What is the remainder of $|e-\sum_{j=0}^n{1\over j!}|$?

Question

I have to find the smallest $n$ such that $|e-\sum_{j=0}^n{1\over j!}|<0.001$, but I want to do it with the remainder. I know that it is ${e^c\over (n+1)!}$ where $0<c<1$, but how do you get that? Furthermore, how, with this $e^c$ standing in the middle, can you find this $n$? Thank you very much.

Answer 1

The remainder term is obtained by expanding the Taylor's polynomial of $\exp$ about $x_0 = 0$ and applying Taylor's Theorem at $x = 1$. If you desire $|\frac{e^c}{(n+1)!}| < 0.001$, you may ask for the sufficient condition $\frac{e}{(n+1)!} < 0.001$, since $c < 1$ and $\exp$ is increasing. By trial and error, $$6! = 720 < \frac{e}{0.001} < 5040 = 7!$$ Therefore, taking $n = 6$ you can make sure the remainder term is small enough.