## Euler Sums, Part II: A Symmetry Formula

A post from a few months ago gave a proof that $\displaystyle \sum_{x=1}^\infty \sum_{y = 1}^x \frac{1}{x^2 y} = 2 \zeta(3).$  In today’s post I’d like to prove a general symmetry formula for Euler sums like this one.  Define

$\displaystyle \zeta_N(a) = \sum_{x=1}^N \frac{1}{x^a},$

$\displaystyle \zeta(a) = \sum_{x=1}^{\infty} \frac{1}{x^a},$

$\displaystyle \zeta_N(a,b) = \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^a y^b},$ and

$\displaystyle \zeta(a,b) = \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{x^a y^b}.$

(Notice the upper index on the second sum is $x-1$, not $x$ like we used in the previous post.)  The formula we will be proving is

$\displaystyle \zeta(a,b) + \zeta(b,a) = \zeta(a) \zeta(b) - \zeta(a+b)$.

We will show this by proving a more precise result:

$\displaystyle \zeta_N(a,b) + \zeta_N(b,a) = \zeta_N(a) \zeta_N(b) - \zeta_N(a+b).$

In other words, the symmetry result is true in both the finite and infinite versions; there’s no error term when you truncate the sum at a fixed N.

Here we go:

$\displaystyle \zeta_N(a,b) + \zeta_N(b,a) = \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^a y^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a}$

$= \displaystyle \sum_{y=1}^N \sum_{x=y+1}^N \frac{1}{x^a y^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a}$, by swapping the order of summation on the first sum

$= \displaystyle \sum_{x=1}^N \sum_{y=x+1}^N \frac{1}{x^b y^a} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a}$, by relabeling variables on the first sum

$\displaystyle = \sum_{x=1}^N \sum_{y=x}^N \frac{1}{x^b y^a} - \sum_{x=1}^N \frac{1}{x^{a+b}} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a},$

$= \displaystyle \sum_{x=1}^N \sum_{y=1}^N \frac{1}{x^b y^a} - \zeta_N(a+b)$

$= \displaystyle \zeta_N(a) \zeta_N(b) - \zeta_N(a+b).$

Taking the limit as $N \to \infty$, we get

$\displaystyle \zeta(a,b) + \zeta(b,a) = \zeta(a) \zeta(b) - \zeta(a+b)$.

(This argument also appears in my post on the evaluation of a triple Euler sum on math.SE.)