Séminaire Lotharingien de Combinatoire 55 (2005), Article B55a

COMPUTER-FREE EVALUATION OF AN INFINITE DOUBLE SUM VIA EULER SUMS

ALOIS PANHOLZER AND HELMUT PRODINGER

Abstract. A short and computer-free proof (using Euler sums and multiple zeta functions) is provided for a double sum that was recently computed by Pemantle and Schneider using the computer software Sigma.

1. Introduction

The evaluation of the double sum

S := ∑_{j,k=1}^{∞} \frac{H_{j}(H_{k+1} − 1)}{jk(k + 1)(j + k)} = −ζ(2) − 2ζ(3) + 4ζ(2)ζ(3) + 2ζ(5),  (1)

(where H_{n} := ∑_{k=1}^{n} \frac{1}{k} denote harmonic numbers) appears in [4] and was obtained using Carsten Schneider’s software Sigma. Here, we will give a short proof that is completely computer-free. It uses Euler sums and multiple zeta functions.

The second order harmonic numbers which appear in the sequel are denoted by H_{n}^{(2)} := ∑_{k=1}^{n} \frac{1}{k^{2}} .

We split the sum S and apply partial fraction decomposition:

S = ∑_{k≥1} \frac{H_{k+1} − 1}{k(k + 1)} ∑_{j≥1} \frac{H_{j}}{j(j + k)} = ∑_{k≥1} \frac{H_{k+1} − 1}{k^{2}(k + 1)} ∑_{j≥1} H_{j}(\frac{1}{j} − \frac{1}{j + k} ).

The inner sum is simplified as follows:

∑_{j≥1} H_{j}(\frac{1}{j} − \frac{1}{j + k} ) = ∑_{j≥1} (\frac{1}{j} − \frac{1}{j + k} ) ∑_{l=1}^{j} \frac{1}{l} = ∑_{l≥1} \frac{1}{l} ∑_{j≥l} (\frac{1}{j} − \frac{1}{j + k} ).

The second sum here telescopes, and only k summands remain:

∑_{l≥1} \frac{1}{l} ∑_{j≥l} (\frac{1}{j} − \frac{1}{j + k} ) = ∑_{l≥1} \frac{1}{l} ∑_{j=0}^{k−1} \frac{1}{l + j} = ∑_{l≥1} \frac{1}{l^{2}} + ∑_{j=1}^{k−1} ∑_{l≥1} \frac{1}{l(l + j)}

= ζ(2) + ∑_{j=1}^{k−1} \frac{1}{j} ∑_{l≥1} (\frac{1}{l} − \frac{1}{l + j} ) [partial fraction decomposition]

= ζ(2) + ∑_{j=1}^{k−1} \frac{H_{j}}{j} [again by telescoping]

= ζ(2) + \frac{H_{k}^{2} + H_{k}^{(2)}}{2} − \frac{H_{k}}{k} .

Thus the task reduces to evaluate the following single sum:

S = ∑_{k=1}^{∞} \frac{H_{k+1} − 1}{k^{2}(k + 1)}(ζ(2) + \frac{H_{k}^{2} + H_{k}^{(2)}}{2} − \frac{H_{k}}{k} ).  (2)

After partial fraction expansion (and shifting the index if necessary), sum S can be written as follows:

S = −2ζ(2) + \frac{1}{2} ∑_{k≥1} \frac{H_{k}H_{k}^{(2)}}{k^{2}} + \frac{1}{2} ∑_{k≥1} \frac{H_{k}^{3}}{k^{2}} − ∑_{k≥1} \frac{H_{k}^{2}}{k^{3}} + (ζ(2) − 1) ∑_{k≥1} \frac{H_{k}}{k^{2}} − 2 ∑_{k≥1} \frac{1}{k^{2}} .

For the final computation of S we require the following evaluations of Euler sums via values of the Riemann zeta function:

∑_{k≥1} \frac{H_{k}H_{k}^{(2)}}{k^{2}} = ζ(5) + ζ(2)ζ(3),  (3a)

∑_{k≥1} \frac{H_{k}^{3}}{k^{2}} = 10ζ(5) + ζ(2)ζ(3),  (3b)

∑_{k≥1} \frac{H_{k}^{2}}{k^{3}} = \frac{7}{2}ζ(5) − ζ(2)ζ(3),  (3c)

∑_{k≥1} \frac{H_{k}}{k^{2}} = 2ζ(3),  (3d)

from which equation (1) then follows.

Equations (3c) and (3d) can be found explicitly in [3]. In [3] one also finds the identities

∑_{k≥1} \frac{H_{k}^{3}}{(k + 1)^{2}} = \frac{15}{2}ζ(5) + ζ(2)ζ(3),  (4a)

∑_{k≥1} \frac{H_{k}}{k^{4}} = 3ζ(5) − ζ(2)ζ(3),  (4b)

and due to

∑_{k≥1} \frac{H_{k}^{3}}{k^{2}} = ∑_{k≥1} \frac{H_{k}^{3}}{(k + 1)^{2}} + 3 ∑_{k≥1} \frac{H_{k}^{2}}{k^{3}} − 3 ∑_{k≥1} \frac{H_{k}}{k^{4}} + ∑_{k≥1} \frac{1}{k^{5}} ,

we obtain equation (3b) as well, using (4a), (4b) and (3c). To show (3a) we will apply Theorem 2 of [2], which gives

∑_{k≥1} \frac{H_{k−1}H_{k−1}^{(2)}}{k^{2}} = ζ(2, 1, 2) + ζ(2, 2, 1) + ζ(2, 3),

where the multiple zeta functions are defined by

ζ(a_{1}, a_{2}, …, a_{m}) = ∑_{k_{1}>k_{2}>···>k_{m}≥1} \frac{1}{k_{1}^{a_{1}}k_{2}^{a_{2}} ⋯ k_{m}^{a_{m}}} .

Since

∑_{k≥1} \frac{H_{k}H_{k}^{(2)}}{k^{2}} = ∑_{k≥1} \frac{H_{k−1}H_{k−1}^{(2)}}{k^{2}} + ∑_{k≥1} \frac{H_{k−1}^{(2)}}{k^{3}} + ∑_{k≥1} \frac{H_{k−1}}{k^{4}} + ∑_{k≥1} \frac{1}{k^{5}}

= ∑_{k≥1} \frac{H_{k−1}H_{k−1}^{(2)}}{k^{2}} + ∑_{k>l≥1} \frac{1}{k^{3}l^{2}} + ∑_{k>l≥1} \frac{1}{k^{4}l} + ∑_{k≥1} \frac{1}{k^{5}} ,

we obtain

∑_{k≥1} \frac{H_{k}H_{k}^{(2)}}{k^{2}} = ζ(2, 1, 2) + ζ(2, 2, 1) + ζ(2, 3) + ζ(3, 2) + ζ(4, 1) + ζ(5).

Using the following evaluations of the multiple zeta function given in [1] resp. [2]:

ζ(2, 1, 2) = ζ(2, 3) = \frac{9}{2}ζ(5) − 2ζ(2)ζ(3),
ζ(2, 2, 1) = ζ(3, 2) = −\frac{11}{2} + 3ζ(2)ζ(3),
ζ(4, 1) = 2ζ(5) − ζ(2)ζ(3),

equation (3a) follows.

References

[1] D. Borwein, J. Borwein and R. Girgensohn, Explicit evaluation of Euler sums, Proceedings of the Edinburgh Mathematical Society 38, 277–294, 1995.

[2] J. Borwein and R. Girgensohn, Evaluation of triple Euler sums, Electronic Journal of Combinatorics 3, research paper #23, 1996.

[3] P. Flajolet and B. Salvy, Euler sums and contour integral representations, Experimental Mathematics 7, 15–35, 1998.

[4] R. Pemantle and C. Schneider, When is 0.999 … equal to 1?, preprint, 2004.

Institut für Diskrete Mathematik und Geometrie, Technische Universität Wien, Wiedner Hauptstraße 8–10/104, A-1040 Wien, Austria.

  E-mail address: Alois.Panholzer@tuwien.ac.at

Helmut Prodinger, Mathematics Department, Stellenbosch University, 7602 Stellenbosch, South Africa.

  E-mail address: hproding@sun.ac.za