exp(x) monotonic and convergent to 0 for negative x

Question

How is it possible that exp(x) for negative x is always close to 0 and even monotonic? I mean, look on the series representation of it: $\sum_{k=0}^\infty\frac{x^k}{k!}$

it makes sense that for positive x it behaves like that, but for huge negative numbers it's like "10 trillions - 43535 gazillions + 4435435 fartillions -...". You know what I mean, how is that not only convergent to 0 as $x\to -\infty$, but also monotonic ?

Answer 1

As $\exp(x)\exp(-x)=1$, $\exp(-x)$ must be monotone (decreasing) since $\exp(x)$ is monotone (increasing). It's also clear that $\lim_{x \rightarrow \infty}\left[\exp(-x)\right]=\lim_{x \rightarrow\infty}\left[\frac{1}{\exp(x)}\right]=0$.

Answer 2

Empirical observations:

Looking at the successive partial sums, you observe that for the even degrees, the polynomial has a single minimum and remains positive. Hence the next partial sum (also the antiderivative), which has an odd degree, is monotonic and has a single root. That root must be to the left of the minimum, and for higher degrees these keep moving towards $-\infty$.

In the limit, you get a positive function with a root at $-\infty$. Positiveness of the function explains monotonicity, as it is its own derivative.

There are two mysteries:

• how the minima of the even polynomials remain positive,

• why the minima/roots escape to infinity instead of converging to a finite value.

---

Additional remark:

For a fixed $x$, the general term $\dfrac{x^n}{n!}$ increases as long as $n<x$, then decreases. This somehow explains why the minima escape, possibly at unit speed.