9
$\begingroup$

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.

$\endgroup$
3
  • 5
    $\begingroup$ What about showing $$\sum_{n \ge 0} \frac{(-x)^{n}}{n!} = \frac{1}{\sum_{n \ge 0} \frac{x^{n}}{n!}}$$ and then taking limits of both sides. $\endgroup$ Dec 6, 2018 at 4:29
  • $\begingroup$ Ah, yes of course, which is a purely algebraic fact if you move the denominator over to the left. $\endgroup$
    – zoidberg
    Dec 6, 2018 at 4:32
  • $\begingroup$ Please replace $\infty$ with $\color{red}{+}\infty$. $\endgroup$ Dec 6, 2018 at 15:40

1 Answer 1

10
$\begingroup$

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, ...

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .