The sum of the series sum_{n=1}^{infty} sin l | vedclass

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(C) We know that $\sin(k\pi) = 0$ for any integer $k$.The given series is $E = \sum_{n=1}^{\infty} \sin \left(\frac{n! \pi}{720}\right)$.Expanding the

Vedclass · Accepted Answer

$\sin \left(\frac{\pi}{6}\right)+\sin \left(\frac{\pi}{30}\right)+\sin \left(\frac{\pi}{120}\right)+\sin \left(\frac{\pi}{360}\right)+\sin \left(\frac{\pi}{720}\right)$

Vedclass · Answer

(C) We know that $\sin(k\pi) = 0$ for any integer $k$.The given series is $E = \sum_{n=1}^{\infty} \sin \left(\frac{n! \pi}{720}\right)$.Expanding the terms:$n=1: \sin \left(\frac{1! \pi}{720}\right) = \sin \left(\frac{\pi}{720}\right)$$n=2: \sin \left(\frac{2! \pi}{720}\right) = \sin \left(\frac{2 \pi}{720}\right) = \sin \left(\frac{\pi}{360}\right)$$n=3: \sin \left(\frac{3! \pi}{720}\right) = \sin \left(\frac{6 \pi}{720}\right) = \sin \left(\frac{\pi}{120}\right)$$n=4: \sin \left(\frac{4! \pi}{720}\right) = \sin \left(\frac{24 \pi}{720}\right) = \sin \left(\frac{\pi}{30}\right)$$n=5: \sin \left(\frac{5! \pi}{720}\right) = \sin \left(\frac{120 \pi}{720}\right) = \sin \left(\frac{\pi}{6}\right)$$n=6: \sin \left(\frac{6! \pi}{720}\right) = \sin \left(\frac{720 \pi}{720}\right) = \sin(\pi) = 0$For all $n \ge 6$,$n!$ is a multiple of $720$,so $\frac{n! \pi}{720}$ is a multiple of $\pi$,making $\sin \left(\frac{n! \pi}{720}\right) = 0$.Thus,the sum is $\sin \left(\frac{\pi}{720}\right) + \sin \left(\frac{\pi}{360}\right) + \sin \left(\frac{\pi}{120}\right) + \sin \left(\frac{\pi}{30}\right) + \sin \left(\frac{\pi}{6}\right)$.

The sum of the series $\sum_{n=1}^{\infty} \sin \left(\frac{n! \pi}{720}\right)$ is

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