Limit with a summation and sine: how to calculate $\lim_{n\to \infty} n^2\sum_{k=0}^{n-1} \sin\left(\frac{2\pi k}n\right)$? 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.
 A: 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.
A: $\sum_{k=0}^{n-1}\sin(2\pi k/n)$ 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
$$\lim_{n \to \infty}n^2 \sum_{k=0}^{n-1}\sin(2\pi k/n) = \lim_{n \to \infty} (n^2 \cdot 0) = \lim_{n \to \infty} 0 = 0$$
A: 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.
$$\forall n\implies\sum_{k=0}^{n-1}\sin\left(\frac{2k\pi}n\right)=0$$
