Is $-\log (1-z) = \sum_{n=1}^{\infty}\frac{z^n}{n}$ for $z \in \mathbb{C}, \|z\|=1, z \neq 1$?

Question

Is

$-\log (1-z) = \sum_{n=1}^{\infty}\frac{z^n}{n}$

for $z \in \mathbb{C}, \|z\|=1, z \neq 1$ ?

In any case, why?

Answer 1

Yes, they are equal. @MisesEnForce showed the convergence. The equality follows from Abel's theorem on power series. http://en.wikipedia.org/wiki/Abel%27s_theorem and the continuity of $\log$.

Example: $- \log 2 = \sum_{n \ge 1} \frac{(-1)^n}{n}$. Try for $z\ne 1$ a root of $1$.

A conclusion from AbeI's theorem: if on at some points $z$ on the circle of convergence $\sum_n a_n z^n$ is convergent and the function $f$ extends holomorphically past that point then the value at $z$ is the sum $\sum_n a_n z^n$

Answer 2

By asking for the equality, I understand that you ask if $-\ln ( 1 -z)$ defined by the convergent series for $|z|<1$ could see its series definition extended to $|z| = 1$ and $z\not = 1$. Right ?

Thanks to Abel's transform, you can show the following assertion :

1) if the partial sums $B_N = \sum_{n=0}^N b_n$ form a bounded sequence (independently of $N$ of course)

2) if $\sum_{n=0}^\infty |a_{n+1} - a_n| < \infty$

3) if $(a_n)_n$ converges to zero

then $S_N = \sum_{n=0}^N a_n b_n$ is a convergent series.

Here $a_n = \frac{1}{n}$ tends to zero indeed, and if $b_n = z^n$ for $|z| = 1$ and $z\not = 1$, then $\sum_{n=0}^N z^n = \frac{1-z^{N+1}}{1-z}$ is bounded indeed in absolute value, by $\frac{2}{|1-z|}$. Finally $|a_{n+1} - a_n | = \frac{1}{n(n+1)} \leq \frac{1}{n^2}$ so that $\sum_{n=0}^\infty |a_{n+1} - a_n| < \infty$

Don't hesitate to tell if you need more info on Abel's transform, or how you prove the assertion that I use from Abel's transform.