maclaurin series of $f(x)=xe^{-x}$

Question

I'm not sure how to do: $$f(x)=xe^{-x}$$I did this one so far: $$ f(x) = x^2 \cos x = x^2 \left( 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} +\dfrac{x^8}{8!} \dots \right) = x^2 - \dfrac{x^4}{2!} + \dfrac{x^6}{4!} - \dfrac{x^8}{6!} +\dfrac{x^{10}}{6!} \dots $$

i know that the maclaurin series for $f(x)=e^x= 1 + x+ \dfrac{x^2}{2!} + \dfrac{x^3}{3!} - \dfrac{x^4}{4!}\dots$

Answer 1

From $$e^x= 1 + x+ \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+\cdots$$ one gets that $$ f(x)=e^{-x}= 1 - x+ \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+\cdots $$ giving $$ f(x)=xe^{-x}= x - x^2+ \dfrac{x^3}{2!} - \dfrac{x^4}{3!} + \dfrac{x^5}{4!}+\cdots $$ or

$$ f(x)=xe^{-x}=\sum_{n=0}^\infty(-1)^n \frac{x^{n+1}}{n!}. $$