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  1. The series for $\log (1+x)$ is convergent(by estimating radius of convergence) only with $|x| <1$. Then how is it still true for $x = 1$?

  2. How is it still apparently true when $x$ is a roots of unity other than $-1$? Such as in the answers to this question: Summing up the series $a_{3k}$ where $\log(1-x+x^2) = \sum a_k x^k$

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Perhaps you might add the series $\log \left( 1+x\right) =\displaystyle\sum_{n=1}^{\infty }\dfrac{(-1)^{n-1}x^{n}}{n}$, if that is what you mean. –  Américo Tavares Nov 24 '10 at 22:49
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Knowing the radius of convergence doesn't, by itself, tell you anything about convergence on the boundary; you have to test for that separately. –  Qiaochu Yuan Nov 24 '10 at 23:03

2 Answers 2

Abel's theorem is the key to why you can say that the limit as you approach the boundary comes out to the same as evaluating the series at the boundary. Sivaram's answer addresses why the series converges at those boundary points.

Explicitly, Abel's theorem implies that

$$\lim_{r\nearrow 1}\log(1+zr)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}z^n$$

for each complex number $z$ with $|z|=1$ such that the series converges. By continuity of the logarithm, this will be $\log(1+z)$ in such cases.

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Look up alternating series test or generalized alternating series test (Also known as Dirichlet's test as George pointed out).

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Rather, for roots of unity, Dirichlets' test: en.wikipedia.org/wiki/Dirichlet_test –  user1119 Nov 24 '10 at 22:43
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@George: Some people call in generalized alternating series test. Anyway I have edited it. I don't want to take credit away from Dirichlet :-). –  user17762 Nov 24 '10 at 22:52

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