<s>[INST] Use $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$ to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$ Is it possible to use the fact that $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$ to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$?
 [/INST] Hint:
Let 
$$E=\sum_{n=1, n=even}^\infty \frac{1}{n^4}$$
$$O=\sum_{n=1, n=odd}^\infty \frac{1}{n^4}$$
You are given that $E+O=\frac{\pi^4}{90}$ and want to find $E-O$.
To do this note that
$$E=\sum_{n=1, n=even}^\infty \frac{1}{n^4}=\sum_{k=1}^\infty \frac{1}{(2k)^4}=\frac{1}{16}\sum_{k=1}^\infty \frac{1}{k^4}=\frac{1}{16}\frac{\pi^4}{90}$$
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