Finding the first 5 terms of the Maclaurin Series of $\frac{\sin x}{e^x}$ Question.
Find the first 5 terms of the Maclaurin Series of $\frac{\sin x}{e^x}$.

I'm trying to find it using division. Is there possibly a shorter way because every time I try to do it, I get the wrong answer:
$$
x-2x^3/3+\cdots$$
I don't really like polynomial long division.
This is how I set it up:
$$
\frac{x - x^3/3! + x^5/5! - x^7/7!+\cdots}{  1 + x + x^2/2! + x^3/3! + x^4/4!+\cdots}
$$
I'm still getting the same answer.

Here are the first two terms:
$x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots$
$=x\left(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+ \dfrac{x^5}{5!}+ \dfrac{x^6}{6!}\cdots\right) -x^2 -\dfrac{2x^3}{3}-\dfrac{x^4}{6}-\dfrac{x^5}{30}-\dfrac{x^6}{120}-\dfrac{x^7}{5040}-\cdots$
$= (x-x^2)\left(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+ \dfrac{x^5}{5!}+ \dfrac{x^6}{6!}\cdots\right) +\dfrac{x^3}{3}+\dfrac{x^4}{3}+\dfrac{2x^5}{15}+\dfrac{x^6}{30}+\dfrac{17x^7}{2520}+\cdots$
and you can continue from there, taking the first term of the residual of the second part as the next term to put into the multiplication in the first part.  
