Rate and order of convergence of $\sum_{k=0}^n \frac{x^k}{k!}$ to $\operatorname{exp}(x)$ as $n \rightarrow \infty$ Since $\lim_{n\rightarrow\infty} \sum_{k=0}^n \frac{x^k}{k!} = \operatorname{exp}(x)$, I was wondering how fast this series converges to $\operatorname{exp}(x)$. I'm suggesting that this series converges linearly to its limit. However, I'm struggling with it's proof.
We defined the rate and order of convergence as the constants $\mu$ and $\rho$, respectively by $$\lim_{n \rightarrow \infty}\frac{|\bar x-x_{n+1}|}{|\bar x - x_{n}|^\rho}=\mu>0,$$
where $\bar x$ denotes the limit of $x_n$.
With this definition, it is clear that in this case, the rate and order of convergence can be evaluated by considering the remainders $R_{n+1}(x)$ and $R_n(x)$ of the exponential series. Thus we get $$\lim_{n \rightarrow \infty}\frac{|R_{n+1}(x)|}{|R_n(x)|^\rho}=\mu.$$ Now, I'm supposing that the order of convergence should be $1$ which simplifies the equation to $$\lim_{n \rightarrow \infty}\frac{R_{n+1}(x)}{R_n(x)}=\mu.$$ My problem is, that I'm unable to proof that $lim_{n\rightarrow\infty}\frac{R_{n+1}(x)}{R_n(x)} \in (0,\infty)$.
Thanks for all your help!
 A: The order of convergence cannot be $1$ when $x$ is positive. Indeed, 
when $x > 0$  one can write
$$
R_{n+1}(x)=\sum_{k=n+2}^{\infty} \frac{x^k}{k!}
=\sum_{k=n+2}^{\infty} \frac{x^{k-1}}{(k-1)!}\frac{x}{k}
\leq \sum_{k=n+2}^{\infty} \frac{x^{k-1}}{(k-1)!}\frac{x}{n+2}=
\frac{x}{n+2} R_n(x).
$$
So $0 \leq \frac{R_{n+1}(x)}{R_n(x)} \leq \frac{x}{n+2}$, and hence $(\frac{R_{n+1}(x)}{R_n(x)}) \to 0$.
A: This sequence doesn't play well with the notion of order of convergence since the remainder behaves essentially like a factorial.
The integral form of the remainder of the Taylor series is
$$
R_n(x) = \frac{1}{n!} \int_0^x e^t (x-t)^n\,dt.
$$
When $0 \leq t \leq x$ we have $1 \leq e^t \leq e^x$, so that
$$
\frac{x^{n+1}}{(n+1)!} \leq R_n(x) \leq \frac{x^{n+1} e^x}{(n+1)!}.
$$
We then have
$$
\frac{R_{n+1}(x)}{R_n(x)^{1+\epsilon}} \geq \frac{\left(\frac{x^{n+2}}{(n+2)!}\right)}{\left(\frac{x^{n+1} e^x}{(n+1)!}\right)^{1+\epsilon}} = \frac{xe^{-(1+\epsilon)x}}{n+2} \left(\frac{(n+1)!}{x^{n+1}}\right)^\epsilon,
$$
so that, if $\epsilon > 0$,
$$
\lim_{n \to \infty} \frac{R_{n+1}(x)}{R_n(x)^{1+\epsilon}} = \infty.
$$
