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

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Homework Help Overview

The discussion revolves around the summation of the series \(\sum\frac{1}{(2j-1)^2}\) from \(j=1\) to infinity. Participants are exploring how to calculate the convergence of this series and its relation to known sums, particularly the sum of the inverse squares.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the sum of odd terms and the sum of all terms in the series of inverse squares. Questions arise about the nature of the series and the methods to approach the summation, including the use of complex analysis and the Riemann zeta function.

Discussion Status

There is ongoing exploration of the relationship between the sums of odd and even terms in the series. Some participants have provided hints and guidance on how to approach the problem, while others express uncertainty about the methods and calculations involved.

Contextual Notes

Participants are navigating assumptions about the series and the definitions of odd and even terms. There is mention of the difficulty of proving certain sums and the constraints of not using geometric sequences for this particular problem.

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[tex]\sum[/tex][tex]\frac{1}{(2j-1)^2}[/tex]

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
 
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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?
 
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?
 
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?
 
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>?
 
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.
 
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?
 
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??
 
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...
 
  • #10
mathsgeek said:
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??
 
  • #11
werll 1/4 sum 1/j^2 for odds numbers equals the sum 1/j^2 for all numbers
 
  • #12
mathsgeek said:
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?
 
  • #13
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?
 
Last edited:
  • #14
mathsgeek said:
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.
 
  • #15
29wwk5j.jpg
 
  • #16
That looks very nice.
 

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