# Summing $\sum_{n=1}^{\infty} \sin\frac{n!\pi}{120}$

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

-
A little experimentation would go a long way. –  Jonas Meyer Jun 1 '12 at 6:51
add comment

## 1 Answer

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

-
add comment