Bounding the error for the remainder of $\log(x)$

Question

We are asked to bound the error given by the remainder of the Taylor series of $\log(x)$ about some point $a>0$. Using the remainder as: $$R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x - a)^{n+1} $$ and taking the $n^{th}$ derivative of $\log (x)$ give: $$f^{(n)} = (-1)^{n+1} \frac{(n-1)!}{x^n} $$ we can write the remainder as: $$R_n = (-1)^{n+2}\frac{n!}{\xi^{n+1}}\frac{1}{(n+1)!}(x - a)^{n+1} $$ So to bound this we need to find the maximum of $1/\xi^{n+1}$. My question is how can we bound this remainder? I don't think it works for $\xi < 1$. Do we require $\xi < a$ ? If so then that would solve the problem, but if not I'm not sure. Any help is appreciated thanks!

Answer 1

I think that you have a mix of $a$ and $x$ in your formulae.

The expansion is $$\log(x)=\log(a)+\sum_{n=1}^p \frac{(-1)^{n+1}}{n\, a^n} (x-a)^{n}+O((x-a)^{p+1})$$