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][1]. Numerical computation suggested that $T$ is extremely close to $3$.

It is not hard to show that
$$T = \zeta(3)^{-1} \sum \frac{1}{\max(i,j) i j} = \zeta(3)^{-1} \left( \sum_{k} \frac{1}{k^3} + 2 \sum_{i<j} \frac{1}{i j^2} \right) = 1 + 2 \zeta(3)^{-1} S,$$
by breaking into cases according to whether $i<j$, $i=j$ or $i>j$.
So $T=3$ iff $S=\zeta(3)$.

As I describe in the above linked thread, numerical computations suggest that the sums agree to $20$ digits of accuracy. What is going on?


  [1]: http://mathoverflow.net/questions/54731/sums-of-fractional-parts-of-linear-functions-of-n