Laurent series for $\cot(\pi z)/z^2$

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I'm looking for the series for $\frac{cot(\pi z)}{z^2}$ using the residue theory, where the function denotes a circle about the origin with a radius of $k+\frac{1}{2}$

I found that the residues of $\frac{cot(\pi z)}{z^2}$ are:

$R(f;0)= \frac{-\pi}{3}$

$R(f;n)=\frac{1}{\pi n^2}$, where $n$ are positive integers.

I know that the residue theory states that the integral of the function f(z) is $2 \pi i$ $\sum Residues$ however I am unsure how to implement the fact that the radius is $k+\frac{1}{2}$

Because from my residue result I would assume the answer would be:

$2\pi i[\frac{-\pi}{3}+\frac{1}{\pi}\sum_{n=1}^k{\frac{1}{n^2}}]$

However the answer is

$2\pi i[\frac{-\pi}{3}+\frac{2}{\pi}\sum_{n=1}^k{\frac{1}{n^2}}]$

I'm not sure where the 2 before the sum cam from.

Thank you!