Let \(n \ge 6\). Define \(\displaystyle f(x) = \cos\left(\frac{2 \pi x}{n}\right) - \frac{x}{n}\) for \(0 \le x \le \frac{n}{6}\). Note that \(\displaystyle 0 \le \frac{2 \pi x}{n} \le \frac{\pi}{3}\) on this interval. Then
\[f'(x) = -\frac{2\pi}{n}\sin\left(\frac{2\pi x}{n}\right)-\frac{1}{n} = -\frac{1}{n}\left(2\pi\sin \left(\frac{2\pi x}{n}\right)+1 \right) < 0 \]
and so \(f\) is strictly decreasing on the interval \(0 \le x \le \frac{n}{6}\). This means that \[f(x) \gt f\left(\frac{n}{6}\right) = \cos\left(\frac{\pi}{3}\right) - \frac{1}{6} = \frac{1}{2}-\frac{1}{6} \gt 0 \Rightarrow \cos\left(\frac{2 \pi x}{n}\right) \gt \frac{x}{n}\] for \(0 \le x \le \frac{n}{6}\). Therefore \[\sum\limits_{k = 1}^{\left\lfloor {n/4} \right\rfloor } \cos \left( \frac{2\pi k}{n} \right) \ge \sum\limits_{k = 1}^{\left\lfloor {n/6} \right\rfloor } \cos \left( \frac{2\pi k}{n} \right) \ge \sum\limits_{k = 1}^{\left\lfloor {n/6} \right\rfloor } \frac{ k}{n} = \frac{1}{n}\cdot\frac{\left\lfloor {\frac{n}{6}} \right\rfloor \left(\left\lfloor {\frac{n}{6}} \right\rfloor + 1\right)}{2} \gt \frac{\left(\frac{n}{6}-1\right)\left(\frac{n}{6}\right)}{2n} = \frac{n-6}{72}\]
Now \[1 +\sum\limits_{k = 1}^{\left\lfloor {n/4} \right\rfloor } \cos \left( \frac{2\pi k}{n} \right) \gt 1+ \frac{n-6}{72} = \frac{n+66}{72} \] and so \[r_n = \frac{1}{{2 \; \left( {1 + \sum\limits_{k = 1}^{\left\lfloor {n/4} \right\rfloor } {\cos \left( {\frac{{2\pi k}}{n}} \right)} } \right)}} < \frac{36}{n+66}.\]
One consequence of this inequality is that \(\displaystyle \lim_{n \rightarrow \infty} r_n = 0\).
