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Suppose $F$ is holomorphic outside a disk and $F(z)=L+O(1/z)$ uniformly as $z\to \infty$. Thenwith counterclockwise large circles. The conventional residue at infinity has the opposite sign from the coefficient extracted by that counterclockwise circle.

To check the sign, use $w=1/z$. The differential becomes $-F(1/w)dw/w$, whose residue at zero is $-L$. Alternatively, $F(z)/z=L/z+O(z^{-2})$, and the large-circle contour estimate makes the error integral vanish. For the constant function $F=1$, the integral is $2\pi i$ but the residue at infinity is $-1$.