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](https://mathoverflow.net/questions/54731/sums-of-fractional-parts-of-linear-functions-of-n). 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?



Does this sum equal zeta(3)?
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](http://en.wikipedia.org/wiki/Multiple_zeta_function). 


