Lets consider the sum $$1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}-...=\sum_{i=1}^{\infty } \frac{(-1)^{i+1}}{(2 i-1)^3}=\frac{\pi^3}{32}$$
And as we know there is no known closed form neither for the $\zeta(3)$, nor for the $$1+\frac{1}{3^3}+\frac{1}{5^3}+\frac{1}{7^3}+...=\sum_{i=1}^{\infty } \frac{1}{(2 i-1)^3}=\frac{7\zeta(3)}{8}$$
This contrast is really strange, and here just asking to share the thoughts about the possible reason, why the first series is so easy to map to simple relation, while the second one is not.
Just an addition: $$1-\frac{1}{3^5}+\frac{1}{5^5}-\frac{1}{7^5}-...=\sum_{i=1}^{\infty } \frac{(-1)^{i+1}}{(2 i-1)^5}=\frac{5\pi^5}{1536}$$
$$1-\frac{1}{3^7}+\frac{1}{5^7}-\frac{1}{7^7}-...=\sum_{i=1}^{\infty } \frac{(-1)^{i+1}}{(2 i-1)^7}=\frac{61\pi^7}{184320}$$
and so on.
$$1+\frac{1}{3^5}+\frac{1}{5^5}+\frac{1}{7^5}+...=\sum_{i=1}^{\infty } \frac{1}{(2 i-1)^5}=\frac{31\zeta(5)}{32}$$
$$1+\frac{1}{3^7}+\frac{1}{5^7}+\frac{1}{7^7}+...=\sum_{i=1}^{\infty } \frac{1}{(2 i-1)^7}=\frac{127\zeta(7)}{128}$$