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If $ \sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6} $ then $ \sum _{n=1}^{\infty} \frac 1 {(2n -1)^2} $

Dont know what kind of series is this. Please educate. How to do such problems?

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$$\sum_{n=1}^\infty \frac1{n^2} =\sum_{n=1}^\infty \frac1{(2n)^2} + \sum_{n=1}^\infty \frac1{(2n-1)^2}=\frac14 \sum_{n=1}^\infty \frac1{n^2} + \sum_{n=1}^\infty \frac1{(2n-1)^2}$$

So, $\sum\limits_{n=1}^\infty \frac1{(2n-1)^2}= \frac34 \sum\limits_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}{8}$.

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  • $\begingroup$ Ok.. How you understood that? Now it looks obvious. But how you got that intuition? $\endgroup$
    – N S
    Mar 17 '15 at 15:16
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    $\begingroup$ I say this is the secret of even numbers : They are twice the natural numbers ! $\endgroup$ Mar 17 '15 at 15:18
  • $\begingroup$ Oh yes yes... A lil bit of explaination would have helped.. Not so fluent in maths.. $\endgroup$
    – N S
    Mar 17 '15 at 15:19
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    $\begingroup$ shouldnt be $ \frac 3 4 * \frac 1 6 $ be $ \frac 1 8 $ $\endgroup$
    – N S
    Mar 17 '15 at 15:22
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you can use the relation $$\sum_{n=1}^\infty \frac{1}{(2n-1)^k}=(1-\frac{1}{2^k})\sum_{n=1}^\infty\frac{1}{n^k}$$ when the $k$ even integer number($k\geq2) $

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  • $\begingroup$ this should work for $k > 1$, integer or not $\endgroup$
    – MCT
    Mar 17 '15 at 17:40
  • $\begingroup$ Is it true would you please provide some proof $\endgroup$
    – Sushil
    Mar 17 '15 at 17:41
  • $\begingroup$ @Soke this for $k=2,4,6,.....$ $\endgroup$
    – E.H.E
    Mar 17 '15 at 18:10

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