# 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! =)

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 $$\lim\limits_{x\rightarrow 0}\frac{\sin{x}}{x}=1$$.

• Omg! The dirichlet kernel... That's basically where I also got the idea from regarding the use of the geometric series. I should have just rewritten everything ^^' Awesome, great simplification =) – Flammable Maths Jan 16 at 9:29

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

• Like I said, this is exactly my approach^^ – Flammable Maths Jan 16 at 9:30
• Then why did you have to ask the question then? – user150203 Jan 16 at 9:38
• To find a simplified approach. – Flammable Maths Jan 16 at 10:33
• @FlammableMaths - Okay, where did you get stuck? – user150203 Jan 16 at 10:37

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 $$\begin{eqnarray} \lim\limits_{n \rightarrow\infty}\frac{\pi}{2n}\sum\limits_{k=1}^{n}\cos\left(\frac{\pi}{2n}k\right)&=&\lim\limits_{n \rightarrow\infty}\frac{\pi}{4n\sin\frac{\pi}{4n}}\left(\sin\frac{2n+1}{4n}\pi-\sin\frac{1}{4n}\pi\right)=\sin\frac{\pi}{2}=1. \end{eqnarray}$$