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

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?

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

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

• Ok.. How you understood that? Now it looks obvious. But how you got that intuition?
– N S
Mar 17 '15 at 15:16
• I say this is the secret of even numbers : They are twice the natural numbers ! Mar 17 '15 at 15:18
• Oh yes yes... A lil bit of explaination would have helped.. Not so fluent in maths..
– N S
Mar 17 '15 at 15:19
• shouldnt be $\frac 3 4 * \frac 1 6$ be $\frac 1 8$
– N S
Mar 17 '15 at 15:22

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($k\geq2)$

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