Let $P_n(x)$ be the $n$-th Taylor polynomial about $x = 0$ for the exponential function $e^x$. Find the least $n$ such that $\left|e − P_n(1)\right| < 10^{-5}$.
My answer is $9$. I used $e-P_n(1)=\frac{e^a}{(n+1)!}$ with some $a \in [0,1]$ where I put $n=9$. I just want to confirm whether it is correct.
If it's not correct, then any hints/solution will be appreciated.
We have that $$0<a_n:=e-P_n(1)=e-\sum_{k=0}^n\frac{1}{k!}$$ and therefore $a_n$ is a strictly decreasing sequence which tend to $0$. It turns out that $$a_8< 2.718282-\frac{109601}{403200}<10^{-5}\quad\mbox{and}\quad a_7> 2.71828-\frac{685}{252}> 0.2 \cdot 10^{-4}$$ where we used the fact that $2.71828< e<2.718282$ (see HERE).