Complex analysis, infinite series expansion of a function

I'm studying from my class notes and I don't understand a step. The function $\displaystyle \log (x+1)$ is analytic over $\displaystyle \mathbb{C}-\mathbb{R}^{\leq 1}$.

Furthermore $\displaystyle \log (x+1)=\sum _{n=1}^{+\infty} \frac{ f^{n} (0) z^n}{n!}=\sum _{n=1}^{\infty} \frac{(-1)^{n-1}z^n}{n}$. From this I get that the radius of convergence of the function is 1, ok but my problem is how could they write the function log(x+1) as an infinite series whose "center" is 0 while the function isn't analytic there. I know I'm not expressing myself well. I mean why do they used $\displaystyle f^{n}(0)$ instead of say any number greater than 2 (since the radius of convergence is 1 and the function isn't analytic for any real number lesser or equal than 1).

So it would mean that any function, regardless of analytic or not, can be written in terms of infinite series? It seems to contradict a theorem that I can't find at the moment.

Any help is greatly appreciated, as usual.

Re: Complex analysis, infinite series expansion of a function

Try the old standard trick. We know that log(x+1) can be derived from the geometric series 1/1-(-x)) = sum n=0 to inf (1/(1-(-x)^n)_ (HELP, where are symbols like sums, integrals etc. hiding?) The sum converges absolutely and thus can be integrated term-by-term yielding a new sum which represents -1log(1-x), after a little tiny bit ofalgebra, you're done.

Is there a drop down panel somewhere with necessary symbols to use in posts? They must be available as I see them in others' posts, beautifully expressed! Can someone show me how to generate such expressions as are seen in the original above. It sure would make for easier responses to

folks!