Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 7 / d / ii / Solution

For $|z|<1$, multiply the absolutely convergent Taylor series of the sine by the geometric series:The product is odd, so all even coefficients vanish. Collecting the coefficient of $z^{2m+1}$ givesFor example, the first terms are $z+5z^3/6+101z^5/120+\cdots$. The denominator vanishes at $z=\pm 1$, and the numerator is nonzero there, so these are genuine simple poles. Therefore $\boxed{R=1}$ is the radius of convergence.