I know the series $$\sum_{k=1}^\infty \frac{(\frac{1}{10})^{k+1}}{k(k+1)}$$ is equal to $$\frac{1}{10}-\frac{9}{10}\ln \Big(\frac{10}{9}\Big)$$ but how would I use the ratio test to show this is true?
I also know that: $$ (1-x)\ln(1-x)=\sum^\infty_{k=1}\frac{x^{k+1}}{k}-\sum^\infty_{k=1}\frac{x^k}{k} =\sum_{k=1}^\infty \frac{x^{k+1}}{k(k+1)}$$