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I ran into this question:

Prove that:

$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\frac{3}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}$$

Thank you very much in advance.

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2 Answers 2

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
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@user76508 No problem, I am loyal to the Workers. –  Ron Ford May 12 '13 at 22:23

Hints:

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

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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. –  Une Femme Douce May 17 '13 at 5:06
    
please help me to understand. –  Une Femme Douce May 17 '13 at 5:07
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@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? –  Une Femme Douce May 17 '13 at 8:00

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