The series for $\log (1+x)$ is convergent(by estimating radius of convergence) only with $|x| <1$. Then how is it still true for $x = 1$?
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 $\log(1-x+x^2) = \sum a_k x^k$
Convergence for log 2
3
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real-analysis
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0Perhaps 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. – 2010-11-24
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9Knowing the radius of convergence doesn't, by itself, tell you anything about convergence on the boundary; you have to test for that separately. – 2010-11-24