Does the series converge to pi^2/8?

[Blitzy89]

Homework Statement

Homework Equations

[PLAIN]http://img577.imageshack.us/img577/6756/physforumsquestion.png

The Attempt at a Solution

I am pretty much stuck guys. Any help will be greatly appreciated!

 

[SammyS]

How are those two things related ?

 

[tiny-tim]

Hi Blitzy89!

(have a pi: π and a sigma: ∑ )

It's half of ∑1/n² - ∑1/(-n)² … does that help?

 

[Char. Limit]

Hi Blitzy89!

(have a pi: π and a sigma: ∑ )

It's half of ∑1/n² - ∑1/(-n)² … does that help?

Something tells me that can't be right... doesn't the difference of sums you mentioned always equal 0?

 

[tiny-tim]

oops!

oops!

maybe that should have been ∑1/n² - ∑1/(2n)²

thanks char! ​

(have i got it right this time? )

 

[uart]

How are those two things related ?

The idea is that if you can find (or perhaps just happen to know) a function with that particular Fourier series then you can integrate it to find the sum.

There's a way to do it using contour integration and residues. I can post more details if you want to go that way.

 

[uart]

oops!

maybe that should have been ∑1/n² - ∑1/(2n)²

thanks char! ​

(have i got it right this time? )

Doesn't that just give \Sigma \frac{3}{4 n^2}? Oh well there might be something in it as numerically it gives the right answer.

 

[dextercioby]

If you're allowed to use \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^{2}}{6} without proving it, then the hint in post #5 easily solves the problem.

If you must prove your hypothesis, then search the forums for a solution. There was a discussion on this a few weeks ago.

 

[Deadstar]

Doesn't that just give \Sigma \frac{3}{4 n^2}? Oh well there might be something in it as numerically it gives the right answer.

If you look at OPs question again, you'll see that it's basically asking you to find the sum of the squares of the odd integers. All tiny tim did was to rewrite this the sum of all integers minus the sum of all even integers.
Now since we know what the sum of the square of all integers is, the question becomes pretty straightforward.

 

[uart]

If you look at OPs question again, you'll see that it's basically asking you to find the sum of the squares of the odd integers. All tiny tim did was to rewrite this the sum of all integers minus the sum of all even integers.
Now since we know what the sum of the square of all integers is, the question becomes pretty straightforward.

Thanks. I could see that it gave the right answer, 3/4 \times \pi^2/6, I just wasn't sure why

 

[uart]

BTW. If you want to do it the hard way I can still do it with a contour integral.

 

[Char. Limit]

oops!

maybe that should have been ∑1/n² - ∑1/(2n)²

thanks char! ​

(have i got it right this time? )

Yeah, that's right. And it gives the right answer too, very easily in fact.