Find the Maclaurin series for $f(x) = e^{-2x}$, and find the interval of convergence for the series.
I got the maclaurin series to be this
$e^x = 1 – 2\frac{x}{1!} + 2^2\frac{x^2}{2!} + \cdots$
$$\sum_{n=0}^{\infty} \frac{(-1)^n (2)^n x^n}{n!}$$
Using the ratio test to find the interval of convergence:
$$\lim_{n\to\infty} \left| \frac{(-1)^{n+1}2^{n+1}x^{n+1}}{(n+1)!} \frac{n!}{(-1)^n 2^n x^n} \right| = 2|x| \lim_{n\to\infty} \frac{1}{n+1} = 2|x|(0) < 1$$
Therefore for any value of x is in the interval of convergence. Is this right?
Best Answer
The series representation for the exponential function converges everywhere (uniformly as well), as so any related sum will also converge everywhere.