How would you calculate this limit? $\lim\limits_{n \rightarrow\infty}\frac{\pi}{2n}\sum\limits_{k=1}^{n}\cos\left(\frac{\pi}{2n}k\right)$

According to this question

$$1 + \sum\limits_{k=1}^n \cos{(k \theta)}=\frac{1}{2}+\frac{\sin\left[\left(n+\frac{1}{2}\right)\theta\right]}{2\sin\left(\frac{\theta}{2}\right)}$$

As a result

$$\lim\limits_{n \rightarrow\infty}\frac{\pi}{2n}\sum\limits_{k=1}^{n}\cos\left(\frac{\pi}{2n}k\right)= \lim\limits_{n \rightarrow\infty}\frac{\pi}{2n}\left(\frac{1}{2}+\frac{\sin\left[\left(n+\frac{1}{2}\right)\frac{\pi}{2n}\right]}{2\sin\left(\frac{\frac{\pi}{2n}}{2}\right)}-1\right)=\\ \lim\limits_{n \rightarrow\infty}\frac{\pi}{2n}\left(\frac{\sin\left(\frac{\pi}{2}+\frac{\pi}{4n}\right)}{2\sin\left(\frac{\pi}{4n}\right)}-\frac{1}{2}\right)= \lim\limits_{n \rightarrow\infty}\frac{\pi}{2n}\left(\frac{\cos\left(\frac{\pi}{4n}\right)}{2\sin\left(\frac{\pi}{4n}\right)}\right)=\\ \frac{\lim\limits_{n \rightarrow\infty}\cos\left(\frac{\pi}{4n}\right)}{\lim\limits_{n \rightarrow\infty} \frac{\sin\left(\frac{\pi}{4n}\right)}{\frac{\pi}{4n}}}=\frac{1}{1}=1$$ using the fact that $\lim\limits_{x\rightarrow 0}\frac{\sin{x}}{x}=1$.

HINTS:

(1) \begin{equation} \cos\left(\frac{\pi}{2n}\cdot k\right) = \Re\left[\exp\left(\frac{\pi}{2n}\cdot k i \right) \right] \end{equation}

(2) \begin{equation} \exp\left(\frac{\pi}{2n}\cdot k i \right) = a^k, \quad a = \exp\left(\frac{\pi}{2n}i \right) \end{equation}

(3) \begin{equation} \sum_{k = 1}^{n} a^k = \frac{a\left(a^{n} - 1\right)}{a - 1} \end{equation}

Here's an approach without Euler's formula using telescoping. With the help of the sum-product formula, we can see $$\begin{eqnarray} \sin\frac{\theta}{2}\sum_{k=1}^n\cos k\theta &=&\frac{1}{2}\sum_{k=0}^n\left(\sin\frac{2k+1}{2}\theta-\sin\frac{2k-1}{2}\theta\right)\\ &=&\frac{1}{2}\left(\sin\frac{2n+1}{2}\theta-\sin\frac{1}{2}\theta\right). \end{eqnarray}$$ This gives $$\begin{eqnarray} \lim\limits_{n \rightarrow\infty}\frac{\pi}{2n}\sum\limits_{k=1}^{n}\cos\left(\frac{\pi}{2n}k\right)&=&\lim\limits_{n \rightarrow\infty}\frac{\pi}{4n\sin\frac{\pi}{4n}}\left(\sin\frac{2n+1}{4n}\pi-\sin\frac{1}{4n}\pi\right)=\sin\frac{\pi}{2}=1. \end{eqnarray}$$