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while doing the Integration problem using Limit of a sum approach i have a doubt how $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln2$$

by infinite geometric series we have

$$1-x+x^2-x^3+x^4-x^5+\cdots =\frac{1}{1+x}$$ for $|x| \lt 1$ Integrting both sides we get

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots $$ for $|x| \lt 1$

Since this series is valid only for $|x| \lt 1$ how can we substitute $x=1$ and conclude $$\ln2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots $$

Jack D'Aurizio
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Umesh shankar
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    You should omit the $\infty$-sign at the end of the series. – Peter Apr 29 '16 at 13:12
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    See here; it converges for $-1<x\le 1$. – Dietrich Burde Apr 29 '16 at 13:14
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    @nayrb You only changed the title; the other four $\infty$'s are still there. – Dietrich Burde Apr 29 '16 at 13:20
  • @DietrichBurde I answered this question giving the taylor series, but received only negative feedback. Is Abel's theorem mentioned in the duplicate ? – Peter Apr 29 '16 at 13:21
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    @Peter You didn't show that the Taylor series converges to $\ln 2$ for $x=1$. – David C. Ullrich Apr 29 '16 at 13:22
  • @Peter, yes, see the answer of Akhil Mathew and the comment of Andrés Caicedo. – Dietrich Burde Apr 29 '16 at 13:27
  • According to Andres Caicedo it is when $x \to 1$, the series tends to $ln2$. am i right? – Umesh shankar Apr 29 '16 at 13:46
  • @Umeshshankar The fact that the series tends to $\ln 2$ as $x\to1$ is trivial, just because the logarithm is continuous. This is not the question - the question is why the series actually equals $\ln 2$ when $x$ equals $1$. – David C. Ullrich Apr 29 '16 at 13:55
  • ok actually i am not aware of complex analysis, so unable to understand Abel's Theorem. Can you explain in a different way of how $x=1$ substitution is valid – Umesh shankar Apr 29 '16 at 14:00
  • @Umeshshankar (i) If you want to understand power series you need to learn some complex analysis. (ii) On the other hand, the proof of Abel's theorem at the link in my answer doesn't use any complex analysis. In fact you can get a proof not even involving complex numbers at all, by taking that proof and applying it to a power series that converges on $(-1,1)$. – David C. Ullrich Apr 29 '16 at 14:05

1 Answers1

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Abel's Theorem shows that you can substitute $x=1$.

David C. Ullrich
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    Use Euler's criterion (sequence is decreasing in absolute value and alternating) to see that the series $\sum_n \frac{(-1)^n}{n}$ converges. Then Abel's theorem says that this is indeed the value for $x=1$. – Henno Brandsma Apr 29 '16 at 13:26