Finding Function of Series: $e^{-kx}$

Question

If the series representation of $e^{-x}$ is:

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

Then what is for $e^{-kx}$?

Answer 1

You should use a different index of summation than variables used outside the scope of the summation. $$ e^{-kx}=\sum_{j=0}^\infty\frac{(-kx)^j}{j!} $$

Answer 2

Consider the summation $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \text{.}$$ You demonstrated that you already know that you're allowed to replace $x$ with $-x$ to obtain the summation $$e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} \text{.}$$

Intuitively you should be able to throw a k in there as well: $$e^{-kx} = \sum_{n=0}^{\infty} \frac{(-kx)^n}{n!} \text{.}$$