What's the residue of $xe^{\frac{1}{x}}$ in $x=0$?

Question

I've been trying to get the residue of $xe^{\frac{1}{x}}$ in $x = 0$. I know the result is 1/2 but I don't know how to get to that result. The most I've done is use L'Hôpital and got $\frac{1}{\frac {e^{\frac{-1}{x}}}{x^2}}$. From here I don't know where to go. Any help?

Answer 1

Notice that $e^x= \sum_{n\geqslant 0} x^n/(n!) \Rightarrow e^{(1/x)} = \sum_{n\geqslant 0} 1/(x^n \cdot n!)$.

Therefore, $x e^\left({1/x}\right) = \sum_{n\geqslant 0} 1/(x^{n-1} \cdot n!)$. Then, residuo is $a_{-1} = 1/2! = 1/2$