# Limit of a series

• May 23rd 2011, 09:56 PM
sparkyboy
Limit of a series
Hi guys, I am currently doing my homework in limits and derivatives, and there is one question that I'm stuck at. I think the correct answer is 0 but still want to ask for you guys' advice and thoughts (Clapping)

http://i617.photobucket.com/albums/t..._1610/lim2.jpg

I am quite skeptical about the functions inside the square brackets ... somehow I think they resemble a series, am I right? I wonder if that series has anything to do with this kind of limit. Any suggestions is highly appreciated. Thanks everyone.
• May 23rd 2011, 10:07 PM
Prove It
According to Wolfram, $\displaystyle \displaystyle \sum_{k = 1}^{n}{\sin{\left(\frac{k\pi}{n}\right)}} = \cot{\left(\frac{\pi}{2n}\right)}$.

What happens as $\displaystyle \displaystyle n \to \infty$?
• May 23rd 2011, 10:42 PM
CaptainBlack
Quote:

Originally Posted by sparkyboy
Hi guys, I am currently doing my homework in limits and derivatives, and there is one question that I'm stuck at. I think the correct answer is 0 but still want to ask for you guys' advice and thoughts (Clapping)

http://i617.photobucket.com/albums/t..._1610/lim2.jpg

I am quite skeptical about the functions inside the square brackets ... somehow I think they resemble a series, am I right? I wonder if that series has anything to do with this kind of limit. Any suggestions is highly appreciated. Thanks everyone.

As n goes to infinity (think Riemann sums here):

(1/n)[sin(pi/n)+sin(2pi/n+...+sin(npi/n)] -> integral sin(pi x) dx from 0 to 1

so:

sin(pi/n)+sin(2pi/n+...+sin(npi/n) = O(n)

That is the thing diverges.

(In fact sin(pi/n)+sin(2pi/n+...+sin(npi/n) ~= 2n/pi for large n)

CB
• May 23rd 2011, 11:02 PM
chisigma
Quote:

Originally Posted by sparkyboy
Hi guys, I am currently doing my homework in limits and derivatives, and there is one question that I'm stuck at. I think the correct answer is 0 but still want to ask for you guys' advice and thoughts (Clapping)

http://i617.photobucket.com/albums/t..._1610/lim2.jpg

I am quite skeptical about the functions inside the square brackets ... somehow I think they resemble a series, am I right? I wonder if that series has anything to do with this kind of limit. Any suggestions is highly appreciated. Thanks everyone.

The first step is the computation of...

$\displaystyle \sigma_{n}= \sum_{k=1}^{n} \sin (k \frac{\pi}{n}) = \frac{1}{2i}\ \sum_{k=1}^{n} (e^{i k \frac{\pi}{n}} - e^{-i k \frac{\pi}{n}})=$

$\displaystyle = \frac{1}{i}\ (\frac{e^{i \frac{\pi}{n}}}{1-e^{i \frac{\pi}{n}}}- \frac{e^{-i \frac{\pi}{n}}}{1-e^{-i \frac{\pi}{n}}}) = \frac{1}{i}\ (\frac{1}{1-e^{i \frac{\pi}{n}}}- \frac{1}{1-e^{-i \frac{\pi}{n}}}) =$

$\displaystyle = \frac{1}{2} \frac{e^{i \frac{\pi}{2n}} + e^{-i \frac{\pi}{2n}}} {\sin \frac{\pi}{2n}}} = \frac {\cos \frac{\pi}{2n}}{\sin \frac{\pi}{2n}}= \cot \frac{\pi}{2n}}$ (1)

The second step [just a bit more confortable...] is to evaluate the $\displaystyle \lim_{n \rightarrow \infty} \sigma_{n}$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$