How is $1- \frac{1}{2} +\frac{1}{3}...$ till infinity $= \log_e2$? I know that $e^x = 1 + \frac{1}{1!} + \frac{1}{2!}...$ but how is this result gotten? I don't have a very extensive knowledge of Euler's number, I just need to learn how to do this because it is part of the solution of a question I'm doing for self-study. Would someone please give me a simple explanation?
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6This question has answers here which I hope aren't too complicated Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ – michaelhowes May 28 '18 at 05:44
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1also you can take a look here or search in google for "mercator series" for some more extensive bibliography – Masacroso May 28 '18 at 05:49
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It comes from the Maclaurin series for $\ln(1+x)$: $$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots =\sum_{n=1}^\infty(-1)^{n-1}\frac{x^n}n.$$ This is valid for $-1<x\le1$.
Angina Seng
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The solution uses the Euler formula of complex numbers:
$\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}}{n}=-\sum\limits_{n=1}^\infty\dfrac{e^{in\pi}}{n}$ where $i^2=-1$
Expand the fraction by $i\pi$ and realize the following:
$-i\pi\sum\limits_{n=1}^\infty\dfrac{e^{in\pi}}{i\pi n}=-i\pi\int\sum\limits_{n=1}^\infty e^{in\pi x} dx_{x=1}=-i\pi\int \sum\limits_{n=1}^\infty e^{in\pi x} dx _{x=1}$
Sum the geometic series we get:
$-i\pi\int \dfrac{e^{i\pi x}}{1-e^{i\pi x}}dx$ using the $t=e^{i\pi x}$ substitution and $t=-1$ at $x=1:$
$-\int \frac{1}{1-t}dt= - \big(-ln(1-t)\big)_{t=-1}=ln2$
JV.Stalker
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