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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.

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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.

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    $\begingroup$ Why the downvote? The OP indicates that they know the series for $\log(1+x)$, so this seems a good answer. $\endgroup$
    – robjohn
    Mar 18 '15 at 10:09
  • $\begingroup$ Doh, I'm an idiot and sleep deprived. Thanks :). gave this one an upvote and the other one the check $\endgroup$
    – Alan
    Mar 18 '15 at 10:56
  • $\begingroup$ @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 :-) $\endgroup$ Mar 18 '15 at 11:00
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Using the Taylor expansion for $$\log (1-z).$$ So $$\log \frac{1}{1-z}=-\log (1-z)=-\left(-\sum_{k=1}^{+\infty}\frac{z^k}{k} \right)=\sum _{{k=1}}^{+\infty}\frac{z^k}{k}.$$

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

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

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

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