[Math] Expressing limit of sum definite integral

Question

Evaluate limit by expressing it as a definite integral.

$$\lim_{n\to\infty}\frac{\pi}{2n}\left[\cos\left(\frac{\pi}{2n}\right)+\cos\left(\frac{\pi}{n}\right)+\cos\left(\frac{3\pi}{2n}\right)+\cdots+\cos\left(\frac{(n-1)\pi}{2n}\right)\right]$$

I do not know how to write this formula out first as a sum formula, any help would be appreciated!

Answer 1

Note that

$$\cos\left(\frac{n\pi}{2n}\right)=\cos\left(\frac{\pi}{2}\right)=0$$

So

$$\cos\left(\frac{\pi}{2n}\right)+\cdots+\cos\left(\frac{(n-1)\pi}{2n}\right)=\cos\left(\frac{\pi}{2n}\right)+\cdots+\cos\left(\frac{(n-1)\pi}{2n}\right)+\cos\left(\frac{n\pi}{2n}\right)$$

And hence

$$\lim_{n\to\infty}\frac{\pi}{2n}\sum_{k=1}^{n-1}\cos\left(\frac{k\pi}{2n}\right)=\lim_{n\to\infty}\frac{\pi}{2n}\sum_{k=1}^{n}\cos\left(\frac{k\pi}{2n}\right)$$

and then

$$\lim_{n\to\infty}\frac{\pi}{2n}\sum_{k=1}^{n}\cos\left(\frac{k\pi}{2n}\right)=\frac{\pi}{2}\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\cos\left(\frac{k\pi}{2n}\right)=\frac{\pi}{2}\int_0^1\cos\left(\frac{x\pi}{2}\right)dx$$