How does the Maclaurin series for $e^x$ converge for negative values? It really doesn't seem to do so. It just oscillates wildly while increasing in magnitude. wtf?
$e^{-\infty }$ should approach 0, but that doesn't seem to be the case here.
https://www.wolframalpha.com/input/?i=sum+from+0+to+111+of+%28-10000%29%5En%2Fn%21
The problem is to know how many terms you need to sum for a given accuracy.
In other words, adding $(n-1)$ terms you want to know $n$ such that
$$\frac {x^n}{n!}  &lt; 10^{-k}\implies n!&gt;x^n\, 10^k$$
If you look at this question of mine, you will see a magnificent approximation by @robjohn for the inverse of the factorial (have a look here).
$$\bbox[5px,border:2px solid #C0A000]{n\sim ex\exp\left(\operatorname{W}\left(\frac k{ex}\log(10)-\frac1{2ex}\log(2\pi x)\right)\right)-\frac12}$$ where $W(\cdot)$ is Lambert function.
Using this approximation for $x=10^4$ and $k=6$ would give $n=27190.6084224$ while the exact solution is $n=27190.6084239$
Checking
$$\frac{(10^4)^{27190}}{27190!}=1.84\times 10^{-6} &gt; 10^{-6}$$
$$\frac{(10^4)^{27191}}{27191!}=6.76\times 10^{-7} &lt; 10^{-6}$$
The Taylor series around zero is $e^z=\sum_{n\ge0}\dfrac {z^n}{n!} $.
By Cauchy-Hadamard, the radius of convergence is  $r=1/\limsup_{n\to\infty}\sqrt [n]{1/n!}=1/0=\infty$.