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$$g(z)=\dfrac {\sin z-z+\dfrac{z^{3}}{6}}{\cos z-1}$$ What is the radius of convergence of the Taylor series of $g$ centered at 0.

My thought was to use the Cauchy-Hadamard formula to calculate it. But I don't know how to expand the $g(x)$ into power series.

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In this Wikipedia article it is stated that

The radius of convergence of a power series $f$ centered on a point $a$ is equal to the distance from $a$ to the nearest point where $f$ cannot be defined in a way that makes it holomorphic.

This can be proven using Cauchy's Integral Formula.

Since the closest non-removable singularity of $g$ to $z=0$ is $z=2\pi$, the radius of convergence should be $2\pi$.

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