Having trouble with a Taylor's series error bound problem

Question

Problem: Suppose you know that \begin{equation} f^{(n)}(4)=\dfrac{(-1)^{n}.n!}{3^{n}(n+1)} \end{equation} and the Taylor's series of f centered at 4 converges to $f(x)$ for all x in the interval of convergence. Show that the $5^{th}$ degree Taylor polynomial approximates $f(5)$ with error < 0.0002.
This prob is from Stewart's Calculus, The solution uses Alternating Test Error Estimation,but I have tried to find and upper bound for $|f^{(6)}(x)|$, (i.e M) to use Taylor's inequality and I could only found $M=0.4825$ to obtain $|R_{5}(x)| \leq M\dfrac{(x-4)^{6}}{6!}\leq \dfrac{M}{6!} = 0.00067$, which is > 0.0002.
Someone help me with this? thanks in advanced!

Answer 1

You have to think that the coefficients of the Taylor series (i.e. $\frac{f^{(n)}(a)}{n!}$) are successive refinements to your estimate, like tiny nudges. Stopping at $k=5$ means that the next biggest nudge you'd have done is proportional to: $$\frac{f^{(6)}(4)}{6!}=\frac{(-1)^6\cdot6!}{3^6\cdot(6+1)\cdot6!}=\frac{1}{3^6\cdot7}\approx1.96\cdot10^{-4}$$ which, at $x\in[4,5]$, is smaller than $2\cdot10^{-4}$ and you can take it as upper bound $M$ as needed.