Given that $$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$$ find $$\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$$ The answer should be $\frac{-7\pi^4}{720}$.
Is it possible to use the fact that $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$ to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$?
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