Calculate $\sum^\infty _{k=1} \frac{10^{-k-1}}{k(k+1)}$ using the Taylor series of $(1-x)\ln(1-x)$

Question

For the Maclaurin series of $(1-x)\ln(1-x)$ I have obtained $$\sum^\infty_{k=1}\frac{x^{k+1}}{k}-\sum^\infty_{k=1}\frac{x^k}{k}$$ I am unsure on how to use this to obtain the value of the sum: $$ \sum ^\infty _{k=1} \frac{\left (\frac{1}{10} \right )^{k+1}}{k(k+1)}$$

Answer 1

If $|x|<1$, then $$ (1-x)\ln(1-x)=\sum^\infty_{k=1}\frac{x^{k+1}}{k}-\sum^\infty_{k=1}\frac{x^k}{k} =\sum^\infty_{k=1}\frac{x^{k+1}}{k}-\sum^\infty_{k=0}\frac{x^{k+1}}{k+1}= \sum_{k=1}^\infty\frac{x^{k+1}}{k(k+1)}-x $$ Hence $$ \sum_{k=1}^\infty \frac{10^{-k-1}}{k(k+1)}=\frac{1}{10}+\frac{9}{10}\ln \Big(\frac{9}{10}\Big) $$