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I'm trying to prove that

$$\sum_{n=0}^N \cos(\pi n/N) = 0$$

when $N$ is large. I could make an integral that basically does the same thing:

$$\int_0^N \cos(\pi n / N) dn$$

$$=-\frac{N}{\pi}\left[ \sin(\pi n /N) \right]^N_0$$

$$=0$$

I'm just not sure how to go from the sum to the integral. Or do I even need to?

user2162498
  • 101

1 Answers1

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If $N$ is even, then your sum will contain summand $\cos\pi/2$ which is equal to zero, so we reduced the problem to the case of odd $N$. In this case each summand $\cos(\pi n/N)$ with $n<N/2$ will be killed by $\cos (\pi m/N)$ with $m=N-n$, becase $\cos (\pi-x)=-\cos x$. So the resulting sum is always zero.

Norbert
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