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This is the limit: $$\lim_{n\to \infty} n^2\sum_{k=0}^{n-1} \sin\left(\frac{2\pi k}n\right)$$

I found that

  • $k/n <1$

  • if $n=2k$ the term of the summation is $0$

  • until $n=4$ the summation is $0$

  • $ \sin\left(\frac{2\pi k}n\right)= 2\sin\left(\frac{\pi k}n\right)\cos\left(\frac{\pi k}n\right)$

I also tried to increase or decrease the summation with an integral but I think I can do it only if the term in the summation is monotonous.

I totally don't know how to deal with this kind of exercise, I'm looking for a general approach

Thanks! Sorry for english.

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3 Answers 3

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$\sum_{k=0}^{n-1}\sin(2\pi k/n)$ is the imaginary part of $\sum_{k=0}^{n-1}e^{i2\pi k/n}$. We can evaluate the latter sum directly since it's a truncated power series: $$\sum_{k=0}^{n-1}z^k = \frac{z^n - 1}{z-1}$$ provided that $z \neq 1$, so $$\sum_{k=0}^{n-1}e^{i2\pi k/n} = \frac{e^{i 2\pi} - 1}{e^{i 2\pi / n} - 1} = \frac{1 - 1}{e^{i 2\pi / n} - 1} = 0$$ Hence also $$\sum_{k=0}^{n-1}\sin(2\pi k/n) = 0$$ for every positive integer $n$. So your limit is $$\lim_{n \to \infty}n^2 \sum_{k=0}^{n-1}\sin(2\pi k/n) = \lim_{n \to \infty} (n^2 \cdot 0) = \lim_{n \to \infty} 0 = 0$$

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  • $\begingroup$ Darn how do you write it up so fast? $\endgroup$ Jul 21, 2017 at 22:56
  • $\begingroup$ Can I deal whit every sum with only sine or cosine introducing the complex exponential? Or maybe this is only a case? $\endgroup$
    – pter26
    Jul 21, 2017 at 23:11
  • $\begingroup$ Another question, is the immaginary part 0 because of the entire complex number is 0 and so the immaginary part? Or maybe is it an increase? $\endgroup$
    – pter26
    Jul 21, 2017 at 23:36
  • $\begingroup$ @user411485 You can certainly try using the complex exponential, but in general it probably won't help much unless the argument of the sine or cosine is of the form $ax+b$. Yes, the imaginary part is zero because the entire complex number is zero. $\endgroup$
    – user169852
    Jul 22, 2017 at 1:55
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If $\zeta_n=e^{\frac{2\pi i}{n}}$, then for every integer $n>1$ we have $$ \sum_{k=0}^{n-1}\zeta_n^k=\frac{\zeta_n^n-1}{\zeta_n-1}=0$$

And since $$\zeta_n^k=e^{\frac{2\pi ik}{n}}=\cos\Big(\frac{2\pi k}{n}\Big)+i\sin\Big(\frac{2\pi k}{n}\Big)$$ taking imaginary parts shows that your limit is zero.

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Note that

$$\sin(x)=-\sin(2\pi-x)$$

And,

$$\sin\left(\frac{2\pi k}n\right)=-\sin\left(2\pi-\frac{2\pi k}n\right)=-\sin\left(\frac{2\pi(1-k)}n\right)$$

Thus, the sum is symmetric about $\pi$.

If $n$ is even, all terms cancel.

If $n$ is odd, all terms cancel except $\sin(\pi)$, which is zero.

$$\forall n\implies\sum_{k=0}^{n-1}\sin\left(\frac{2k\pi}n\right)=0$$

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