I'm trying to prove or disprove the absolute convergence of the sum $\sum_\limits{k,l=1}^{\infty}\frac{k-l}{k^4+l^4} $.
Every series that I found that bounds the absolute values of my original series from above diverges and every series that bounds it from below converges, so I'm stuck with respect to the comparison test.
We have that $$ S := \sum_{k,l = 1}^{\infty} \frac{|k-l|}{k^4 + l^4} = \sum_{n=2}^{\infty} \sum_{k,l;\, k+l=n} \frac{|k-l|}{k^4 + l^4} \le \sum_{n=2}^{\infty} n \sum_{k,l;\, k+l=n} \frac{1}{k^4 + l^4}.$$ Now, notice that for any $a, b \in \mathbb{R}$ we have $(a + b)^2 \le 2a^2 + 2b^2$. Iterating, $$ (a+b)^4 \le [2a^2 + 2b^2]^2 = 4(a^2 + b^2)^2 \le 8(a^4 + b^4).$$ Hence, we can write $$ S \le \sum_{n=2}^\infty n \sum_{k,l;\, k+l=n} \frac{8}{(k+l)^4} = 8\sum_{n=2}^\infty \frac{1}{n^2} < \infty.$$