3
$\begingroup$
  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 $\log(1-x+x^2) = \sum a_k x^k$

  • 0
    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.2010-11-24
  • 9
    Knowing 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

2 Answers 2