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 $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$?

Solution 1:

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

Solution 2:

Note

$$\sum_{n=1}^\infty\frac{1}{n^4}+\sum_{n=1}^\infty\frac{(-1)^n}{n^4}=2\left(\frac{1}{2^4}+\frac{1}{4^4}+\dots+\right)=\frac{1}{8}\left(\frac{1}{1^4}+\frac{1}{2^4}+\dots\right)=\frac{1}{8}\sum_{n=1}^\infty\frac{1}{n^4}$$

Solution 3:

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 $$E=\sum_{n=1, n=even}^\infty \frac{1}{n^4}=\sum_{k=1}^\infty \frac{1}{(2k)^4}=\frac{1}{16}\sum_{k=1}^\infty \frac{1}{k^4}=\frac{1}{16}\frac{\pi^4}{90}$$