Taylor series of ln(1/(1-z)) around 0

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.

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.

• Why the downvote? The OP indicates that they know the series for $\log(1+x)$, so this seems a good answer.
– robjohn
Mar 18 '15 at 10:09
• Doh, I'm an idiot and sleep deprived. Thanks :). gave this one an upvote and the other one the check
– Alan
Mar 18 '15 at 10:56
• @Alan. You are not an idiot since you asked the question and, for me, there is no stupid question (stupid answers, yes). If I had be given one dollar each time I missed something obvious, I probably should be a billionaire ! Don't worry. Cheers :-) Mar 18 '15 at 11:00

Using the Taylor expansion for $$\log (1-z).$$ So $$\log \frac{1}{1-z}=-\log (1-z)=-\left(-\sum_{k=1}^{+\infty}\frac{z^k}{k} \right)=\sum _{{k=1}}^{+\infty}\frac{z^k}{k}.$$

Hint : $\displaystyle \int (1-x)^{-1}\,dx =-\log(1-x)$ .Now use the series expansion of $(1-x)^{-1}$.

$$f(z)=\log\frac{1}{1-z} \implies f(0) = 0\\ f'(z) = \frac{1}{1-z} \implies f'(0) = 1\\ f^{(2)}(z) = \frac{1}{(1-z)^2} \implies \frac{f^{(2)}(0)}{2!} = \frac{1}{2}\\ f^{(3)}(z) = \frac{2!}{(1-z)^3} \implies \frac{f^{(3)}(0)}{3!} = \frac{1}{3}\\ \dots\\ f^{(n)}(z) = \frac{(n-1)!}{(1-z)^n} \implies \frac{f^{(n)}(0)}{n!} = \frac{1}{n}\\$$

• $f'(z)$ is not calculated correctly...
– 5xum
Mar 18 '15 at 10:10
• $f'(z) = \frac{1}{\frac{1}{1-z}}\frac{-1}{(1-z)^2}(-1)=\frac{1}{1-z}$ Mar 18 '15 at 10:11
• You're missing one minus somewhere.
– 5xum
Mar 18 '15 at 10:12
• It looks OK to me. I think @5xum misses a minus somewhere. Mar 18 '15 at 10:13
• We have the technology. Mar 18 '15 at 10:40