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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1This question is similar to: Evaluating a summation of inverse squares over odd indices. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – peterwhy Sep 15 '24 at 23:07
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Hint: $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}+\sum_{n=1}^{\infty}\frac{1}{(2n)^2}=\sum_{n=1}^{\infty}\frac{1}{n^2}$$
Julian Kuelshammer
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Kortlek
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Hints:
$$n\in\Bbb N:=\{1,2,\ldots\}:\;\;\;\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n)^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac14\sum_{n=1}^\infty\frac1{n^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}\ldots$$
DonAntonio
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I simply not understanding how the right and left side are equal of the series. – Myshkin May 17 '13 at 05:06
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2@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 07:20
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$1^2+{1\over 2^2}+{1\over 3^2}+\dots+{1\over n^2}+\dots={1\over (2n)^2}+{1\over (2n-1)^2}+\dots$ am I right? – Myshkin May 17 '13 at 08:00
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