I must evaluate $$\lim_{n \to \infty} n^2 \sum_{k=0}^{n-1} \sin \left (2\pi \frac{k}{n}\right)$$ It reminds me a Riemann sum, so I'm trying to arrive to that $$\lim_{n \to \infty} n^2 \sum_{k=0}^{n-1} \sin \left (2\pi \frac{k}{n}\right)=\lim_{n \to \infty}\frac{n^3}{2\pi} \sum_{k=0}^{n-1} \frac{2\pi}{n} \sin \left (2\pi \frac{k}{n}\right)=$$ $$=\left(\lim_{n \to \infty}\frac{n^3}{2\pi}\right) \left(\lim_{n \to \infty} \sum_{k=0}^{n-1} \frac{2\pi}{n} \sin \left (2\pi \frac{k}{n}\right)\right)=\left(\lim_{n \to \infty}\frac{n^3}{2\pi}\right) \int_0^{2\pi} \sin x \text{d}x=\left(\lim_{n \to \infty}\frac{n^3}{2\pi}\right) \cdot 0=$$ $$=\lim_{n \to \infty} 0 =0$$ But I'm afraid that I can't separate the limit in the product of two limits, even if maybe it is correct because $\frac{n^3}{2\pi} \to \infty$ as $n \to \infty$ and the other one tends to the integral (which is a number, but it happens to be $0$; so I don't know if it can be considered as an indeterminate form $\infty \cdot 0$).
Thanks.
The sum of the nth roots of unity is always equal to zero.
$s=\sum_{k=0}^{n-1} e^\frac{i2\pi k}{n}=0$
$e^{i\theta}=cos(\theta)+isin(\theta)$
$Im(s)=0 \implies \sum_{k=0}^{n-1} sin(\frac{2\pi k}{n})=0 $
Since the sum is zero for all values of n, the limit is also equal to zero.