I found a proof of $$5 \sum_{m=1}^{\infty} \frac{1}{m^4} = 2 \left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right)^2$$ by rearranging sums and wrote it up. The argument is just 1.5 pages, the other 4.5 are explanations and context.
Here is a summary using divergent sums; see the write up for a correct version. Set $$h(m,n) = \begin{cases} \frac{1}{m^3 (n-m)} & m \neq n,\ m \neq 0 \\ 0 & m=n \ \mbox{or} \ m=0 \end{cases}.$$$$h(m,n) = \begin{cases} \frac{1}{m^3 (n-m)} & m \neq n,\ m \neq 0, \\ 0 & m=n \ \text{or} \ m=0. \end{cases}$$ Then we should have $$\sum_{(m,n) \in \mathbb{Z}^2} h(m,n) - h(n,2n-m) =0$$ as every value occurs twice with opposite signs. So $\sum_{(m,n) \in \mathbb{Z}^2} g(m,n)=0$ where $$g(m,n) := h(m,n) - h(n,2n-m) = \begin{cases} \frac{m^2+mn+n^2}{m^3 n^3} & m \neq n,\ m, n \neq 0 \\ - \frac{1}{m^4} & n=0,\ m \neq 0 \\ - \frac{1}{n^4} & m=0,\ n \neq 0 \\ 0 & m=n \end{cases}$$ Group together the terms where $(|m|, |n|)$ have a common value; we get $\sum_{(m,n) \in \mathbb{Z}_{\geq 0 }^2} f(m,n) =0$ where $$f(m,n) = \begin{cases} \frac{4}{m^2 n^2} & m \neq n,\ m,n >0 \\ - \frac{2}{m^4} & m>n=0 \\ - \frac{2}{n^4} & n>m=0 \\ - \frac{2}{m^4} & m=n>0 \\ 0 & m=n=0\\ \end{cases}. $$$$f(m,n) = \begin{cases} \frac{4}{m^2 n^2} & m \neq n,\ m,n >0, \\ - \frac{2}{m^4} & m>n=0, \\ - \frac{2}{n^4} & n>m=0, \\ - \frac{2}{m^4} & m=n>0, \\ 0 & m=n=0. \end{cases} $$ Writing this out, $4\zeta(2)^2 - 6 \zeta(4) - 4 \zeta(4)=0$, as desired.
Has anyone seen this? If this is new, I'm thinking of sending it to the Monthly.
