Integral on semicircular arc of a function with a simple pole at $z=0$ Let $C(\epsilon)$ denote the semicircular arc about $z=0$. Suppose $f(z)$ is a meromorphic function that has a simple pole at $z=0$. I'm trying to show that $$\lim_{\epsilon\to 0}\int_{C(\epsilon)} f(z) = -\pi i \mathrm{Res}_{z=0}(f(z))$$ First, I expand $f(z)$ as a Laurent series $f(z) = \frac{b_{1}}{z} + \sum_{k=0}^\infty a_kz^k$ and then I integrate about the specified region: $$\int_{C(\epsilon)} f(z) = -\pi i b_1 + \sum_{k=1}^\infty \frac{a_k\epsilon^{k+1}}{k+1}(1+(-1)^k).$$ Now I need to justify that $$\lim_{\epsilon\to 0} \sum_{k=1}^\infty \frac{a_k\epsilon^{k+1}}{k+1}(1+(-1)^k)=0.$$ but I have an issue with interchanging the sum and the limit. I was thinking of using the ratio test combined with the bound on $a_k$ provided by Taylor's Theorem, but is there a simpler way?

It's not hard to show that sum tends to $0$ with $\epsilon$. But there's no reason it has to come up. Instead of starting with the Laurent series start with $b_{-1}/z + g(z)$, where $g$ is holomorphic near the origin, hence bounded in some neighborhood of the origin.