Infinite series

$1 + \frac{1}{3^2} + \frac{1}{5^2} +.... = (1+\frac{1}{2^2} + \frac{1}{3^2} + ....) - (\frac{1}{2^2} + \frac{1}{4^2} +\frac{1}{6^2}+...)$

= $\frac{\pi^2}{6} - \frac{1}{2^2} (\frac{\pi^2}{6} )= \frac{\pi^2}{6} - \frac{1}{4} (\frac{\pi^2}{6}) = \frac{3}{4}(\frac{\pi^2}{6}) = \frac{\pi^2}{8} = \boxed{1.23}$ .

Observe that

$S = \zeta(2) - \sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \zeta(2) - \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} = \zeta(2) - \frac{1}{4} \zeta(2) = \frac{3}{4} \zeta(2) \approx 1.2337$