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

Community
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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

3 Answers3

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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.

Jonas Meyer
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  • Logarithm is for sure continuous but function term serieses aren't always. The Fourier series of the function $f: \mathbb{R} \rightarrow \mathbb{C}, , f(x) = \sum_{k=-\infty}^\infty \chi_{[-L/4,L/4]} , (x-kL)$ is a counterexample. It is continuous on the interval $(-L/4,L/4)$ but has a discountinuity at $\pm L/4$. – Juha-Matti Vihtanen Jan 26 '15 at 16:52
  • @Juha-Matti: I don't see how your comment is relevant to this Q&A. – Jonas Meyer Jan 26 '15 at 16:54
  • I try to say, that the left hand side is continuous, but the right hand side is not necessarily. – Juha-Matti Vihtanen Jan 26 '15 at 16:56
  • @Juha-Matti: Right-hand side of what? If you have an equation, both sides are equal. The example you gave is of a discontinuous function, so the left-hand side and right-hand side are discontinuous. But more generally you will not find a counterexample relevant to this question, because Abel's theorem tells us that when convergence of a power series on the boundary occurs, the extension to the boundary is continuous. – Jonas Meyer Jan 26 '15 at 16:59
  • The right hand side $\sum_{n=1} ^\infty \frac{(-1)^{n+1}}{n}z^n$ of the equation, that has it as its right hand side. I remember only that Abel's theorem said that if a power series with center $a$ converges at $z_0$ then it converges on $|z-a| < |z_0-a|$ and if it diverges at $z_1$ then it diverges on $|z-a| > |z_1-a|$. It was taught to me that convergence on the boundary has to be analyzed separately. The function $f$ is continuous on $[-L/4,L/4]$. – Juha-Matti Vihtanen Jan 26 '15 at 17:16
  • @Juha-Matti: I think you should look up Abel's theorem rather than rely on memory. There is a link in my answer to a page where you can read on it. You have included some true statements in your comment, but they are not supporting any point really. – Jonas Meyer Jan 26 '15 at 17:20
  • I'm sorry about the confusion. You referred to Abel's second theorem when I referred to Abel's first theorem. – Juha-Matti Vihtanen Jan 26 '15 at 18:49
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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: http://en.wikipedia.org/wiki/Dirichlet_test –  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 :-). –  Nov 24 '10 at 22:52
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Here is a way of seeing this result directly without using Abel's theorem. We have $$ \log(1+x)=\int_{1}^{x+1}\frac{1}{t} \, dt=\int_{0}^{x}\frac{1}{t+1} \, dt \, . $$ Using the fact that for all $n\in\mathbb N$, $$ \frac{1}{1+t}=1-t+t^2-t^3+\dots+(-1)^n\frac{t^n}{1+t} \, , $$ we obtain the following expression for $\log(1+x)$, with the remainder term expressed as an integral: $$ \int_{0}^{x}\frac{1}{1+t} \, dt=x-\frac{x^2}{2!}+\frac{x^3}{3!}-\frac{x^4}{4!}+\dots+(-1)^{n-1}\frac{x^{n-1}}{(n-1)!}+(-1)^n\int_{0}^{x}\frac{t^n}{1+t} \, dt $$ Now note that $$ 0\le\int_0^x\frac{t^n}{1+t} \, dt\le\int_{0}^{x}t^{n-1} \, dt=\frac{x^n}{n} $$ and so in the case $x=1$, the remainder term tends to $0$ as $n\to\infty$. Hence the power series for $\log2$ does indeed converge to $\log 2$.

Joe
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