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I ran into this question:

Prove that:


Thank you very much in advance.

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

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thank you very much – user76508 May 12 '13 at 14:28
@user76508 No problem, I am loyal to the Workers. – Kortlek May 12 '13 at 22:23
@RobFord I thought you were libertarian? – Anonymous - a group Oct 9 '14 at 16:05


$$n\in\Bbb N:=\{1,2,\ldots\}:\;\;\;\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n)^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac14\sum_{n=1}^\infty\frac1{n^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}\ldots$$

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thank you very much – user76508 May 12 '13 at 14:27
I simply not understanding how the right and left side are equal of the series. – Un Chien Andalou May 17 '13 at 5:06
please help me to understand. – Un Chien Andalou May 17 '13 at 5:07
@Tsotsi: just divide the sum of the reciprocals of the squared naturals in even ones $\,(2n)^{-2}\,$ and odd ones $\,(2n-1)^{-2}\,$ , but then $\,(2n)^{-2}=2^{-2}\cdot n^{-2}\,$ and etc. – DonAntonio May 17 '13 at 7:20
$1^2+{1\over 2^2}+{1\over 3^2}+\dots+{1\over n^2}+\dots={1\over (2n)^2}+{1\over (2n-1)^2}+\dots$ am I right? – Un Chien Andalou May 17 '13 at 8:00

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