \sum_{n=2019}^\infty\frac{1}{\binom{n}{2019}}=\ &\displaystyle\frac{2019}{2018}\sum_{n=2019}^{\infty}\left(\frac{1}{\binom{n-1}{2018}}-\frac{1}{\binom{n}{2018}}\right)\\=\ &\frac{2019}{2018}\cdot\frac{1}{\binom{2018}{2018}}\\=\ &\boxed{\frac{2019}{2018}}\end{align*}$$