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Reading an solution I get stuck at this step.

$$ \frac{1}{2i} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \left( e^{2in} - e^{-2in} \right) \\ = - \frac{1}{2i} \left( \ln(1 + e^{2i}) - \ln(1- e^{-2i}) \right) $$

Could someone explain to me what is done here? How can $\sum$ be replace with $\ln$? What is really done here? Is it possible to divide this into more steps?

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    $\begingroup$ $$\forall |z|<1 : -\ln(1+z)=\sum_{n=1}^\infty\frac{(-1)^n}{n}z^n$$ $\endgroup$ Oct 4, 2014 at 6:41
  • $\begingroup$ Where does this come from? $\endgroup$
    – iveqy
    Oct 4, 2014 at 6:44
  • $\begingroup$ It's a Taylor series expansion. en.wikipedia.org/wiki/Taylor_series $\endgroup$ Oct 4, 2014 at 6:47

1 Answer 1

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Hints:

$$\ln (1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{n}}{n} $$

Where $x\in (-1,1]$

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  • $\begingroup$ this is not different from what ethan has written in his comments.... please consider editing this.. $\endgroup$
    – user87543
    Oct 4, 2014 at 6:50
  • $\begingroup$ @PraphullaKoushik: I've edited it. $\endgroup$
    – Paul
    Oct 4, 2014 at 6:55
  • $\begingroup$ What have you edited? It looks same. $\endgroup$
    – user170039
    Oct 4, 2014 at 7:06

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