I have got Abel's Theorem in this form: If a power series converges at one of the ends of the partition of convergence, its sum is continuous at this point (one-sided).
And I have got an example
$\ln (1+x)=x-\frac{x^2}{2}+...+(-1)^{n+1} \frac{x^n}{n}+...$ then
$\ln2=1-\frac{1}{2}+ \frac{1}{3}-...+(-1)^{n+1}\frac{1}{n}+...$
How is this theorem used in this example?
On
Consider the "artificial function" $$s(x):=\sum_{k=1}^\infty {(-1)^{k-1}\over k}x^k=x-{x^2\over2}+\ldots\qquad(-1<x<1)\ .$$Since the series converges also for $x=1$ we can extend the domain of $s(\cdot)$ to $\ ]{-1},1]$ by putting $$s(1):=\sum_{k=1}^\infty {(-1)^{k-1}\over k}\ .$$ Abel's theorem then tells us that $s(\cdot)$ is continuous at $1$, which implies $$\sum_{k=1}^\infty {(-1)^{k-1}\over k}=s(1)=\lim_{x\to 1-}s(x)=\lim_{x\to1-}\log(1+x)=\log 2\ ,$$ because we know from other sources that $\log$ is continuous at $2$.
The power series
$$f(x) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n$$
converges for $x=1$ (by Leibinz' alternating series criterion, for example). Hence by Abel's theorem, we have
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} = \lim_{x\to 1^-} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n = \lim_{x\to 1^-} \ln (1+x) = \ln 2.$$
Abel's theorem gives the first equality above, which allows to determine the sum of the alternating harmonic series. It is not a priori clear that the equality holds, since the uniform convergence of the series which is used to deduce the continuity of the sum is only shown to hold on compact subintervals of the open interval of convergence by the usual convergence tests (root or ratio test, typically), and thus these yield only the continuity on the open interval of convergence. Abel's theorem extends that to the endpoints of the interval where the series still converges.