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Theorem 1.4: Let $f$ be periodic and piecewise differentiable. Then at each point $\theta$ the symmetric partial sums

$$

S_N(\theta) = \sum_{n=-N}^Na_ne^{in\theta}

$$

converge to $\frac{1}{2}\left[f(\theta)+f(-\theta)\right]$; if $f$ is continuous at $\theta$, they converge to $f(\theta)$.

So taking this series:

$$

2\sum_{n =1}^{\infty}\left[\frac{1}{2n-1}\sin(2n-1)\theta - \frac{1}{2n}\sin 2n\theta\right]

$$

What am I supposed to do?