One more taylor/maclurian series problem to which I know the answer of, I just have no idea how to get there (This is as a formal power series, so convergence is not an issue)
$$\log \left(\frac 1 {1-z}\right)=\sum _{k=1}^\infty \frac 1 kz^k$$.
I've tried playing around with rewriting $\frac 1 {1-z}$ as $1+\frac z {1-z}$ and using the taylor expansion for $\log (1+z)$, but I can't seem to figure out what to do with all those powers in the denominator that show up.
Best Answer
Hint
$$\log \left(\frac 1 {1-z}\right)=-\log(1-z)$$ Now consider the series for $\log(1+x)$ and make $x=-z$ in the result and you will be done.