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?

We know that $\text{log}$ is the inverse function to $g(y) = e^y$. We have that $g$ takes only positive values and is continuous, so $g$ is bounded away from $0$ on $[-T, T]$ for all $T > 0$. Consequently, the inverse function $\log$ must be unbounded in any neighborhood of $0$, and so the power series for $\log(1 + x)$ can not converge to a continuous function on any neighborhood of $x = -1$. Thus, our best hope is to have a series converging to $\log(1 + x)$ for $\left|x\right| < 1$. In fact, it will be continuous on $(-1, 1]$, though we will not worry about $x = 1$.
Now, since$$g'(y) = g(y) > 0$$for all $y$, we have that $g$ is strictly increasing, so if $g(y_n) \to g(y_0)$, we must have $y_n \to y_0$. Thus, if $g(y_0) = x_0$, $h = g^{-1}$, and $g(y_n) = x_n$ with $y_n \to y_0$, we have$$\lim_{n \to \infty} \left({{h(x_n) - h(x_0)}\over{x_n - x_0}}\right)\left({{g(y_n) - g(y_0)}\over{y_n - y_0}}\right) = 1.$$Since$${{g(y_n) - g(y_0)}\over{y_n - y_0}} \to g(y) \neq 0,$$we get$$\lim_{n \to \infty} {{h(x_0) - h(x_0)}\over{x_n - x_0}} = {1\over{g(y)}} = {1\over{x_0}}.$$This is true for any sequence $x_n \to x_0$, so we conclude $g'(x_0) = 1/x_0$. It is clear that $g(y)  \to \infty$ as $y \to \infty$, so $g(-y) \to 0$ as $y \to \infty$ by the functional equation. But $g(y) >0$ for all $y \in \mathbb{R}$, so by the Intermediate Value Theorem, we conclude $g$ has range $(0, \infty)$, and $g$ is injective as $g' > 0$ and the Mean Value Theorem imply $g$ is $1$-$1$. So $h = g'$ is defined on $(0, \infty)$, and we see $h'(x) = 1/x$ for all $x$. We have that $h(1) = 0$, so$$h(x) = \int_1^x {{dt}\over{t}}.$$Thus,$$f(x) = \int_1^x {{dt}\over{1 + t}}.$$Suppose $\left|x\right| < 1$. Then on $[-\left|x\right|,\left|x\right|]$, the power series for $1/(1 + t)$ converges uniformly, so we may interchange limits, as follows:$$f(x) = \left(\int_0^x \left(\lim_{n \to \infty} \sum_{k=0}^n (-t)^k\right)dt\right) + f(0)$$$$= \left(\lim_{n \to \infty} \int_0^x \left(\sum_{k=0}^n (-t)^k\right) dt\right) + 0$$$$= \lim_{n \to \infty} \left(-{{\sum_{k=1}^{n+1} (-x)^k}\over{k}}\right)$$$$= -\sum_{k=1}^\infty (-x)^k,$$and the series converges to $\log(x)$ for $\left|x\right| < 1$. Note that uniform convergence was needed to change the order of the limit and the integral.
