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 > So if $n \to \infty$, then all terms above containing $c^n \to 0$.

This is false. It's true that as $n \to \infty$, $c^n \to 0$ (for $0 < c < 1$). But a term like $\frac{s}{c^n}$ will diverge instead of converging to $0$.

You haven't done yourself any favors by dividing everything by $c^n$. Just use the fact that $c^n \to 0$ (and similarly $w^n \to 0$) with the original expression, go slowly, and you should get the right answer.