Taylor's Polynomial with Lagrange's form of remainder.

Question

Suppose our function be written as $$f(x)=P_n(x)+R_n(x)$$ where $P_n(x)$ is n-th Taylor's Polynomial about $x_0$ and $R_n(x)$ is associated remainder/error term both of which are given as $$P_n(x)= \sum _ { k = 0 } ^ { n } \frac { f ^ { ( k ) } \left( x _ { 0 } \right) } { k ! } \left( x - x _ { 0 } \right) ^ { k }$$ $$R _ { n } ( x ) = \frac { f ^ { ( n + 1 ) } ( \xi ( x ) ) } { ( n + 1 ) ! } \left( x - x _ { 0 } \right) ^ { n + 1 }\quad \text{for } \xi(x)\in [x,x_0]$$

Suppose $\lim\limits_{n\to\infty}R_n(x)=0$ or in other words $\lim\limits_{n\to\infty}P_n(x)=f(x)$

Define $$g(n)=\sup |R_n(x)| = \sup |f(x)-P_n(x)|$$

Can it be proven that $g(n)=\sup |R_n(x)|$ is decreasing function of $n$

Answer 1

Consider $f(x) = 1/(1-x)$ on $[-2,0]$. Expand it at $x =0$, then $$ f^{(n)}(0) = n!(1-x)^{-(n+1)}|_{x=0} = n!, $$ and the Taylor formula is $$ f(x) = \sum_0^n x^j + (1- \xi)^{-(n+2)} x^{n+1}. $$ Now $$ \sup |R_n(x)| = \sup\left| \frac 1{1-x}- \sum_0^n x^j\right| = \sup\left|\frac {1-(1-x^{n+1})}{1-x}\right| = \sup \left|\frac{x^{n+1}} {1-x}\right| =\sup \frac {(-x)^{n+1}} {1-x} \geqslant \frac {2^{n+1}} {1+2} \nearrow +\infty, $$ so the sequence $g(n)$ cannot be decreasing.

UPDATE

Inspired by @zhw., if the interval is $(-1,1)$ instead, then $$ g(n) \geqslant \frac {(1-1/n)^{n+1}}{1 - 1+1/n} \xrightarrow{n\to \infty} +\infty, $$ and $g(n)$ still cannot be decreasing.