Given $\frac{1}{1^4}+\frac{1}{2^4}...\infty=\frac{\pi^4}{90}$

Question

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.]

Answer 1

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.

Answer 2

HINT:

$$\sum_{n=1}^{2N} a_n=\sum_{n=1}^N (a_{2n-1}+a_{2n})$$

Answer 3

$$\sum_{n=1}^{\infty}\frac{1}{n^4}=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^4}+\sum_{n=1}^{\infty}\frac{1}{(2n)^4}$$ $$\sum_{n=1}^{\infty}(\frac{1}{n^4}-\frac{1}{16n^4})=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^4}$$ $$(1-\frac{1}{16})\sum_{n=1}^{\infty}\frac{1}{n^4}=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^4}$$

Answer 4

\begin{eqnarray*} S_{even}=\sum_{i=1}^{\infty} \frac{1}{(2i)^4} = \frac{1}{16} S_{All} \\ S_{Odd}=\sum_{i=1}^{\infty} \frac{1}{(2i-1)^4} \end{eqnarray*} and of course $S_{even}+S_{Odd}=S_{All}$.