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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.2012-06-01

1 Answers 1

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Note that $\sin \left(\dfrac{n! \pi}{120} \right) = 0$ for all $n \geq 5$. Hence, $\sum_{n=1}^{\infty} \sin \left(\dfrac{n! \pi}{120} \right) = \sin \left(\dfrac{1! \pi}{120} \right) + \sin \left(\dfrac{2! \pi}{120} \right) + \sin \left(\dfrac{3! \pi}{120} \right) + \sin \left(\dfrac{4! \pi}{120} \right)$