I'm currently stuck on the question regarding the Maclaurin series for
$e^x +2e^{-x}$
I've found that the power series representation for it is $$\sum_{n=0}^\infty \dfrac{x^n + 2(-x)^n}{n!}$$ however, I do not know how to find the interval of convergence; I've tried the ratio test and have been unable to complete it.
Any help would be appreciated, thanks!
Best Answer
Well, you haven't really made this as simple as possible. (Also, you've made an error, and your indices are not matching.)
$$\begin{align}e^x+2e^{-x} &= \sum_{n=0}^\infty\frac1{n!}x^n+2\sum_{n=0}^\infty\frac1{n!}(-x)^n\\ &= \sum_{n=0}^\infty\frac1{n!}x^n+\sum_{n=0}^\infty\frac{2(-1)^n}{n!}x^n\\ &= \sum_{n=0}^\infty\frac{1+2(-1)^n}{n!}x^n\\ &= \sum_{n=0}^\infty a_nx^n,\end{align}$$ where $$a_n=\begin{cases}\frac3{n!} & \text{if }n\text{ is even}\\\frac{-1}{n!} & \text{if }n\text{ is odd.}\end{cases}$$
Observing that $\left|\frac{a_{n+1}}{a_n}\right|\le\frac{3}{n+1}$ for all $n,$ we see that the series converges everywhere. Alternately, note that the radius of convergence of a finite sum of MacLaurin series will be the minimum of the radii of convergence of the MacLaurin series being added together. (It's a nice exercise to prove this fact.) Since the series of $e^x$ and $2e^{-x}$ both converge everywhere, then so does the series for $e^x+2e^{-x}$.