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!

$cos(\frac\pi2)=0$ so you could formally incorporate in in the sum. — Mirko, Nov 26 '14 at 05:23
Related : http://math.stackexchange.com/questions/465075/find-lim-limits-n-to-infty-frac1n-sum-limits2n-r-1-fracr-sq — lab bhattacharjee, Nov 26 '14 at 05:26

Aditya Hase · Accepted Answer · 2014-11-26T05:36:33.430

2

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

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

edited Nov 26 '14 at 05:36

answered Nov 26 '14 at 05:23

Aditya Hase

8,851

So, if I properly understood, the result should be $1$. I think it could be $0$. Where am I wrong ? – Claude Leibovici Nov 26 '14 at 07:23
This is precisely my problem ! I don't find anything wrong in your post. Since I had a doubt, I checked numerically. What if we write that $\cos\left(\frac{(n-1)\pi}{2n}\right)=\sin \left(\frac{\pi }{2 n}\right)$ ? – Claude Leibovici Nov 26 '14 at 07:29
Interesting, isn't it ? Cheers :-) – Claude Leibovici Nov 26 '14 at 07:34

Expressing limit of sum definite integral

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