How would you calculate this limit? $\lim\limits_{n \rightarrow\infty}\frac{\pi}{2n}\sum\limits_{k=1}^{n}\cos\left(\frac{\pi}{2n}k\right)$ I decided to calculate $\int_{0}^{\pi/2}cos(x)dx$ using the sum definition of the integral. Obviously the answer is $1$ . I managed to calculate the resulting limit using the geometric series, taking the real part of the complex exponential function and several iterations of l'hopital's rule. Are you able to simplify this absolute mess, i.e. find a better way of arriving at the desired answer?
$$\lim\limits_{n \rightarrow\infty}\frac{\pi}{2n}\sum\limits_{k=1}^{n}\cos\left(\frac{\pi}{2n}k\right)$$
Every answer is highly appreciated =)
PS: If you want to see my solution, feel free to tell me! =)
 A: 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 $\lim\limits_{x\rightarrow 0}\frac{\sin{x}}{x}=1$.
A: 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}
A: 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
$$\begin{eqnarray}
\lim\limits_{n \rightarrow\infty}\frac{\pi}{2n}\sum\limits_{k=1}^{n}\cos\left(\frac{\pi}{2n}k\right)&=&\lim\limits_{n \rightarrow\infty}\frac{\pi}{4n\sin\frac{\pi}{4n}}\left(\sin\frac{2n+1}{4n}\pi-\sin\frac{1}{4n}\pi\right)=\sin\frac{\pi}{2}=1.
\end{eqnarray}$$
