Suppose that we have the following relationship:
$\sum_1^\infty\log(1-x_i) = \log(y)$
where $y\in(0,1)$ and $(x_i)_{i=1}^\infty$ $\subset(0,1)$ is a sequence. If we want the find the elements of $(x_i)$, would we do a Taylor series expansion for $\log(1-x_i)$ and $\log(y)$?
Could we construct what $(x_i)$ will look like from the above equality?
Take any convergent series such that $\sum_{n=1}^\infty s_n=S$ with $s_n<0$. Define $x_n=1-e^{s_n}$ and $y=e^S$. It is easy to verify that this choice satisfies the relationship $\sum_{n=1}^\infty\log(1-x_n)=\log y$.