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 $n$