I wish to approximate the value of $e$ and show this is accurate to at least the sixth digit. I am wondering if in doing so I need to use the Taylor Series for $e^x$ about $1$ or does it suffice to use $\Sigma_{n=0}^{\infty}\frac{x^n}{n!}$ with $x=1$.
From here I believe I then would need to solve $|R_n(x)|<|\frac{f^{n+1}(c)}{(n+1)!}|=|\frac{3}{(n+1)!}|<.0000005$ which would work for $n=10$. Then just calculate $P_{10}(x)$. I have already shown that $e^x<3$ which is how I bounded the remainder function.
Does this work or should I redo the process with the Taylor Series centered about the value which we wish to approximate?
This works. It was possible, because you know the values of $e^x$ and its derivatives, at 0 (they all equal 1). If you centre the series at 1 instead, then you need to know the value of $e^x$ at $x=1$, which is the very thing that you hope to approximate !