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.)

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