$\eta(1) = \ln(2)$ proof using Abel's Theorem

Question

Hi I was just wondering how does one justify $\eta(1) = \ln(2)$. Looking at the power series for $\ln(1+x)$ we have

\begin{equation} \ln(1+x)= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^{n}}{n} \end{equation}

This expansion is only valid for $|x| <1$. Can we use Abels theorem here? How would I go about this?

$\log(1+x)$ is a continuous function in a neighbourhood of $x=1$, hence $\eta(1)=\log(2)$ by Abel's lemma. — Jack D'Aurizio, Jul 02 '15 at 15:58
Due to the functions convergence, we can say that the $\lim _ {x \rightarrow 1^{-}} \frac{(-1)^{n+1}x^{n}}{n} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}?$ — Sean, Jul 02 '15 at 15:59

score 3 · Answer 1 · answered Aug 20 '17 at 20:42

By the alternating series test, we can see that the sum converges and

$$\frac12<\sum_{n=1}^\infty\frac{(-1)^{n+1}}n<1$$

By Abel's theorem,

$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\lim_{x\to1^-}\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}n=\lim_{x\to1^-}\ln(1+x)=\ln(2)$$

1 Answers1