Help with Summation Proof: \sum\frac{1}{(2j-1)^2}

[mathsgeek]

\sum\frac{1}{(2j-1)^2}

This fgoes from j=1 to infinity. I was just wondering if somebody could calculate and show all working to show the value that this function converges to as i have no idea of how to do this? Thanks for your help

 

[Dick]

No one is just going to work it out for you with no input from you. Do you know how to sum 1/j^2?

 

[mathsgeek]

I never said i wasnt going to input, I am happy to put input in, i just don't know how to start it. And the sum from 1/j^2 is 2 using geometric sequences. Where do i go from here tho?

 

[Dick]

The sum of 1/j^2 isn't 2 and it isn't geometric. The sum is pi^2/6. Proving that is tricky, you have to use complex analysis or worse. My guess is that you were given that and you are expected to use that to find the sum 1/(2j-1)^2. Hint: those are the odd terms in the series 1/1^2+1/2^2+1/3^2+1/4^2+... Can you find the sum of the even terms?

 

[mathsgeek]

Just a q, is doing the sum 1/j^2 going to help in the sum i want to calculate. Also, what do you mean by even terms as i used it when j=1,2,3, what do you specifically mean by even terms? Thanks for ur help so far, i know complex numbers and all that so i could always use complex analysis but is it easier this way, or how would i start it using complex analysis>?

 

[Dick]

I wouldn't try computing sum 1/j^2 from scratch, it's too hard. The value is zeta(2)=pi^2/6 (where zeta is the Riemann zeta function). If you want some samples of proofs google for "zeta(2)" or "sum of inverse squares". On the other hand figuring sum 1/(2j-1)^2 is not that hard if you already know sum 1/j^2. What I mean by even and odd is split the 1/j^2 series into terms where j is even and j is odd. Your sum 1/(2j-1)^2 is the odd ones.

 

[mathsgeek]

Well, i know the odd series is 1+1/9+1/25+... but what process would you add them all together, like what method would you use?

 

[Dick]

If you add the sum of the odds and the sum of the evens, you get pi^2/6. Can you figure out the sum of the evens??

 

[mathsgeek]

The thing is i don't know how to add them because am i not supposed to use geometric sequence cause it isn't one. The sequence for evens is 1/4+1/16+1/36...

 

[Dick]

The thing is i don't know how to add them because am i not supposed to use geometric sequence cause it isn't one. The sequence for evens is 1/4+1/16+1/36...

Factor 1/4 out of all those terms. Does what's left over look familiar??

 

[mathsgeek]

werll 1/4 sum 1/j^2 for odds numbers equals the sum 1/j^2 for all numbers

 

[Dick]

werll 1/4 sum 1/j^2 for odds numbers equals the sum 1/j^2 for all numbers

You mean evens, right? I'm hoping you also mean (1/4)*sum of all=sum of evens. So if sum 1/j^2 for all numbers is pi^2/6, what is that? Can you use that to deduce the sum of the odds?

 

[mathsgeek]

I got pi^2/8, so therefore, sum of j=1 to infinity of 1/(2j-1)^2 is basically pi^2/8 because theyre are equal, this is correct isn't it?

 

[Dick]

I got pi^2/8, so therefore, sum of j=1 to infinity of 1/(2j-1)^2 is basically pi^2/8 because theyre are equal, this is correct isn't it?

That's not a very clear description of what you did, but yes, the sum of the odds is pi^2/8.

 

[mathsgeek]

 

[Dick]

That looks very nice.