Tornheim-Like Series, Harmonic Numbers, and Zeta Values

Ilham A. Aliev and Ayhan Dil Department of Mathematics Akdeniz University 07070 Antalya Turkey

                 ialiev@akdeniz.edu.tr
                  adil@akdeniz.edu.tr

Abstract

We obtain explicit evaluations of the Tornheim-like multiple series involving harmonic numbers. We give a new relationship between harmonic numbers and ζ(2). We also present closed-form formulas of some multiple series in terms of zeta values.

1 Introduction

The Riemann zeta function is defined by

ζ(z) = ∑_{k=1}^{∞} \frac{1}{k^{z}} , (Re z > 1).

For even positive integers, one has the well-known relationship between zeta values and Bernoulli numbers:

ζ(2n) = (−1)^{n+1} \frac{(2π)^{2n}}{2(2n)!} B_{2n}.  (1)

Here

B_{0} = 1, B_{2} = \frac{1}{6}, B_{4} = −\frac{1}{30}, B_{6} = \frac{1}{42}, ….

This result was proved for the first time by Euler in 1740.

For odd positive integers, no simple expression like (1) is known. Roger Apéry [1] proved the irrationality of ζ(3), and after that ζ(3) was named Apéry’s constant. Rivoal [17] showed that infinitely many of the numbers ζ(2n + 1) must be irrational. Zudilin [23] showed that at least one of the numbers ζ(5), ζ(7), ζ(9) and ζ(11) is irrational.

For a positive integer n and an integer m the nth partial sum of ζ(m) is called the nth generalized harmonic number of order m, and is denoted by H_{n}^{(m)}, i.e.,

H_{n}^{(m)} = ∑_{k=1}^{n} \frac{1}{k^{m}} .

The special case m = 1 is the classical harmonic number

H_{n} = ∑_{k=1}^{n} \frac{1}{k}, H_{0} = 0.

It is well known that there are deep relationships between Tornheim-like series, harmonic numbers and zeta values. The Tornheim double series [19] (or the so-called Witten zeta function [22]) is defined by

S(a, b, c) := ∑_{m,n=1}^{∞} \frac{1}{m^{a}n^{b}(m + n)^{c}} .  (2)

The following equation is a simple and nice example of a connection between Tornheim-like series, harmonic numbers, and zeta values (see [4, 5, 10, 11, 16]):

∑_{n,m=1}^{∞} \frac{1}{nm(n + m)} = ∑_{m=1}^{∞} \frac{H_{m}}{m^{2}} = 2ζ(3).  (3)

Tornheim-like series have attracted increasing attention in recent years and they have proven to be a powerful tool for finding many interesting relationships between various zeta values ([2, 3, 4, 5, 6, 10, 11, 13, 15, 18, 20]).

Boyadzhiev [7, 8] described a simple method for evaluating double series of the form (2) in terms of zeta values. Kuba [13] considered the following general sum:

V = ∑_{j,k=1}^{∞} \frac{H_{j+k}^{(u)}}{j^{r}k^{s}(j + k)^{t}} .

This sum includes the Tornheim double series (2) as a special case. Kuba [13] proved that whenever w = r + s + t + u is even, for r, s, t, w ∈ ℕ , the series V can be explicitly evaluated in terms of zeta functions.

On the other hand, Xu and Li [20] used the Tornheim type series for the evaluation of non-linear Euler sums. Among other results they obtained

∑_{m=1}^{∞} \frac{H_{m+k}}{m(m + k)} = \frac{H_{k}^{2} + H_{k}^{(2)}}{k} , k ∈ ℕ = \{1, 2, 3, …\} .  (4)

From (3) and (4) it is easy to see that the value of the series

a(k) = ∑_{m=1}^{∞} \frac{H_{m+k}}{m(m + k)}, k ∈ ℕ∪ \{0\}

is irrational for k = 0 and rational for every k ∈ ℕ. Hence the following questions naturally arise: for integers s ∈ ℕ∪ {0}, are the values of the double series

∑_{n,m=1}^{∞} \frac{H_{n+m+s}}{nm(n + m + s)},

and more generally, the multiple series

A_{n}(s) = ∑_{k_{1}=1}^{∞} ⋯ ∑_{k_{n−1}=1}^{∞} \frac{H_{k_{1}+···+k_{n−1}+s}}{k_{1} ⋯ k_{n−1}(k_{1} + ⋯ + k_{n−1} + s)}

rational or irrational numbers? This question is studied in the second section. Namely, in the case when n is odd, we have solved this question exactly. If n is even, we give a partial solution depending on the odd zeta values.

In the third section, some new relationships between harmonic numbers and ζ(2) are given and explicit evaluation formulas for some double series via zeta values are established.

2 Explicit evaluations of the Tornheim-like multiple series involving harmonic numbers

Theorem 1. Consider the double series

A(s) = ∑_{n,m=1}^{∞} \frac{H_{n+m+s}}{nm(n + m + s)}, s ∈ ℕ∪ \{0\} .

For any s ∈ ℕ the value of A(s) is rational but A(0) is irrational. More precisely,

A(s) = \begin{cases}6ζ(4), & if s = 0; \\ 6 ∑_{j=0}^{s−1}(−1)^{j}\binom{s−1}{j} \frac{1}{(j+1)^{4}} , & if s ≥ 1. \end{cases}

Proof. Using telescoping series, we have

1 + \frac{1}{2} + \frac{1}{3} + ⋯ + \frac{1}{n + m + s} = ∑_{k=1}^{∞}(\frac{1}{k} − \frac{1}{k + n + m + s})

= (n + m + s) ∑_{k=1}^{∞} \frac{1}{k(k + n + m + s)}.

It then follows that

A(s) = ∑_{n,m=1}^{∞} \frac{1}{nm(n + m + s)}(1 + \frac{1}{2} + \frac{1}{3} + ⋯ + \frac{1}{n + m + s})

= ∑_{n,m,k=1}^{∞} \frac{1}{nmk(n + m + k + s)}

= ∑_{n,m,k=1}^{∞} (∫_{0}^{1} x^{n−1}dx) (∫_{0}^{1} y^{m−1}dy) (∫_{0}^{1} z^{k−1}dz) (∫_{0}^{1} t^{n+m+k+s−1}dt)

= ∫_{0}^{1} t^{s+2} [∫_{0}^{1} (∑_{n=1}^{∞}(xt)^{n−1}) dx ∫_{0}^{1} (∑_{m=1}^{∞}(yt)^{m−1}) dy ∫_{0}^{1} (∑_{k=1}^{∞}(zt)^{k−1}) dz] dt

= ∫_{0}^{1} t^{s+2} [∫_{0}^{1} \frac{1}{1 − xt}dx ∫_{0}^{1} \frac{1}{1 − yt}dy ∫_{0}^{1} \frac{1}{1 − zt}dz] dt.

Since

∫_{0}^{1} \frac{1}{1 − ut}du = −\frac{1}{t} ln(1 − t),

we have

A(s) = − ∫_{0}^{1} t^{s−1} ln^{3}(1 − t)dt = − ∫_{0}^{1}(1 − t)^{s−1} ln^{3} tdt.  (5)

Setting s = 0, it follows that

A(0) = − ∫_{0}^{1} \frac{1}{1 − t} ln^{3} tdt = − ∑_{j=0}^{∞} ∫_{0}^{1} t^{j} ln^{3} tdt

= − ∑_{j=0}^{∞}(−\frac{6}{(j + 1)^{4}} ) = 6ζ(4) = \frac{π^{4}}{16}.

On the other hand, if s ≥ 1, then utilizing the formulas

(1 − t)^{s−1} = ∑_{j=0}^{s−1}(−1)^{j}\binom{s − 1}{j}t^{j}

and

∫_{0}^{1} t^{j} ln^{3} tdt = −\frac{3!}{(j + 1)^{4}} ,

(5) can be computed explicitly as

A(s) = − ∫_{0}^{1}(1 − t)^{s−1} ln^{3} tdt

= 3! ∑_{j=0}^{s−1}(−1)^{j}\binom{s − 1}{j}\frac{1}{(j + 1)^{4}} .

This proves the stated result.

In the same way as in Theorem 1, by making use of the formulas

(1 − t)^{s−1} = ∑_{j=0}^{s−1}(−1)^{j}\binom{s − 1}{j}t^{j} and ∫_{0}^{1} t^{j} ln^{k} tdt = (−1)^{k} \frac{k!}{(j + 1)^{k+1}} ,

one can prove the following more general result.

Theorem 2. Let

A_{n}(s) = ∑_{k_{1}=1}^{∞} ⋯ ∑_{k_{n−1}=1}^{∞} \frac{H_{k_{1}+···+k_{n−1}+s}}{k_{1} ⋯ k_{n−1}(k_{1} + ⋯ + k_{n−1} + s)}, s ∈ ℕ∪ \{0\} , n ≥ 2.

Then

A_{n}(s) = \begin{cases}n!ζ(n + 1), & if s = 0; \\ n! ∑_{j=0}^{s−1}(−1)^{j}\binom{s−1}{j} \frac{1}{(j+1)^{n+1}} , & if s ≥ 1. \end{cases}  (6)

Two special cases of the theorem are as follows:

A_{2}(s) = ∑_{k=1}^{∞} \frac{H_{k+s}}{k(k + s)} = \begin{cases}2!ζ(3), & if s = 0; \\ 2! ∑_{j=0}^{s−1}(−1)^{j}\binom{s−1}{j} \frac{1}{(j+1)^{3}} , & if s ≥ 1, \end{cases}

and

A_{4}(s) = ∑_{k,m,n=1}^{∞} \frac{H_{k+m+n+s}}{kmn(k + m + n + s)} = \begin{cases}4!ζ(5), & if s = 0; \\ 4! ∑_{j=0}^{s−1}(−1)^{j}\binom{s−1}{j} \frac{1}{(j+1)^{5}} , & if s ≥ 1. \end{cases}

Remark 3. It can be easily seen from (6) that the expression A_{n}(s) is a rational number for all s ≥ 1 and n ≥ 2. However A_{2}(0) = 2!ζ(3) is irrational (Apéry). If n ≥ 4 and even, it is not known whether the numbers A_{n}(0) = n!ζ(n + 1) are irrational or not. On the other hand, for any odd n ∈ ℕ we have A_{n}(0) = n!ζ(n + 1) = r_{n}π^{n+1} (see (1)), which is irrational because r_{n} is rational and π^{n+1} is irrational. Notice that, as is well known, the irrationality of π^{n} is a consequence of the transcendence of π.

Remark 4. There is an interesting connection between the multiple series A_{n}(s) and the Bell polynomials b_{n}(x). For a given sequence x = (x_{1}, x_{2}, …), the Bell polynomials b_{n}(x) = b_{n}(x_{1}, x_{2}, …) are defined by the generating function (see [9] or [14])

∑_{n=0}^{∞} b_{n}(x)\frac{t^{n}}{n!} = exp \{∑_{k=1}^{∞} x_{k} \frac{t^{k}}{k!}\} .

As a result of this definition one has the following explicit expression [9]:

b_{n}(x) = ∑_{σ(n)} \frac{n!}{a_{1}!a_{2}! ⋯ a_{n}!} (\frac{x_{1}}{1} )^{a_{1}} (\frac{x_{2}}{2} )^{a_{2}} ⋯ (\frac{x_{n}}{n} )^{a_{n}} ,

where the summation ranges over the set σ(n) of all partitions of n. Using this explicit expression we can give a list of the first few Bell polynomials as

b_{0}(x) = 1,  b_{1}(x) = x_{1},  (7)
b_{2}(x) = x_{1}^{2} + x_{2},  b_{3}(x) = x_{1}^{3} + 3x_{1}x_{2} + 2x_{3}.

Considering the well-known harmonic number identity (see [21, Eq. (3.56)] or [12, Corollary 2.2])

i!m\binom{m + n}{n} ∑_{k=0}^{n}(−1)^{k}\binom{n}{k}\frac{1}{(m + k)^{i+1}}

= b_{i}(H_{m+n} − H_{m−1}, H_{m+n}^{(2)} − H_{m−1}^{(2)}, …, H_{m+n}^{(i)} − H_{m−1}^{(i)})

we have

n!s ∑_{j=0}^{s−1}(−1)^{j}\binom{s − 1}{j}\frac{1}{(j + 1)^{n+1}} = b_{n}(H_{s}, H_{s}^{(2)}, …, H_{s}^{(n)}),  (8)

which corresponds to (6) for s ≥ 1. Hence we can restate A_{n}(s) in terms of the Bell polynomials as

A_{n}(s) = \begin{cases}n!ζ(n + 1), & if s = 0; \\ \frac{1}{s}b_{n}(H_{s}, H_{s}^{(2)}, …, H_{s}^{(n)}), & if s ≥ 1. \end{cases}

Thanks to this formula, considering (7) we can write A_{n}(s) as a finite combination of the harmonic and generalized harmonic numbers. For instance, for n = 2 we have

A_{2}(s) = \begin{cases}2ζ(3), & if s = 0; \\ \frac{1}{s}((H_{s})^{2} + H_{s}^{(2)}), & if s ≥ 1, \end{cases}

which coincides with (4). For n = 3 we have

A_{3}(s) = \begin{cases}6ζ(4), & if s = 0; \\ \frac{1}{s}((H_{s})^{3} + 3H_{s}H_{s}^{(2)} + 2H_{s}^{(3)}), & if s ≥ 1. \end{cases}

3 Explicit evaluations of some double series in zeta val-

ues

The next theorem gives a new relationship between harmonic numbers and ζ(2). Theorem 5. Let O_{m} = ∑_{k=1}^{m} \frac{1}{2k−1} = H_{2m} − \frac{1}{2}H_{m}. Then

∑_{m=1}^{∞} \frac{O_{m}}{2m(2m + 1)} = \frac{1}{4}ζ(2).  (9)

Proof. Let

A = ∑_{m,n=0}^{∞} \frac{1}{(2m + 1)(2n + 1)(2m + 2n + 1)}

and

B = ∑_{m,n=1}^{∞} \frac{1}{(2m + 1)(2n + 1)(2m + 2n + 1)}.

From the equation

∑_{k=0}^{∞} \frac{1}{(2k + 1)^{2}} = \frac{π^{2}}{8} ,

we have

A = \frac{π^{2}}{4} − 1 + B.  (10)

Further, using telescoping series, we have

B = ∑_{m=1}^{∞} \frac{1}{2m + 1} \frac{1}{2m} ∑_{n=1}^{∞}(\frac{1}{2n + 1} − \frac{1}{2n + 1 + 2m})

= ∑_{m=1}^{∞} \frac{1}{2m(2m + 1)}(\frac{1}{3} + \frac{1}{5} + ⋯ + \frac{1}{2m + 1})

= ∑_{m=1}^{∞} \frac{O_{m} − \frac{2m}{2m+1}}{2m(2m + 1)}

= ∑_{m=1}^{∞} \frac{O_{m}}{2m(2m + 1)} − ∑_{m=1}^{∞} \frac{1}{(2m + 1)^{2}}

= ∑_{m=1}^{∞} \frac{O_{m}}{2m(2m + 1)} − \frac{π^{2}}{8} + 1.

Hence we obtain that

B = ∑_{m=1}^{∞} \frac{O_{m}}{2m(2m + 1)} + 1 − \frac{3}{4}ζ(2).  (11)

Now let us evaluate A.

A = ∑_{m,n=0}^{∞} (∫_{0}^{1} x^{2m}dx) (∫_{0}^{1} y^{2n}dy) (∫_{0}^{1} t^{2m+2n}dt)

= ∫_{0}^{1} (∫_{0}^{1} ∑_{m=0}^{∞}(xt)^{2m}dx ∫_{0}^{1} ∑_{n=0}^{∞}(yt)^{2n}dy) dt

= ∫_{0}^{1} (∫_{0}^{1} \frac{1}{1 − (xt)^{2}} dx ∫_{0}^{1} \frac{1}{1 − (yt)^{2}} dy) dt

= \frac{1}{4} ∫_{0}^{1} \frac{1}{t^{2}} ln^{2} (\frac{1 + t}{1 − t}) dt.

The substitution \frac{1+t}{1−t} = u immediately leads to the following equality:

A = \frac{1}{2} ∫_{1}^{∞} \frac{1}{(1 − u)^{2}} ln^{2} udu.

Integration by parts gives

A = ∫_{1}^{∞} \frac{1}{u(u − 1)} ln u du = ∫_{1}^{∞} \frac{1}{1 − \frac{1}{u}} \frac{ln u}{u^{2}}du

= ∑_{k=0}^{∞} ∫_{1}^{∞} u^{−k−2} ln u du = ∑_{k=0}^{∞} \frac{1}{(k + 1)^{2}} = ζ(2) = \frac{π^{2}}{6} .  (12)

Similarly, from (10), (11) and (12) we have

\frac{π^{2}}{6} = \frac{π^{2}}{4} − 1 + ∑_{m=1}^{∞} \frac{O_{m}}{2m(2m + 1)} − \frac{3}{4}ζ(2) + 1

and as a result

∑_{m=1}^{∞} \frac{O_{m}}{2m(2m + 1)} = \frac{1}{4}ζ(2).

Remark 6. Considering (11) we have

B = ∑_{m=1}^{∞} \frac{H_{2m+1} − 1 − \frac{1}{2}H_{m}}{2m(2m + 1)} .  (13)

Now we use (10) and (12) to conclude that

∑_{m=1}^{∞} \frac{2H_{2m+1} − H_{m}}{2m(2m + 1)} = 2(2 − ln 2) − ζ(2).  (14)

In the following theorem we obtain interesting relationships between some special double series and zeta values ζ(2) and ζ(3). Theorem 7. We have the following series evaluations:

(a) ∑_{m,n=0}^{∞} \frac{1}{(m+ \frac{1}{2} )(n+ \frac{1}{2} )(m+n+ \frac{1}{2} )(m+n+1)} = 16ζ(2) − 14ζ(3).

(b) ∑_{m,n=0}^{∞} \frac{1}{(m+ \frac{1}{2} )(n+ \frac{1}{2} )(m+n+1)(m+n+ \frac{3}{2} )} = 14ζ(3) − 8ζ(2) .

(c) ∑_{m,n=0}^{∞} \frac{1}{(m+ \frac{1}{2} )(n+ \frac{1}{2} )(m+n+ \frac{1}{2} )(m+n+1)(m+n+ \frac{3}{2} )} = 24ζ(2) − 28ζ(3). Proof. (a) We have

∑_{m,n=0}^{∞} \frac{1}{(2m + 1)(2n + 1)(2m + 2n + 2)}

= ∑_{m,n=0}^{∞} (∫_{0}^{1} x^{2m}dx) (∫_{0}^{1} y^{2n}dy) (∫_{0}^{1} t^{2m+2n+1}dt)

= ∫_{0}^{1} (∫_{0}^{1} ∑_{m=0}^{∞}(xt)^{2m}dx ∫_{0}^{1} ∑_{n=0}^{∞}(yt)^{2n}dy) tdt

= \frac{1}{4} ∫_{0}^{1} \frac{1}{t} ln^{2} (\frac{1 + t}{1 − t}) dt.

Here the substitution \frac{1+t}{1−t} = u leads to the following equality:

\frac{1}{4} ∫_{0}^{1} \frac{1}{t} ln^{2} (\frac{1 + t}{1 − t}) dt = \frac{1}{2} ∫_{1}^{∞} \frac{1}{(u^{2} − 1)} ln^{2} udu = \frac{1}{2} ∫_{1}^{∞} \frac{1}{u^{2}} \frac{1}{(1 − \frac{1}{u^{2}} )} ln^{2} udu

= \frac{1}{2} ∑_{k=0}^{∞} ∫_{1}^{∞} u^{−2k−2} ln u du.

After integration by parts we get

\frac{1}{2} ∑_{k=0}^{∞} ∫_{1}^{∞} u^{−2k−2} ln u du = ∑_{k=0}^{∞} \frac{1}{(2k + 1)^{3}} = \frac{7}{8}ζ(3).

Hence

∑_{m,n=0}^{∞} \frac{1}{(2m + 1)(2n + 1)(2m + 2n + 2)} = \frac{7}{8}ζ(3).  (15)

On the other hand, according to the formula (12),

A = ∑_{m,n=0}^{∞} \frac{1}{(2m + 1)(2n + 1)(2m + 2n + 1)} = ζ(2).  (16)

Now, from (15) and (16) it follows that

ζ(2) − \frac{7}{8}ζ(3) = ∑_{m,n=0}^{∞} \frac{1}{(2m + 1)(2n + 1)(2m + 2n + 1)}

− ∑_{m,n=0}^{∞} \frac{1}{(2m + 1)(2n + 1)(2m + 2n + 2)}

= ∑_{m,n=0}^{∞} \frac{1}{(2m + 1)(2n + 1)(2m + 2n + 1)(2m + 2n + 2)}

and this proves (a).

(b) By the same method in the proof of (a), we have

∑_{m,n=0}^{∞} \frac{1}{(2m + 1)(2n + 1)(2m + 2n + 3)}

= ∑_{m,n=0}^{∞} (∫_{0}^{1} x^{2m}dx) (∫_{0}^{1} y^{2n}dy) (∫_{0}^{1} t^{2m+2n+2}dt)

= ∫_{0}^{1} (∫_{0}^{1} \frac{1}{1 − (xt)^{2}} dx ∫_{0}^{1} \frac{1}{1 − (yt)^{2}} dy) t^{2}dt

= \frac{1}{4} ∫_{0}^{1} ln^{2} (\frac{1 + t}{1 − t}) dt

= \frac{1}{2} ∫_{1}^{∞} \frac{1}{(u + 1)^{2}} ln^{2} udu = −\frac{1}{2} ∫_{1}^{∞} ln^{2} ud(\frac{1}{u + 1})

= ∫_{1}^{∞} \frac{1}{u(u + 1)} ln u du = ∑_{k=2}^{∞}(−1)^{k} ∫_{1}^{∞} u^{−k} ln u du

= ∑_{k=2}^{∞}(−1)^{k} \frac{1}{(k − 1)^{2}} = \frac{π^{2}}{12}.

Thus

∑_{m,n=0}^{∞} \frac{1}{(2m + 1)(2n + 1)(2m + 2n + 3)} = \frac{π^{2}}{12} = \frac{1}{2}ζ(2).  (17)

Now, from (15) and (17) we have

\frac{7}{8}ζ(3) − \frac{1}{2}ζ(2) = ∑_{m,n=0}^{∞} \frac{1}{(2m + 1)(2n + 1)(2m + 2n + 2)(2m + 2n + 3)}

and this proves (b).

(c) Finally, formula (c) can be obtained by subtracting the formula (b) from the formula (a).

4 Acknowledgments

The authors gratefully acknowledge the many helpful suggestions of the anonymous referee.

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2010 Mathematics Subject Classification: Primary 11M32; Secondary 40B05. Keywords: Tornheim series, harmonic number, Riemann zeta value.

(Concerned with sequences \underline{A000110}, \underline{A001008}, and \underline{A263633}.)

Received December 6 2021; revised versions received December 9 2021; April 5 2022; April 6 2022. Published in Journal of Integer Sequences, June 20 2022.

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