Maclaurin series for $(1+x)\cdot e^{-x}$ I’m having problems with writing Maclaurin series for $$(1+x)\cdot e^{-x}$$ as I need to write it in the form of power series, but I have struggles with using derivatives here.
 A: Write down the power series for $e^{-x}$. Multiply by $x$ to find the power series of $xe^{-x}$. Finally, take the power series for $e^{-x}$ and the power series for $xe^{-x}$, and add them together, and you have the power series of $e^{-x} + xe^{-x} = (1+x)e^{-x}$.
A: We have $$e^x=\sum_{i=0}^\infty\frac{x^i}{i!}\implies e^{-x}=\sum_{i=0}^\infty\frac{(-1)^ix^i}{i!}$$ so $$(1+x)\cdot e^{-x}=e^{-x}+xe^{-x}=\quad ?$$
A: We have $f(x)=(1+x)e^{-x}$. The power series expansion of $e^x$ is $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=1+x+\frac12 x^2+\frac16x^3+\cdots\\\implies (1+x)e^{-x}=(1+x)\sum_{n=0}^\infty\frac{(-x)^n}{n!}=1+\sum_{n=1}^\infty \frac{(-x)^n}{n!}-\sum_{n=1}^\infty \frac{(-x)^{n}}{(n-1)!}\\=1+\sum_{n=1}^\infty\frac{(1-n)(-1)^n}{n!}x^n$$
So we have an explicit formula for the coefficient - we can write $(1+x)e^{-x}=\sum_{j=0}^\infty a_j x^j$ with $$a_j=\frac{(1-j)(-1)^j}{j!}$$

Equivalently, 
$$(1+x)e^{-x}=1+\sum_{n=0}^\infty\frac{(1-(n+1))(-1)^{n+1}}{(n+1)!}x^{n+1}=1+\sum_{n=0}^\infty\frac{n(-1)^{n}}{(n+1)n!}x^{n+1}\\=1+\sum_{n=0}^\infty\frac{(-1)^{n}}{(n+1)(n-1)!}x^{n+1}$$
