Power series representation of $e^x$ and $e^{-x}$

Question

The power series representation of $e^x = \sum \limits_{k=0}^{\infty} \frac{x^k}{k!}$. Can I use this approximation for $e^{-x} = 1/e^x = 1/\sum \limits_{k=0}^{\infty} \frac{x^k}{k!}$ instead of $e^{-x} = \sum \limits_{k=0}^{\infty} \frac{(-1)^k x^k}{k!}$. What is the difference between two approximation?

Answer 1

The two are equivalent. If you know about power series multiplication, you can see that

$$ \left(\sum_{k=0}^\infty \frac{x^k}{k!}\right)\cdot \left(\sum_{k=0}^\infty \frac{(-1)^kx^k}{k!}\right)=1. $$

If you expand the product of the two power series, terms will cancel, leaving you with $1$.