Contour integration ambiguity between function and it's series

Question

$\frac{1}{1-z}$ is a shift and flip of $\frac{1}{z}$, whose contour integral over a simple loop around the singularity is then $-2i\pi$. However, the Taylor series of the function is $z^n$ summed over all natural n, which integrated in the same contour equals zero. I though about the possibility of accidently defining the infinite series outside the radius of convergence, but for any path within the interval were you integrate you get the same answer. Why did this contradiction arise?

Answer 1

There is no contradiction. The series $\sum_{n=0}^\infty z^n$ converges to $\frac1{1-z}$ when and only when $\lvert z\rvert<1$. Any closed loop around $1$ has to have points whose absolute value is greater than $1$.