How to prove $\ln(x)$ is analytic everywhere?

Question

When approximating $\ln(x)$ using the Taylor polynomials and remainder function, if we want to show that if we instead expand up to $n=\infty$ i.e. the Taylor series, then we need

$$\frac{n!}{x^{n+1}} \leq M \tag{The $n+1$st derivative of $\ln(x)$}$$

so that

$$\lim_{n \to \infty} \frac{M(x-a)^{n+1}}{(n+1)!} = 0$$

over all $x,a$ relevant to the domain of $\ln(x)$, i.e. $a,x>0$.

For example if we were finding this for $\sin(x)$ or $\cos(x)$, their $n+1$st derivatives are always $\leq 1$ so setting $M=1$ allows that limit to approach $0$, showing that they are analytic everywhere. Likewise for $e^x$ we could pick $e^{\max(x,a)}$ and it would basically become a constant that we could pull out of the limit, and then $e^{\max(x,a)} \cdot 0 = 0$. In other words it's easy if $M$ is not a function of $n$ but rather $x$ and $a$ since the limit won't care about it.

However I am having difficulty with $\ln(x)$ because I can't find a good way to show that the limit goes to $0$ for all $x,a$ for $\ln(x)$.

Answer 1

Please note that, if $n\in\mathbb N$, then$$\log^{(n)}(x)=\frac{(-1)^{n-1}(n-1)!}{x^n}$$and that therefore that the $N$^th Taylor polynomial of $\log$ centered at $a$ is$$\log(a)+\sum_{n=1}^N\frac{(-1)^{n-1}}{na^n}(x-a)^n.$$An upper bound of the absolute value of the remainder is$$\frac1{n+1}\left|\frac{x-a}a\right|^n$$and$$\bigl(\forall x\in(0,2a)\bigr):\lim_{n\to\infty}\frac1{n+1}\left|\frac{x-a}a\right|^n=0.$$

Answer 2

The series $$\log(1+x)=\sum_{k=1}^\infty{(-1)^{k-1}\over k}x^k\qquad\bigl(|x|<1\bigr)\ ,$$ obtained from termwise integration of ${1\over 1+x}=\sum_{k=0}^\infty(-x)^k$, exhibits $\log$ as a real analytic function in the interval $\>]{-1},1[\>$. Given an $a>0$ the functional equation of $\log$ allows to transport this fact to a neighborhood of $a$: $$\log(a+y)=\log a+\log\left(1+{y\over a}\right)=\log a+\sum_{k=1}^\infty{(-1)^{k-1}\over k}\left({y\over a}\right)^k\qquad(-a<y<a)\ .$$ Altogether this shows that the function $\log$ is real analytic in all of ${\mathbb R}_{>0}$.