Series

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

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

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

$=\frac{\pi^2}6-\frac14\frac{\pi^2}6$

$=\frac{\pi^2}8$

$\approx\boxed{1.234}.$

Let

$\begin{aligned} S &= \sum_{n=i}^{\infty} \frac{1}{n^2} = \frac{1}{1^2}+\frac{1}{2^2} + \frac{1}{3^2} + \dots \\ A &= \sum_{n=i}^{\infty} \frac{1}{(2n-1)^2} = \frac{1}{1^2}+\frac{1}{3^2} + \frac{1}{5^2} + \dots \end{aligned}$

Hence,

$\begin{aligned} S-A &= \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots \\ S - A &= \frac{1}{2^2} \left ( \frac{1}{1^2}+\frac{1}{2^2} + \frac{1}{3^2} + \dots \right ) \\ S - A &= \frac{1}{2^2} S \\ A &= \frac{3}{4}S \end{aligned}$

It is known that $S = \frac{\pi ^2}{6}$ , thus $A = \frac{1}{8}\pi ^2 \approx 1.2337$ .