I have an function $f$ with a Taylor expansion $f(x) = \sum_{k=0}^\infty a_k x^k$.
Suppose I write the function in terms of its remainder: $$f(x) = P_{N}(x) + R_{N+1}(x)$$ where $P_N$ contains all the terms in the Taylor series of $O(x^N)$ and below, and remainder $R_{N+1}(x)$ contains all terms above.
Suppose I have an independently obtained bound $|f(x)| \leq \sum_{k=0}^\infty |b_k| x^k$ for all $x>0$.
Can I prove a bound of the form:
$$ |R_{N+1}(x)| \leq \sum_{k=N+1}^\infty |b_k| x^k,$$
(essentially, does the remainder of my upper bound always bound the remainder of the original function)?
Let's try $f(x) = \sin x$. Check these details:
The Maclaurin series is $$ f(x) = x-\frac{1}{6}x^3+O(x^5) $$ It is known that $$ |f(x)| \le 1\qquad\text{for all } x . $$ So we can use $b_0=1$ with $b_k = 0$ for all $k \ge 1$ to achieve $$ |f(x)| \le |b_0|+|b_1| x + |b_2| x^2 + \dots\qquad \text{for all } x. $$ Consider $P_2(x) = x$ and $R_3(x) = \sin x - x$. However, we do not have inequality $|P_3(x)| \le |b_3|x^3+\dots$ That is, we do not have $|\sin x - x| = 0$.
In fact, we do not even have $|R_3(x)| \le 1$.