Problem
Suppose you know that $$f^n(4)=\frac{(-1)^nn!}{3^n(n+1)}$$ and the Taylor series of $f$ centered at $4$ converges to $f(x)$ for all $x$ in the interval of convergence. Show that the fifth-degree Taylor polynomial approximates $f(5)$ with error less than $0.0002$.
I can prove the statement but in order to do so I have to assume that the maximum value of $|f^6(x)|$ on $3\leq x\leq 5$ is at $x=4$. Is there something I am missing where I can conclude the maximum value is at $x=4$ based on the information the problem provides?
You can estimate it directly as follows:
$$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(4)}{n!}(x-4)^n = \sum_{n=0}^{\infty}\frac{(-1)^n}{3^n(n+1)}(x-4)^n$$
has a radius of convergence of $3$.
So,
$$f(5) = \sum_{n=0}^{\infty}\frac{(-1)^n}{3^n(n+1)}$$
The searched for difference to the corresponding value of the $5$th- degree Taylor polynomial is
$$\left|f(5) - \sum_{n=0}^{\color{blue}{5}}\frac{(-1)^n}{3^n(n+1)}\right| = \left|\sum_{n=\color{blue}{6}}^{\infty}\frac{(-1)^n}{3^n(n+1)}\right|$$ $$= \frac{1}{3^6}\left|\sum^{\infty}_{n=6}\frac{(-1)^n}{3^{n-\color{blue}{6}}(n+1)}\right| = \frac{1}{3^6}\left|\sum^{\infty}_{n=\color{blue}{0}}\frac{(-1)^n}{3^n(n+\color{blue}{7})}\right| \leq \ldots$$
Now note that the last series is an alternating one ($S = \sum_{n=0}^{\infty}(-1)^na_n$) with decreasing $a_n \geq 0$. So, you can always estimate $S \leq S_n + a_{n+1}$ where $S_n =\sum_{k=0}^{n}(-1)^ka_k$:
Hence, $$\ldots \leq \frac{1}{3^6}\left( \underbrace{\left(\frac{1}{7} - \frac{1}{3\cdot 8}\right)}_{=S_2} + \underbrace{\frac{1}{3^2\cdot 9}}_{= a_3}\right) \boxed{< 0.0002}$$