<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] (Here I'll assume that $-1^k$ here ought to be $(-1)^k$.) Hint:
$$\sum_{n = 1}^{\infty} \frac{(-1)^n}{n^4} = -\sum_{n = 1}^{\infty} \frac{1}{n^4} + 2 \sum_{n > 1 \text{ even}}^{\infty} \frac{1}{n^4} = - \sum_{n = 1}^{\infty} \frac{1}{n^4} + 2 \sum_{n = 1}^{\infty} \frac{1}{(2 n)^4}.$$

 This is $$- \sum_{n = 1}^{\infty} \frac{1}{n^4} + \frac{1}{8} \sum_{n = 1}^{\infty} \frac{1}{n^4} = - \frac{7}{8} \sum_{n = 1}^{\infty} \frac{1}{n^4}.$$

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