Does: $$\sum_{1 \leq i<j} \frac{1}{i j^2} = \sum_{1 \leq k} \frac{1}{k^3}?$$
Motivation: Call the above sum $S$, and let $$T := \sum_{ GCD(i,j)=1} \frac{1}{\max(i,j) i j}.$$ The sum $T$ came up in a computation on Jim Propp's question here. Numerical computation suggested that $T$ is extremely close to $3$.