How does one evaluate the limit: $$ \lim_{n \to \infty} \frac{1}{n}\sum\limits_{k=1}^{\lfloor{\frac{n}{2}\rfloor}} \cos\Bigl(\frac{k\pi}{n}\Bigr)$$
Yes, i recognize this as soon as i saw the problem: $$\int\limits_{0}^{1}f(x) \ dx = \lim_{n \to \infty} \frac{1}{n} \sum\limits_{k=1}^{n} f\Bigl(\frac{r}{n}\Bigr)$$ but the problem is there is $\lfloor{\frac{n}{2}\rfloor}$.