In the book of analytical combinatorics of Philippe Flajolet Robert Sedgewick pp 259 they affirm the following:
If $f(z)\sim c/(z-a)$ as $z\to a$, then
$\displaystyle\text{Res} (f(z)z^{-n-1}; z=a)=\text{Res}\left(\frac{c}{(z-a)}z^{-n-1}; z=a\right)=\frac{c}{a^{n+1}}$
Is there a way to prove this using the Cauchy residue theorem?
$g(z)=z^{-n-1}$ is holomorphic in a neighborhood of $a$ (which I assume is different from $0$). Therefore $$ g(z)=a^{-n-1}+A(z-a)+B(z-a)^2+\dots $$ and $$ f(z)g(z)\sim \frac{ca^{-n-1}}{z-a} $$ as $z\to a$. Therefore $$ Res\, (fg;\, z=a)=ca^{-n-1}. $$