Seeing that $\lim_{x \to \infty} \sum (-x)^n/n! = 0$

Question

Is there any way to see directly from the power series that $$\lim_{x \to \infty} \sum_{n=0}^\infty \frac{(-x)^n}{n!} = 0$$? I realize that $\displaystyle{\lim_{x \to \infty} e^{-x} = 0}$. That's not what I'm asking.

Answer 1

Just working with series and the Cauchy product we have

$$\sum_{n=0}^\infty \frac{(-x)^n}{n!}\sum_{n=0}^\infty \frac{x^n}{n!} = \sum_{n=0}^\infty\sum_{k=0}^n \frac{x^k}{k!}\frac{(-x)^{n-k}}{(n-k)!} = \sum_{n=0}^\infty\frac{(x+(-x))^n}{n!} = 1$$

and it is easy to show that as $x \to \infty$

$$\sum_{n=0}^\infty \frac{x^n}{n!} \to +\infty$$

Hence, ...