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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 $\frac{-7\pi^4}{720}$.

Git Gud
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Rudolf
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    This has been asked here: http://math.stackexchange.com/questions/1169861/use-sum-n-1-infty-frac1n4-frac-pi490-to-compute-sum-n-1 – N.U. Mar 01 '15 at 23:38

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Hint: $$\sum_{n=1}^\infty \frac{(-1)^n}{n^4} =\sum_{n=1}^\infty \frac{1}{(2n)^4}- \sum_{n=1}^\infty \frac{1}{(2n-1)^4}$$ and $$\sum_{n=1}^\infty \frac{1}{(2n-1)^4} = \sum_{n=1}^\infty \frac{1}{n^4}-\sum_{n=1}^\infty \frac{1}{(2n)^4}$$

graydad
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