TITLE: Does this sum equal zeta(3)?
QUESTION [14 upvotes]: 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 $3$.
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?

REPLY [26 votes]: Hi David,
This is the first example of a multiple zeta identity.  Your sum S is just $\zeta(1,2)$, where the multiple zeta value is defined by:
$$\zeta(s_1, s_2, \ldots, s_k) = \sum_{0 < n_1 < n_2 < \cdots n_k} \left( \prod_{i=1}^k n_i^{-s_i} \right).$$
Your identity $\zeta(1,2) = \zeta(3)$ was discovered by Euler according to Wikipedia.