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Question: The sum of the series \[\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}\] is ...

The sum of the series ∑n=1∞sin⁡(n!π720)\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)} is
a) sin⁡(π180)+sin⁡(π360)+sin⁡(π540)\sin \left( \dfrac{\pi }{180} \right)+\sin \left( \dfrac{\pi }{360} \right)+\sin \left( \dfrac{\pi }{540} \right)
b) sin⁡(π6)+sin⁡(π30)+sin⁡(π120)+sin⁡(π360)\sin \left( \dfrac{\pi }{6} \right)+\sin \left( \dfrac{\pi }{30} \right)+\sin \left( \dfrac{\pi }{120} \right)+\sin \left( \dfrac{\pi }{360} \right)
c) sin⁡(π6)+sin⁡(π30)+sin⁡(π120)+sin⁡(π360)+sin⁡(π720)\sin \left( \dfrac{\pi }{6} \right)+\sin \left( \dfrac{\pi }{30} \right)+\sin \left( \dfrac{\pi }{120} \right)+\sin \left( \dfrac{\pi }{360} \right)+\sin \left( \dfrac{\pi }{720} \right)
d) sin⁡(π180)+sin⁡(π360)\sin \left( \dfrac{\pi }{180} \right)+\sin \left( \dfrac{\pi }{360} \right)

Explanation

Solution

Hint: The value of sinesine function for any argument that is an integral multiple of π\pi is equal to 00 , i.e. sin⁡(nπ)=0\sin (n\pi )=0 where n=1,2,3,4...n=1,2,3,4... and so on.

Before we proceed with the solution , we must know some properties of the sinesine function .
a) The value of sin⁡(π)\sin (\pi ) is equal to 00.
b) sin⁡(π+θ)=−sin⁡θ\sin (\pi +\theta )=-\sin \theta
Now , to find the value of sin⁡(2π)\sin (2\pi ) , we will substitute θ=π\theta =\pi in property (b)(b).
So , we get sin⁡(2π)=sin⁡(π+π)=−sin⁡π=0\sin (2\pi )=\sin \left( \pi +\pi \right)=-\sin \pi =0 .
Similarly , sin⁡(3π)=sin⁡(π+2π)=−sin⁡2π=0\sin \left( 3\pi \right)=\sin \left( \pi +2\pi \right)=-\sin 2\pi =0. So , following the pattern , we can conclude that the value of sinesine function for any argument that is an integral multiple of π\pi is equal to 00, i.e.
sin⁡(nπ)=0\sin (n\pi )=0 where n=1,2,3,4...n=1,2,3,4...and so on.
Now , coming to the question , we need to find the value of the sum ∑n=1∞sin⁡(n!π720)\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}.
First , we will open the summation sign and write it as a sum of a series.
So , we can write ∑n=1∞sin⁡(n!π720)\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)} as ∑n=1∞sin⁡(n!π720)=sin⁡(π720)+sin⁡(2π720)+sin⁡(6π720)+sin⁡(24π720)+sin⁡(120π720)....\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}=\sin \left( \dfrac{\pi }{720} \right)+\sin \left( \dfrac{2\pi }{720} \right)+\sin \left( \dfrac{6\pi }{720} \right)+\sin \left( \dfrac{24\pi }{720} \right)+\sin \left( \dfrac{120\pi }{720} \right)....
Now , when n=6n=6 , we get sin⁡(n!π720)=sin⁡(6!π720)=sin⁡(720π720)=sin⁡π\sin \left( \dfrac{n!\pi }{720} \right)=\sin \left( \dfrac{6!\pi }{720} \right)=\sin \left( \dfrac{720\pi }{720} \right)=\sin \pi .
But , from the property(a)(a) of sinesine function , we know the value of sin⁡(π)\sin (\pi ) is equal to 00.
So , the value of sin⁡(n!π720)\sin \left( \dfrac{n!\pi }{720} \right), when n=6n=6 is equal to 00.
Now , when n=7n=7, we get sin⁡(n!π720)=sin⁡(7!π720)\sin \left( \dfrac{n!\pi }{720} \right)=\sin \left( \dfrac{7!\pi }{720} \right) .
We know , n!=n×(n−1)!n!=n\times (n-1)!.
So , 7!=7×6!=7×7207!=7\times 6!=7\times 720.
So , sin⁡(7!π720)=sin⁡(7×720π720)=sin⁡7π\sin \left( \dfrac{7!\pi }{720} \right)=\sin \left( \dfrac{7\times 720\pi }{720} \right)=\sin 7\pi .
Now , we know , the value of sine function for any argument that is an integral multiple of π\pi is equal to 00.
So , sin⁡7π=0\sin 7\pi =0.
Similarly , for n=8,9,....n=8,9,....and so on , we can see that the value of sin⁡(n!π720)\sin \left( \dfrac{n!\pi }{720} \right) is equal to 00.
So , we can write the sum as ∑n=1∞sin⁡(n!π720)=sin⁡(π720)+sin⁡(2π720)+sin⁡(6π720)+sin⁡(24π720)+sin⁡(120π720)+0+0+0....\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}=\sin \left( \dfrac{\pi }{720} \right)+\sin \left( \dfrac{2\pi }{720} \right)+\sin \left( \dfrac{6\pi }{720} \right)+\sin \left( \dfrac{24\pi }{720} \right)+\sin \left( \dfrac{120\pi }{720} \right)+0+0+0....
So , ∑n=1∞sin⁡(n!π720)=sin⁡(π720)+sin⁡(2π720)+sin⁡(6π720)+sin⁡(24π720)+sin⁡(120π720)\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}=\sin \left( \dfrac{\pi }{720} \right)+\sin \left( \dfrac{2\pi }{720} \right)+\sin \left( \dfrac{6\pi }{720} \right)+\sin \left( \dfrac{24\pi }{720} \right)+\sin \left( \dfrac{120\pi }{720} \right)
Or , ∑n=1∞sin⁡(n!π720)=sin⁡(π720)+sin⁡(π360)+sin⁡(π120)+sin⁡(π30)+sin⁡(π6)\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}=\sin \left( \dfrac{\pi }{720} \right)+\sin \left( \dfrac{\pi }{360} \right)+\sin \left( \dfrac{\pi }{120} \right)+\sin \left( \dfrac{\pi }{30} \right)+\sin \left( \dfrac{\pi }{6} \right)
Or , ∑n=1∞sin⁡(n!π720)=sin⁡(π6)+sin⁡(π30)+sin⁡(π120)+sin⁡(π360)+sin⁡(π720)\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}=\sin \left( \dfrac{\pi }{6} \right)+\sin \left( \dfrac{\pi }{30} \right)+\sin \left( \dfrac{\pi }{120} \right)+\sin \left( \dfrac{\pi }{360} \right)+\sin \left( \dfrac{\pi }{720} \right)
Hence , the value of the sum of the series ∑n=1∞sin⁡(n!π720)\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)} is equal to sin⁡(π6)+sin⁡(π30)+sin⁡(π120)+sin⁡(π360)+sin⁡(π720)\sin \left( \dfrac{\pi }{6} \right)+\sin \left( \dfrac{\pi }{30} \right)+\sin \left( \dfrac{\pi }{120} \right)+\sin \left( \dfrac{\pi }{360} \right)+\sin \left( \dfrac{\pi }{720} \right).
Therefore , option (c) is the correct answer.

Note: Students generally get confused between the values of sin⁡π\sin \pi and cos⁡π\cos \pi . sin⁡π=0\sin \pi =0 and cos⁡π=−1\cos \pi =-1 . Confusion between the values should be avoided as it can lead to wrong answers .