Sum of the given cosine series

Question

How to find the sum of the following series:

$$\sum_{n =-\infty}^{\infty}\cos \left(8n+\frac{2\pi}{3}\right)$$

If it was only in terms of $\pi$, I would have handled it. But I don't know how to deal with $8n$.

Any hints will be appreciated.

Thanks

Answer 1

$$ \cos ( 8n + \frac {2\pi }{3} ) = \cos ( 8n)\cos (\frac {2\pi }{3} )-\sin( 8n)\sin(\frac {2\pi }{3} )$$

$$\sum \cos ( 8n + \frac {2\pi }{3} )=\cos (\frac {2\pi }{3} )\sum\cos ( 8n)-\sin (\frac {2\pi }{3} )\sum\sin ( 8n)$$

$$\sum _{-\infty}^{\infty}\sin ( 8n)=0$$

$$\sum _{-\infty}^{\infty}\cos ( 8n)=1+2\sum _{1}^{\infty}\cos ( 8n)$$

$$\sum _{1}^{\infty}\cos ( 8n) = \cos 8 +\cos 16 +\cos 24 + ...$$ $$= \frac { \cos 8\sin 4 +\cos 16\sin +\cos 24 \sin 4+ ...}{\sin 4}$$

$$=\frac { \sin 12 -\sin 4 +\sin 20 -\sin 12 +\sin 28 - \sin 20+ ...}{2\sin 4}$$

This series does not converge because the general term does not tend to zero.

Thus the final result is $$ \sum _{-\infty}^{\infty}\cos ( 8n + \frac {2\pi }{3} ) $$ does not converge.