Weighted uniform convergence of Taylor series of exponential function 
Is the limit 
  $$
e^{-x}\sum_{n=0}^N \frac{(-1)^n}{n!}x^n\to e^{-2x}  \quad \text{as } \ N\to\infty \tag1
$$
  uniform on $[0,+\infty)$? 

Numerically this appears to be true: see the difference of two sides in (1)  for $N=10$ and $N=100$ plotted  below. But the convergence is very slow (logarithmic  error $\approx N^{-1/2}$ as shown by Antonio Vargas in his answer). In particular, putting $e^{-0.9x}$ and $e^{-1.9x}$ in (1) clearly makes convergence non-uniform. 
One difficulty here is that the Taylor remainder formula is effective only up to $x\approx N/e$, and the maximum of the difference is at $x\approx N$.


The question is inspired by an attempt to find an alternative proof of $\epsilon>0$ there is a polynomial $p$ such that $|f(x)-e^{-x}p|<\epsilon\forall x\in[0,\infty)$. 
 A: Credits should go to Landscape.
Define $$r_n(x)=\sum_{k=n+1}^\infty (-1)^k\frac{x^k}{k!}$$
Note that by Taylor's theorem with Lagrange's form of the remainder we can write $$r_n(x)=(-1)^{n+1}e^{-x'}\frac{x^{n+1}}{(n+1)!}$$
where $x'$ is positive. It follows $$e^{-x}|r_n(x)|\leq e^{-x}\frac{x^{n+1}}{(n+1)!}$$
Easy verification shows the last function has absolute maximum at $x=n+1$. But $$\frac{1}{{(n + 1)!}}{\left( {\frac{{n + 1}}{e}} \right)^{n + 1}} \sim \frac{1}{{\sqrt {2\pi \left( {n + 1} \right)} }}$$ by Stirling, so convergence is indeed uniform. $\quad \Box$
A: Thanks, this was a fun problem.
From the integral representation
$$
\sum_{k=0}^{n} \frac{x^k}{k!} = \frac{1}{n!} \int_0^\infty (x+t)^n e^{-t} \,dt  \tag1
$$
we can derive the expression
$$
e^{-x} \sum_{k=0}^{n} \frac{(-x)^k}{k!} = e^{-2x} - \frac{e^{-2x} (-x)^{n+1}}{n!} \int_0^1 t^n e^{xt}\,dt. \tag2
$$
Now
$$
\int_0^1 t^n e^{xt}\,dt \leq e^x \int_0^1 t^n\,dt = \frac{e^x}{n+1}, \tag3
$$
so that
$$
\begin{align}
\left|\frac{e^{-2x} (-x)^{n+1}}{n!} \int_0^1 t^n e^{xt}\,dt\right| &\leq \frac{e^{-x} x^{n+1}}{(n+1)!} \\
&\leq \frac{e^{-n-1} (n+1)^{n+1}}{(n+1)!} \\
&\sim \frac{1}{\sqrt{2\pi n}}
\end{align} \tag4
$$
by Stirling's formula.

Added by 40 votes for those interested in the derivation of (2) from (1): 
$$
\sum_{k=0}^{n} \frac{(-x)^k}{k!} = \frac{1}{n!} \int_0^x (t-x)^n e^{-t} \,dt  +
\frac{1}{n!} \int_x^\infty (t-x)^n e^{-t} \,dt \tag{A}
$$
Substitute $u=t-x$ in the second integral on the right of (A): 
$$\frac{1}{n!}\int_x^\infty (t-x)^n e^{-t} \,dt
=\frac{1}{n!}\int_0^\infty u^n e^{-u-x} \,dt = e^{-x}
 \tag{B}$$
Substitute $u=1-t/x$ in the first integral on the right of (A), noting that $(t-x)^n=(-x)^n u^n$ and $dt=(-x)du$: 
$$\frac{1}{n!} \int_0^x (-x+t)^n e^{-t} \,dt 
= \frac{(-x)^{n+1}}{n!} \int_0^1 u^n e^{xu-x} \,du = \frac{e^{-x}(-x)^{n+1}}{n!} \int_0^1 u^n e^{xu} \,du \tag{C}
$$
Adding (B) and (C), identity (2) follows. 
A: I'd like to provide another solution which is a mixture of Antonio's and Landscape's. One can also write $$\left|r_n(x)\right|=\int_0^x {e^{-t}}\frac{(x-t)^n}{n!}dt$$
by virtue of Taylor's theorem with the integral remainder. But then again $$\left| {{r_n}(x)} \right| \leqslant \int_0^x {\frac{{{{(x - t)}^n}}}{{n!}}dt}  = \int_0^x {\frac{{{u^n}}}{{n!}}du}  = \frac{x^{n + 1}}{{\left( {n + 1} \right)!}}$$
and Strling does the job.
