Explanation
Given $S = \{0, 1, 2, 3, ...\}$. Also, $\log_e y = x \log_e (\frac{2}{5})$.
This implies $y = (\frac{2}{5})^x$.
Since $x \in S$, $x$ can take values $0, 1, 2, 3, ...$.
The required sum is $1 + (\frac{2}{5})^1 + (\frac{2}{5})^2 + (\frac{2}{5})^3 + ... = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$.