# How To Prove:$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4} = -\frac{7}{{720}}{\pi ^4}$

When I tried to solve this integral: $$\int_0^\infty {\frac{{{x^3}}}{{1 + {e^x}}}} \;{\rm{d}}x$$ I had trouble computing the sieries: $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4}$$ Thanks.

• – Watson Jan 27 '17 at 20:47

$$\sum_{n=1}^{+\infty}\frac{(-1)^n}{n^4} = -\sum_{n=1}^{+\infty}\frac{1}{n^4}+2\sum_{m=1}^{+\infty}\frac{1}{(2m)^4} = \left(-1+\frac{2}{16}\right)\sum_{n=1}^{+\infty}\frac{1}{n^4}=-\frac{7}{8}\zeta(4) = -\frac{7\pi^4}{720}.$$

• And many nice ways to show $\zeta(4) = \frac{\pi^4}{90}$ are given in this post: math.stackexchange.com/questions/28329/… . I find the proof using Parseval's identity especially nice. – Travis Oct 10 '14 at 14:00
• PERFECT solution! – Shine Mic Oct 10 '14 at 14:01

From the definition of the polylogarithm, we have that $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4}=\mathrm{Li}_4(-1).$$

Since $\mathrm{Li}_s(-1)=-\eta(s)$, we then seek to evaluate $-\eta(4)$. This is easy, as

$$\eta(s)=(1-2^{1-s})\zeta(s).$$

This would imply that

$$-\eta(4)=-(1-2^{-3})\zeta(4),$$

and since $\zeta(4) = \pi^4/90$, we have

$$-\eta(4)=-\frac{7\pi^4}{720}.$$

Note that $\eta(s)$ denotes the Dirichlet eta function.

• Another Nice Proof:) – Shine Mic Oct 10 '14 at 14:12