Calculating cos(2nπ/3)/(n²) using the Sum of Series 1/n²

[andrey21]

Using the fact that sum of series 1/n^2 is (pi)^2/6 FInd the following:





cos(2npi/3)/(n^2)



Not sure really where to start on this one any help would be great.

 

[micromass]

What did you already try to solve the problem?
Surely you can start by calculating $$\cos(2\pi n/3)$$...

 

[Hootenanny]

Using the fact that sum of series 1/n^2 is (pi)^2/6 FInd the following:





cos(2npi/3)/(n^2)



Not sure really where to start on this one any help would be great.

What is cos(2*n*pi/3) for integer n?

 

[andrey21]

Yes I thought about calculating cos(2*n*pi/3), but for how many values of n?? until it tends to a figure?

 

[micromass]

Calculate $$\cos(2n\pi/3)$$ until you notice a pattern.
Try to calculate it for n=0,1,2,3,4,5,... A pattern should pop up very quickly...

 

[Hootenanny]

Yes I thought about calculating cos(2*n*pi/3), but for how many values of n?? until it tends to a figure?

HINT: Cosine is periodic (with period 2Pi)

 

[andrey21]

So taking ur advice I calculated cos(2npi/3)/(n^2) as the question states. I can see it converges on a value of 1. Correct? I did n=0,1,2,3,4,5,6,7,8,9,10,100,1000,10000,100000,1000000

 

[micromass]

No, don't calculate $$\cos(2n\pi/3)/n^2$$. Only calculate $$\cos(2n\pi/3)$$. What happens if n=0,1,2,3,4,5?? Can you give me that values?

 

[andrey21]

Oh ok I've done that also I have 1,-0.5,-0.5,1,-0.5 correct??

 

[micromass]

Yes, so you see an easy pattern showing up:

$$\cos(2n\pi/3)=1 ~\text{if}~n=3k~\text{and}~\cos(2n\pi/3)=1/2~\text{otherwise}$$

So, can you now split up your series in two parts?

 

[andrey21]

From what you have said:
$$\cos(2n\pi/3)=1 ~\text{if}~n=3k~\text{and}~\cos(2n\pi/3)=1/2~\text{otherwise}$$

Should it be -1/2?

 

[micromass]

Yes, I'm sorry. It should be a -1/2. Sorry for the confusion

 

[andrey21]

Ok so splitting the series in two. What do u mean here? Use sum of 1/n^2 and claculate some of cos(2npi/3) and sum them together?

 

[micromass]

I mean the following

$$\sum_{n=1}^{+\infty}{\frac{\cos(2\pi n/3)}{n^2}}=\sum_{k=1}^{+\infty}{\frac{1}{(3k)^2}}-\frac{1}{2}\left(\sum_{k=1}^{+\infty}{\frac{1}{(3k+1)^2}}+\sum_{k=1}^{+\infty}{\frac{1}{(3k+2)^2}}\right)$$

 

[andrey21]

Oh ok I understand now, so simply subbing in values for k until it converges on a figure will be the answer?

 

[andrey21]

Am i rite to simply plug k values in and see what value it tends to. For examples:

k = 1 would give

1/9 - 1/2(1/16+1/25)

=0.05986...

 

[andrey21]

is this correct micromass??

 

[micromass]

Well, although it is not incorrect, I doubt that this is what they want you to do. You'll need to find the value of the sum EXACTLY. The thing you suggested will only be an approximation to the sum.

Let's consider the first sum

$$\sum_{n=1}^{+\infty}{\frac{1}{(3n)^2}}$$

How would you evaluate this sum??

 

[andrey21]

Erm well I am not to sure. evaluating it at some values for n i obtain:

1/9 + 1/36 + 1/81 + 1/144 + 1/225...

How do I get the excat value?

 

[micromass]

HINT: use that

$$\sum_{n=1}^{+\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6}$$

The only problem is that you have (3n)² instead of n². How would you solve that?

 

[andrey21]

Would that just be pi squared / 18

 

[micromass]

No, that is incorrect. How did you arrive there?

 

[andrey21]

Sorry I don't know what I was writing above. I havnt got a proof of how 1/n^2 = pi squared /6. The question just states to use this.

 

[andrey21]

Could u show me a link or a proof of 1/n^2 = Pi^2/6 micromass.

 

[micromass]

Hmmm, the proof of this is not so easy. It needs Fourier series and stuff, so I don't think you'll need to know the proof... But here it is: http://en.wikipedia.org/wiki/Basel_problem

 

[andrey21]

Oh i thought i cud just adapt that but use 3n^2 instead of n^2. If not how can I solve 1/3n^2?? I am really struggling with this question.

 

[micromass]

What does $$\frac{1}{(3n)^2}$$ equal? Just bring the 3 from under the square...

 

[andrey21]

So it equals:

1/9n^2 or 9n^(-2)

Is this correct?

 

[micromass]

Allright, so

$$\sum_{n=1}^{+\infty}{\frac{1}{(3n)^2}}=\sum_{n=1}^{+\infty}{\frac{1}{9}\frac{1}{n^2}}$$

Can you calculate the sum now?

 

[andrey21]

So 1/9n^2 = 1/9 . Pi^2/6

gives pi^2/54

 

[micromass]

Yes! So that's the first sum. Now try to evaluate

$$\sum_{n=1}^{+\infty}{\frac{1}{(2n+1)^2}}+\sum_{n=1}^{+\infty}{\frac{1}{(2n+2)^2}}$$

 

[andrey21]

Ye ok is it (2n+2)^2 or (3n+2)^2? Just in the first page it was 3 not 2 before the n.

 

[micromass]

Ow, my apologies, it is 3 instead of 2...

 

[andrey21]

Ok so from the two sums I get:

1/(9n^2+6n+1)

and

1/(9n^2 +12n+4)

 

[micromass]

Hmm, you won't be able to calculate those sums separately...
You'll have to use the following:

$$\sum_{n=1}^{+\infty}{\frac{1}{n^2}}=\sum_{n=1}^{+\infty}{\frac{1}{(3n)^2}}+\sum_{n=1}^{+\infty}{\frac{1}{(3n+1)^2}}+\sum_{n=1}^{+\infty}{\frac{1}{(3n+2)^2}}$$

You know two of the above series...