Taylor expansion of logarithm function. Expand $f(x) = \log(1 + x)$ around $x = 0$ to all orders. More precisely, find $a_n$ such that for any positive integer $N$, we have$$f(x) = \left(\sum_{n=0}^{N-1} a_nx^n\right) + E_N(x) \text{ for all }\left|x\right| < {1\over2},$$where $\left|E_N(x)\right| \le C_N\left|x\right|^N$ for $\left| x\right| \le 1/2$. How does the constant $C_N$ depend on $N$? How do I see that we have an infinite Taylor expansion$$f(x) = \sum_{n=0}^\infty a_nx^n \text{ for all }\left|x\right| < {1\over2}?$$What is the largest interval of validity of this series representation? Can we extend to $\left|x\right| < 1$ or beyond?

Here is an outline of an approach:
First, show (induction works well) that the bound on $E_n$ gives
$a_n = {1 \over n!} f^{(n)}(0)$. In particular, the $a_n$ are unique, so we 
can use any technique to find them.
Note that since $f$ is only defined on $(-1,\infty)$ (in particular, $\lim_{x \downarrow -1} f(x) = -\infty$), the largest radius of convergence can be is bounded by $1$ (since $|0-(-1)| = 1$).
Second, note that $f'(x) = {1 \over 1+x}$, and note that if $|x|<1$, then
$f'(x) = 1-x+x^2-x^3+\cdots$. Furthermore, convergence is uniform on any
compact subset of $(-1,1)$ (Weierstrass M-test), and hence we can exchange
integration and summation to see that
$f(x) = x-{1 \over 2}x^2+{1 \over 3} x^3-{{1 \over 4} x^4 + \cdots}$. It follows from this that
$a_0 = 0, a_n  = (-1)^{n+1}{1 \over n}$, for $n>0$.
Then $$E_N(x) = \sum_{n=N}^\infty (-1)^{n+1}{1 \over n} x^n = x^N \sum_{n=N}^\infty (-1)^{n+1}{1 \over n} x^{n-N} = x^N \sum_{n=0}^\infty (-1)^{n+1+N}{1 \over {n+N}} x^{n}$$
It follows that $C_N = \sup_{|x|< {1 \over 2}} |\sum_{n=0}^\infty (-1)^{n+1+N}{1 \over {n+N}} x^{n}| = \sum_{n=0}^\infty {1 \over {n+N}} {1 \over 2^{n}}$. An immediate closed form expression for $C_N$ is not clear to me,
but it is easy to see the estimate $C_N \le {2 \over N}$.
In fact, if we take $|x| < R$, where $R < 1$, we can form a bound
$C_N \le {K \over N}$, where $K$ is independent of $N$. In particular,
we have
$\sup_{|x|<R} |E_N(x)| \le {K \over N} R^N$
Hence we see that for any $R<1$, we have
$\lim_{N \to \infty} \sup_{|x|<R} |E_N(x)| = 0$, and hence the Taylor series approximation converges uniformly to $f(x)$ for $|x|<R$.
