# Summation of a trigonometric series

1. Oct 21, 2012

### fLambda

1. The problem statement, all variables and given/known data

Find the limit of the series $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} cos (i \theta / n)$, 0≤θ≤π/2

2. Relevant equations

3. The attempt at a solution

I know that the expansion looks like $\cos \theta / n + cos 2 \theta / n + ... + cos \theta$, but I couldn't begin to guess how to state this as a function of θ. It doesn't even look like it converges. To be honest I think I may have derived the series incorrectly, because it should converge on a value. My main question is whether it will converge on a function of θ, and if so, how I might look for it.

2. Oct 21, 2012

### bossman27

I'm no expert, but I'm trying to figure this out and I'm a little confused by your notation. Why is there an 'i' and an 'n' in your series term?

3. Oct 21, 2012

### Dick

You are right. It doesn't converge. You can tell just by looking at it. As n->infinity, θ/n goes to 0 so cos(θ/n) goes to 1. I can say the same thing about 2θ/n. For large n you have an awful lot of terms close to 1 and the rest are positive. It diverges.

Last edited: Oct 21, 2012
4. Oct 21, 2012

### fLambda

I probably did the notation wrong. I was aiming to describe the expansion I gave as a series sum but I wouldn't be close to stunned if I did it wrong (all of my series knowledge is self-taught and scantily so).

What I was aiming at was, "I want to split an angle θ into n equal angles, then get n distinct angles by multiplying θ/n by all the integers i from 1 to n. Then I want to take the cosine of each angle iθ/n and take the sum of all those cosines, then take the limit of that sum as n goes to infinity."

I was also reviewing my notes and I missed a step at the end: once I have this sum of cos iθ/n as n goes to infinity, I want to divide it by n^2. So it actually should converge on a function of θ, I think. Anything you can suggest or even any tutorials anyone knows of on this kind of series would be staggeringly helpful.

5. Oct 21, 2012

### Dick

Its pretty easy to show the sum from 1 to n is less than or equal to n. So if you divide by n^2 it converges all right. To zero. What problem are you actually trying to solve? Is it a Riemann sum?

6. Oct 21, 2012

### bossman27

This is just an off-the-wall reaction, but isn't this series the same as $\int^{\theta}_{0} cos \theta d\theta$
From what I can tell, your splitting θ into infinitesimally small pieces (i.e. dθ's) and then summing them from $\frac{i\theta}{n} = 0$ to $\frac{i\theta}{n} = \theta$.
How is this different than taking the integral over [0, θ]?

7. Oct 21, 2012

### Dick

It would be that if there were a factor of 1/n in front of the sum. You need to multiply the values by the length of each interval.

8. Oct 21, 2012

### fLambda

This actually came out of an analysis of trebuchets I'm doing for physics class. I'm trying to use this series sum (really the integral you've been describing, but I've been avoiding that because my calculus is weak) times a gravitational field strength and a length and mass of a pivoting arm to find the work done by gravity on a point mass on that arm. I'm evidently a little out of my depth with the math, but the rest of it is going fine so I want to press forward with this method. Do you recommend I just brush up my integration or even just consult an integral table?

9. Oct 21, 2012

### Dick

First try and figure out where your physics analysis missed a factor of 1/n. If you can find it, then, sure. It's MUCH easier to work out the integral of cos(t) from 0 to theta than to sum your series and find the limit. You'd need to use complex exponentials and geometric series to do that.

10. Oct 21, 2012

### fLambda

Okay, looking at my physics, it looks like the series expansion should look like $\lim_{n \rightarrow \infty} (1/n) (cos \theta / n + cos 2 \theta / n +...+ cos \theta)$. Does this help? Thanks, by the way, for the help and the patience. I was a little nervous about posting this because I know my notation is shaky and my integration is worse, so thanks for being kind!

11. Oct 21, 2012

### Dick

No problem. With the factor of 1/n it is a Riemann sum approximation to the integral of cos(t) from 0 to theta. So that is the limit as n->infinity. Doing the calculus is much easier than evaluating the limiting sum. That's what calculus is for. If you are doing physics I really think it would be better to take that route.