What degree Maclaurin polynomial of $e^x$ must be taken to guarantee an estimate of $e$ to within $1*10^{-6}$ What degree Maclaurin polynomial of $e^x$ must be taken to guarantee an estimate of $e$ to within $1*10^{-6}$. The Maclaurin polynomial is:
$f(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}...$
So
$f(x)=\Sigma^n_0\frac{1}{k!}$
At this part I applied the error term
$f(x)=\Sigma^n_{k=0}\frac{1}{k!}+1*10^{-6}$
$e-1*10^{-6}=\Sigma^n_{k=0}\frac{1}{k!}$
Solving it I got $n=9$. Howerver that is not the right anwser. In the mark scheme $n=17$. I tried to prove that I am right substituting the values in the error theorem formula.
$R_n$$=\frac{f^{n+1}(c)*x^{n+1}}{(n+1)!}$
$R_9$$=\frac{e}{(10)!}$
$R_9=7.49*10^{-7}$
For $R_{17}$
$R_{17}=7.642*10^{-15}$
However I got that the value of $n$ that most approches $1*10^{-6}$ is $n=8$. Can someone say which anwser is right or solve it? Thanks
 A: The least $n$ such that
$$
    \left|\sum_{k=0}^n \frac1{k!} - e\right| < \frac1{10^6}
$$
is in fact $n=9$.
This can be determined by using the Lagrange formula for the remainder:
$$
   R_k(x) = f^{(k+1)}(\xi) \frac{(x-a)^{k+1}}{(k+1)!}
$$
for some $\xi$ between $a$ and $x$: in this case, $R_9(1) = \frac{e^{\xi}}{10!}$ for some $\xi \in [0,1]$, so in particular $R_9(1) \le \frac{e}{10!} \approx 7.49 \times 10^{-7}$ (choosing $\xi=1$, the value giving the largest error term). On the other hand, $R_8(1) \ge \frac1{9!} \approx 2.75 \times 10^{-6}$ (choosing $\xi=0$, the value giving the smallest error term), which is not good enough.
For this particular function, we could even do better: the error when $n=9$ is 
$$
   \frac1{10!} + \frac1{11!} + \frac1{12!} + \dots = \frac1{10!}\left(1 + \frac1{11} + \frac1{11\cdot 12} + \frac1{11\cdot12 \cdot 13} + \dotsb\right)
$$
and the sum in the parentheses is at least $1$ and at most $1 + \frac1{11} + \frac1{11^2} + \frac1{11^3} + \dots = \frac{11}{10}$. Therefore we have $$\frac{1}{10!} \le R_9(1) \le \frac{11}{10\cdot 10!}$$ 
(placing the error between $2.75 \times 10^{-7}$ and $3.03 \times 10^{-7}$) while, by a similar argument, $$\frac1{9!} \le R_8(1) \le \frac{10}{9\cdot 9!}$$ (placing the error between $2.75 \times 10^{-6}$ and $3.06 \times 10^{-6}$).
Finally, a direct computation shows that $R_9(1) \approx 3.02886\times 10^{-7}$ and another computation shows that $R_8(1) \approx 3.05862 \times 10^{-6}$, so the fancy estimates used above aren't leading us astray.

I guess I should emphasize: in all the approaches above, I'm doing two calculations (an upper bound on $R_9(1)$ and a lower bound on $R_8(1)$) because that's what we need to say that $n=9$ is the least $n$ necessary: we need to show that $n=9$ is enough, and that $n=8$ is not.
