What is the solution to this tricky series riddle?

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Discussion Overview

The discussion revolves around the evaluation of the infinite series \(\sum 1/(2n + 1)^2\) from \(n = 0\) to infinity. Participants explore methods to relate this series to the known result of \(\sum 1/n^2 = \pi^2/6\) and seek to clarify the steps involved in the calculation.

Discussion Character

  • Mathematical reasoning, Homework-related

Main Points Raised

  • One participant states that the answer to the series is \(\pi^2/8\) but expresses confusion about the calculation process.
  • Another participant suggests that knowing the value of \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) can help find \(\sum_{n=1}^{\infty} \frac{1}{(2n)^2}\) and subsequently the desired series.
  • A third participant reiterates the need to calculate \(\sum_{n=1}^{\infty} \frac{1}{(2n)^2}\) and relates it to the original series.
  • One participant humorously admits to being confused due to having had too much to drink, indicating that the problem was not as difficult as initially thought.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the solution, and the discussion includes various approaches and expressions of confusion regarding the calculations.

Contextual Notes

There are unresolved steps in the calculations, and the participants rely on known results without fully detailing the derivations or justifications for their claims.

joris_pixie
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Hey, guys !
I have to calculate:

[tex]\sum 1/(2n + 1)^2[/tex]
n from 0 to infinite

Knowing that 1/n² = (pi)²/6

The answer is (pi)²/8

I don't know why I'm stuck here but I just don't get it...

Greetings
 
Last edited:
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Once you have what

[tex]\sum_{n=1}^{ \infty } \frac{1}{n^2}[/tex] is, you should be able to find what

[tex]\sum_{n=1}^{ \infty } \frac{1}{(2n)^2}[/tex] is pretty easily. And then use both of these to find the series sum that you're looking for
 
It is given ...
Calculate:
[tex]\sum^{\infty}_{0} 1/(2n +1)^2[/tex]

if you know that:

[tex]\sum^{\infty}_{1} 1/n^2 = (pi)^2 /6[/tex]But I can't make it work :(
 
Can you calculate this? [tex] \sum_{n=1}^{ \infty } \frac{1}{(2n)^2}[/tex]

Then what is [tex] \sum_{n=1}^{ \infty } \frac{1}{(2n)^2} + \sum_{n=0}^{ \infty } \frac{1}{(2n+1)^2}[/tex]
 
haha I'm sorry for wasting your time !
It was infact not hard, but had a few drinks too much last night.

http://www.pixie.be/solution.JPG thank you anyway !
 
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