5.3. Find $\frac{S_{1}}{S_{2}}$, where

$S_{1}=\frac{1}{3^{2019}}+\frac{1}{3^{2018}}-\frac{1}{3^{2017}}+\ldots+\frac{1}{3^{3}}+\frac{1}{3^{2}}-\frac{1}{3}, \quad S_{2}=\frac{1}{3}+\frac{1}{3^{2}}-\frac{1}{3^{3}}+\ldots+\frac{1}{3^{2017}}+\frac{1}{3^{2018}}-\frac{1}{3^{2019}}$

(in both sums, the signs of the terms alternate as follows:,,,,,,,,,$++-++-++- \ldots)$
Answer. $-\frac{5}{11} \approx-0.45$.