Find the Maclaurin series for the function using definition of a Mclaurin series. Also, find the radius of convergence.
$\displaystyle f(x) = e^{-5x}$
For power series the simplest way to find the radius of convergence is to use the "ratio test".
A series of positive numbers, $\displaystyle \sum a_n$ will converge as long as the sequence $\displaystyle \frac{a_{n+1}}{a_n}$ has a limit less than 1. (If that limit is larger than 1, it diverges. If it is equal to 1, we don't know.)
Here, $\displaystyle a_n= \frac{-5x^n}{n!}$. Those terms may not be positive but we know that a power series always converges absolutely inside its radius of convergence so we can take the absolute value:
$\displaystyle \left|\frac{-5x^{n+1}}{(n+1)!}\frac{n!}{-5x^n}\right|= \frac{|x|}{n+1}$
What is the limit of that as n goes to infinity? For what values of x is that limit less than 1?
$\displaystyle \lim_{n \to \infty} \frac{|x|}{n+1}$
$\displaystyle \lim_{n \to \infty} \frac{\frac{|x|}{n}}{\frac{n}{n}+\frac{1}{n}}$
$\displaystyle = 0 $ Not sure I'm doing this right?
$\displaystyle e^{-5x} = 1 + (-5x) + \frac{(-5x)^2}{2!} + \frac{(-5x)^3}{3!} + \frac{(-5x)^4}{4!} + ... = \sum_{n=0}^\infty \frac{(-5x)^n}{n!}$
have you used the ratio test to find a radius of convergence before?
$\displaystyle \lim_{n \to \infty} \left| \frac{(-5x)^{n+1}}{(n+1)!} \cdot \frac{n!}{(-5x)^n} \right| < 1$
$\displaystyle \lim_{n \to \infty} \left| \frac{-5x}{n+1} \right| < 1$
$\displaystyle 5|x| \cdot \lim_{n \to \infty} \frac{1}{n+1} < 1$
$\displaystyle 5|x| \cdot 0 < 1$ for all x ... so, what's the radius of convergence?