It converges by the alternating series test. It's like the Maclaurin series for $\ln{(1+x)}$ except it has $(1-s^k)$ in the denominator. (The question arose from this.)
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$$F(r,s)=\sum_{k=1}^\infty\frac{(-1)^{k+1}r^k}{k(1-s^k)}, \text{where $0<r<1$ and $0<s<1$}$$ Use $\frac{1}{1-s^k}=\sum_{j=0}^{\infty}s^{kj}$ $$F(r,s)=\sum_{j=0}^{\infty}\sum_{k=1}^{\infty} (-1)^{k+1}\frac{(rs^j)^{k}}{k}.$$ $$\implies \sum_{j=0}^{\infty} \log[1+rs^j]=\log[\Pi_{j=0}^{\infty}(1+rs^j)]$$ So finally, we have $$F(r,s)=\log[(1+r)(1+rs)(1+rs^2)...]=\log \text{q-Pochhammer symbol}(-r,s)_ \infty.$$
