If $\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$ then $\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}$ is equal to
(A) $\frac{{\pi}^2}{24}$ (B) $\frac{{\pi}^2}{8}$ (C) $\frac{{\pi}^2}{6}$ (D) $\frac{{\pi}^2}{3}$
Try to write the summation as sum of square of reciprocal of odd numbers and even numbers and take the advantage of the infinite sum
$\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$
$\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n)^2} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}= \frac{{\pi}^2}{6} $
$\Rightarrow \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{{n^2}} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{{\pi}^2}{6} $
we know $\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$
So from the above equation we get
Hence $\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{{\pi}^2}{6} - \frac{{\pi}^2}{6\cdot4}$
$\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{{\pi}^2}{8} $
So the correct answer is option B
