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How do I sum $$\sum_{n=1}^{\infty} \sin\frac{n!\pi}{120}$$

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A little experimentation would go a long way. –  Jonas Meyer Jun 1 '12 at 6:51
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Note that $\sin \left(\dfrac{n! \pi}{120} \right) = 0$ for all $n \geq 5$. Hence, $$\sum_{n=1}^{\infty} \sin \left(\dfrac{n! \pi}{120} \right) = \sin \left(\dfrac{1! \pi}{120} \right) + \sin \left(\dfrac{2! \pi}{120} \right) + \sin \left(\dfrac{3! \pi}{120} \right) + \sin \left(\dfrac{4! \pi}{120} \right)$$

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