Given $\frac{1}{1^4}+\frac{1}{2^4}...\infty=\frac{\pi^4}{90}$ If $$\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}...\infty=\frac{\pi^4}{90}$$
then, $$\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}...\infty=?$$
Please provide a hint instead of the complete answer. Thanks. [Please edit the tags for me as I have no idea which sector this problem lies in.]

hint: For the even terms: $\dfrac{1}{(2n)^4} = \dfrac{1}{16}\cdot \dfrac{1}{n^4}$, and you get to solve $S + \dfrac{S}{16} = \dfrac{\pi^4}{90}$, with $S$ is the desire sum.
