<s>[INST] Are there other methods to evaluate $\frac{1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots}$? Are there other methods to evaluate the following series?
 $$\frac{1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots}$$
My attempt is as follows,
\begin{align}
\frac{1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots} &=\frac{x}{y} \\
\frac{1^{-4}+3^{-4}+\cdots+2^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots} &=\frac{x}{y} \\
\frac{y+2^{-4}(1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots)}{y} &=\frac{x}{y} \\
\frac{y+2^{-4}x}{y} &=\frac{x}{y} \\
\frac{x}{y} &= \frac{1}{1-2^{-4}}
\end{align}
 [/INST] Your proof is fine. 
The sigma notation can make some things cleaner. The below is the same argument, just written with different notation: $$x=\sum_{n=1}^\infty \frac{1}{n^4}\\y=\sum_{n=1}^\infty \frac{1}{(2n-1)^4}$$
Then $$x=\sum_{n=1}^{\infty} \left(\frac{1}{(2n-1)^4}+\frac{1}{(2n)^4}\right)=y+\frac{1}{2^4}x$$
and thus $$\frac{x}{y} = 1+\frac{1}{2^4}\frac{x}{y}\implies\\\frac{x}{y}=\frac{1}{1-\frac{1}{2^4}}$$
I suppose that the other approach is to note that every $n$ can be written uniquely as $2^k(2m+1)$. So:
$$\begin{align}x = \sum_{n=1}^\infty \frac{1}{n^4} &= \sum_{k=0}^\infty\sum_{m=1}^{\infty} \frac{1}{(2^k(2m+1))^4}\\
&=\left(\sum_{k=0}^\infty \frac{1}{2^{4k}}\right)\left(\sum_{m=0}^\infty \frac{1}{(2m+1)^4}\right)\\
&=\frac{1}{1-\frac{1}{2^4}}y
\end{align}$$
</s>