Integrals and Polygamma Representations for Binomial Sums

Anthony Sofo School of Engineering and Science Victoria University PO Box 14428 Melbourne City, VIC 8001 Australia

                    anthony.sofo@vu.edu.au

Abstract We consider sums involving the product of reciprocal binomial coefficient and polynomial terms and develop some double integral identities. In particular cases it is possible to express the sums in closed form, give some general results, recover some known results in Coffey and produce new identities.

1 Introduction

In a recent paper Coffey [8] considers summations over digamma and polygamma functions and develops many results, namely two of his propositions, in terms of the Riemann zeta function ζ (·), are respectively equations (59) and (66a)

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

and

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

Coffey [8] also constructs new integral representations for these sums. The major aim of this paper is to investigate general binomial sums with various parameters that then enables one to give more general representations of (1) and (2), thereby generalizing the propositions of Coffey, both in closed form in terms of zeta functions and digamma functions at possible rational values of the argument, and in double integral form. The following definitions will be useful. The Psi, or digamma function ψ (z) , is defined by

ψ (z) = \frac{d}{dz} log Γ (z) = \frac{Γ′ (z)}{Γ (z)} = ∑_{n=0}^{∞} (\frac{1}{n + 1} − \frac{1}{n + z} ) − γ

where γ denotes the Euler-Mascheroni constant and Γ (z) is the Gamma function. Similarly

ψ (z + 1) = ∑_{n=1}^{∞} (\frac{1}{n} − \frac{1}{n + z} ) − γ,  (3)

and

2ψ (2z) = ψ (z) + ψ (z + \frac{1}{2}) + 2 ln 2.  (4)

Sums of reciprocals of binomial coefficients appear in the calculation of massive Feynman diagrams [13] within several different approaches: for instance, as solutions of differential equations for Feynman amplitudes, through a naive ε-expansion of hypergeometric functions within Mellin-Barnes technique or in the framework of recently proposed algebraic approach [12]. There has recently been a renewed interest in the study of series involving binomial coefficients and a number of authors have obtained either closed form representation or integral representation for some particular cases of these series. The interested reader is referred to [1, 2, 3, 4, 5, 6, 9, 11, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27].

2 The main results

The following Lemma and well-known definition will be useful in the proof of the main theorem.

Definition 1. Let |z| ≤ 1, m ≥ 1 and q ≠ −1, −2, −3, .... Then

∑_{r=1}^{∞} \frac{z^{r}}{(q + r)^{m}} = \frac{(−1)^{m−1}}{(m − 1)!} ∫_{0}^{1} \frac{z y^{q} (ln (y))^{m−1}}{1 − zy} dy.  (5)

The next Lemma deals with two infinite sums.

Lemma 2. Let a and r be positive real numbers. Then

∑_{n=1}^{∞} \frac{1}{n (an + r)} = \frac{H(1)_{\frac{r}{a}}}{r} and  (6)

∑_{n=1}^{∞} \frac{(−1)^{n}}{n (an + r)} = \frac{1}{r} \{H(1)_{\frac{r}{2a}} − H(1)_{\frac{r}{a}}\} .  (7)

Proof.

∑_{n=1}^{∞} \frac{1}{n (an + r)} = \frac{1}{r} ∑_{n=1}^{∞} (\frac{1}{n} − \frac{1}{n + \frac{r}{a}} ) and from (3)
= \frac{1}{r} [γ + ψ (\frac{r}{a} + 1)]
= \frac{H(1)_{\frac{r}{a}}}{r} ; hence (6) is attained.

∑_{n=1}^{∞} \frac{(−1)^{n}}{n (an + r)} = \frac{1}{2r} [∑_{n=1}^{∞} (\frac{1}{n} − \frac{1}{n + \frac{r}{2a}} ) − ∑_{n=1}^{∞} (\frac{1}{n} − \frac{1}{n + \frac{r−a}{2a}} ) − ∑_{n=1}^{∞} \frac{1}{n (2n − 1)}]

= \frac{1}{2r} [−γ + ψ (\frac{r}{2a} + 1) + γ − ψ (\frac{r}{2a} + \frac{1}{2}) − 2 ln 2]
= \frac{1}{2r} [ψ (\frac{r}{2a} + 1) − ψ (\frac{r}{2a} + \frac{1}{2}) − 2 ln 2] ,

from the definition (4)

ψ (\frac{r}{2a} + \frac{1}{2}) = 2ψ (\frac{r}{a} + 1) − ψ (\frac{r}{2a} + 1) − 2 ln 2; hence

∑_{n=1}^{∞} \frac{(−1)^{n}}{n (an + r)} = \frac{1}{2r} [2ψ (\frac{r}{2a} + 1) − 2ψ (\frac{r}{a} + 1)]
= \frac{1}{r} [H(1)_{\frac{r}{2a}} − γ − H(1)_{\frac{r}{a}} + γ], therefore (7) follows.

Remark 3. In the following Corollaries and remarks we encounter harmonic numbers at possible rational values of the argument, of the form H(α)_{\frac{r}{a}} where r = 1, 2, 3, ..., k, α = 1, 2, 3, ... and k ∈ ℕ. The polygamma function ψ^{(α)} (z) is defined as:

ψ^{(α)} (z) = \frac{d^{α+1}}{dz^{α+1}} [log Γ (z)] = \frac{d^{α}}{dz^{α}} [ψ (z)] , z ≠ \{0, −1, −2, −3, ...\} .

To evaluate H(α)_{\frac{r}{a}} we have available a relation in terms of the polygamma function ψ^{(α)} (z) , for rational arguments z,

H(α+1)_{\frac{r}{a}} = ζ (α + 1) + \frac{(−1)^{α}}{α!} ψ^{(α)} (\frac{r}{a} + 1)

we also define

H(1)_{\frac{r}{a}} = γ + ψ (\frac{r}{a} + 1), and H_{0}^{(α)} = 0.

The evaluation of the polygamma function ψ^{(α)} (\frac{r}{a}) at rational values of the argument can be explicitly done via a formula as given by Kölbig [15], (see also [14]), or Choi and Cvijovic [7] in terms of the polylogarithmic or other special functions. Some specific values are given as, many others are listed in the book [24]:

ψ^{(n)} (\frac{1}{2}) = (−1)^{n} n! (2^{n+1} − 1) ζ (n + 1)

H(1)_{\frac{1}{4}} = 4 − \frac{π}{2} − 3 ln (2) , H(1)_{\frac{3}{4}} = \frac{4}{3} + \frac{π}{2} − 3 ln (2) ,

H(1)_{\frac{1}{3}} = \frac{3}{2} − \frac{π}{2\sqrt{3}} − \frac{3 ln 3}{2}, and H(1)_{\frac{5}{6}} = \frac{6}{5} + \frac{\sqrt{3}π}{2} − \frac{3 ln (3)}{2} − 2 ln (2) .

We now state the following theorem.

Theorem 4. Let a be a positive real number, |t| ≤ 1 , j ≥ 0, and k ∈ ℕ ∪ {0} . Then

S_{k+1} (a, j, t) = ∑_{n=1}^{∞} \frac{t^{n}}{n^{k+1} ( \begin{matrix}an + j + 1 \\ j + 1 \end{matrix} )}  (8)

= \{ \begin{matrix}\frac{(j+1)t(−1)^{k}}{k!} ∫_{0}^{1} ∫_{0}^{1} \frac{(1−x)^{j} x^{a}(ln(y))^{k}}{1−tx^{a}y} dxdy, for k ≥ 1 \\ at ∫_{0}^{1} \frac{(1−x)^{j+1} x^{a−1}}{1−tx^{a}} dx, for k = 0 \end{matrix} .

= T_{0 a+k+1}F_{a+k} [ \begin{matrix}\overbrace{1, 1, ....., 1}^{(k+2)−terms}, \overbrace{\frac{a + 1}{a}, .. . . . ., \frac{2a − 1}{a}}^{(a−1)−terms} \\ \underbrace{2, 2, ....., 2}_{k−terms}, \underbrace{\frac{a + j + 2}{a}, . . . ..., \frac{a + j + a + 1}{a}}_{a−terms} \end{matrix} | t]  (9)

where _{p}F_{q} [·] is the generalized hypergeometric function,

T_{0} = t (j + 1) B (j + 1, a + 1) ,

and B (·, ·) is the beta function.

Proof. Consider

∑_{n=1}^{∞} \frac{t^{n}}{n^{k+1} ( \begin{matrix}an + j + 1 \\ j + 1 \end{matrix} )} = ∑_{n=1}^{∞} \frac{(j + 1) t^{n} Γ (j + 1) Γ (an + 1)}{n^{k+1} Γ (an + j + 2)}

= (j + 1) ∑_{n=1}^{∞} \frac{t^{n}}{n^{k+1}} B (an + 1, j + 1)

now replacing the beta function with its integral representation, we have

(j + 1) ∑_{n=1}^{∞} \frac{t^{n}}{n^{k+1}} B (an + 1, j + 1) = (j + 1) ∑_{n=1}^{∞} \frac{t^{n}}{n^{k+1}} ∫_{0}^{1} x^{an} (1 − x)^{j} dx.

By a justified changing the order of integration and summation we have,

∑_{n=1}^{∞} \frac{t^{n}}{n^{k+1} ( \begin{matrix}an + j + 1 \\ j + 1 \end{matrix} )} = (j + 1) ∫_{0}^{1} (1 − x)^{j} ∑_{n=1}^{∞} \frac{(tx^{a})^{n}}{n^{k+1}} dx

= \frac{(j + 1) t (−1)^{k}}{k!} ∫_{0}^{1} ∫_{0}^{1} \frac{(1 − x)^{j} x^{a} (ln (y))^{k}}{1 − tx^{a}y} dxdy, for k ≥ 1

upon utilizing Definition 1. The case of k = 0 follows in a similar way so that

S_{1} (a, j, t) = ∑_{n=1}^{∞} \frac{t^{n}}{n ( \begin{matrix}an + j + 1 \\ j + 1 \end{matrix} )}

= at ∫_{0}^{1} \frac{(1 − x)^{j+1} x^{a−1}}{1 − tx^{a}} dx;

hence the integrals in (8) are attained. By the consideration of the ratio of successive terms \frac{U_{n+1}}{U_{n}} where

U_{n} = \frac{t^{n}}{n^{k+1} ( \begin{matrix}an + j + 1 \\ j + 1 \end{matrix} )}

we obtain the result (9).

The following interesting corollaries follow from Theorem 4.

Corollary 5. Let t = 1 and a > 0. Also let j ≥ 0 and k ≥ 1 be integers. Then

S_{k+1} (a, j, 1) = ∑_{n=1}^{∞} \frac{1}{n^{k+1} ( \begin{matrix}an + j + 1 \\ j + 1 \end{matrix} )}

= \frac{(j + 1) (−1)^{k}}{k!} ∫_{0}^{1} ∫_{0}^{1} \frac{(1 − x)^{j} x^{a} (ln (y))^{k}}{1 − x^{a}y} dxdy, for k ≥ 1  (10)

= ∑_{s=0}^{k−1} A_{s} (j + 1)!ζ (k + 1 − s) + ∑_{r=1}^{j+1} (−1)^{r+k+1} (\frac{a}{r})^{k} \binom{j + 1}{r} H(1)_{\frac{r}{a}} (11)

where

A_{s} = lim_{n→0} [\frac{1}{s!} \frac{d^{s}}{dn^{s}} \{\frac{n^{k}}{n^{k} ∏_{r=1}^{j+1} (an + r)}\}] , s = 0, 1, 2, ...k − 1.  (12)

Proof. By expansion,

∑_{n=1}^{∞} \frac{1}{n^{k+1} ( \begin{matrix}an + j + 1 \\ j + 1 \end{matrix} )} = ∑_{n=1}^{∞} \frac{(j + 1)!}{n^{k+1} (an + 1)_{j+2}} = ∑_{n=1}^{∞} \frac{(j + 1)!}{n^{k+1} ∏_{r=1}^{j+1} (an + r)}

= ∑_{n=1}^{∞} \frac{(j + 1)!}{n} [∑_{s=0}^{k−1} \frac{A_{s}}{n^{k−s}} + ∑_{r=1}^{j+1} \frac{B_{r}}{an + r}] ,

where

B_{r} = lim_{n→(− \frac{r}{a})} \{\frac{an + r}{∏_{r=1}^{j+1} (an + r)}\} = \frac{(−1)^{r+k+1} r}{(j + 1)!} (\frac{a}{r})^{k} \binom{j + 1}{r} ,

A_{s} is defined by (12). Hence, after interchanging the sums, we have

∑_{n=1}^{∞} \frac{1}{n^{k+1} ( \begin{matrix}an + j + 1 \\ j + 1 \end{matrix} )}

= (j + 1)! [∑_{s=0}^{k−1} A_{s} ∑_{n=1}^{∞} \frac{1}{n^{k+1−s}} + ∑_{r=1}^{j+1} B_{r} ∑_{n=1}^{∞} \frac{1}{n (an + r)}]

= (j + 1)! ∑_{s=0}^{k−1} A_{s} ζ (k + 1 − s) + ∑_{r=1}^{j+1} (−1)^{r+k+1} (\frac{a}{r})^{k} \binom{j + 1}{r} H(1)_{\frac{r}{a}}

upon utilizing Lemma 2, which is the result (11). The degenerate case, for j = −1, gives the known result

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

The integral (10) follows from the integral in (8).

A similar result is evident for the case t = −1.

Corollary 6. Let t = −1 and a > 0. Also let j ≥ 0 and k ≥ 1 be integers. Then

S_{k+1} (a, j, − 1) = ∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{k+1} ( \begin{matrix}an + j + 1 \\ j + 1 \end{matrix} )}

= \frac{(j + 1) (−1)^{k+1}}{k!} ∫_{0}^{1} ∫_{0}^{1} \frac{(1 − x)^{j} x^{a} (ln (y))^{k}}{1 + x^{a}y} dxdy, for k ≥ 1

= ∑_{s=0}^{k−1} A_{s} (j + 1)! (2^{s−k} − 1) ζ (k + 1 − s)  (13)

+ ∑_{r=1}^{j+1} (−1)^{r+k+1} (\frac{a}{r})^{k} \binom{j + 1}{r} (H(1)_{\frac{r}{a}} − H(1)_{\frac{r}{2a}})

Proof. The proof, uses (7) and follows the same details as that of Corollary 5, and will not be given here.

The addition and subtraction of (11) and (13) gives us the following representations. Remark 7. Let a > 0 and let j ≥ 0 and k ≥ 1 be integers. Then

2^{1−k} ∑_{n=1}^{∞} \frac{1}{n^{k+1} ( \begin{matrix}2an + j + 1 \\ j + 1 \end{matrix} )}

= ∑_{s=0}^{k−1} A_{s} (j + 1)! 2^{s−k} ζ (k + 1 − s) + ∑_{r=1}^{j+1} (−1)^{r+k+1} (\frac{a}{r})^{k} \binom{j + 1}{r} (2H(1)_{\frac{r}{a}} − H(1)_{\frac{r}{2a}})

and

∑_{n=1}^{∞} \frac{1}{(2n − 1)^{k+1} ( \begin{matrix}2an − a + j + 1 \\ j + 1 \end{matrix} )}

= ∑_{s=0}^{k−1} A_{s} (j + 1)! (2 − 2^{s−k}) ζ (k + 1 − s) + ∑_{r=1}^{j+1} (−1)^{r+k+1} (\frac{a}{r})^{k} \binom{j + 1}{r} H(1)_{\frac{r}{2a}}.

We give the following example to illustrate some of the above identities. Example 8. Let k = 4, from (12)

A_{0} = \frac{1}{(j + 1)!}, A_{1} = −\frac{aH_{j+1}^{(1)}}{(j + 1)!}, A_{2} = \frac{a^{2} ((H_{j+1}^{(1)})^{2} + H_{j+1}^{(2)})}{2 (j + 1)!}
A_{3} = −\frac{a^{3} ((H_{j+1}^{(1)})^{3} + 3H_{j+1}^{(1)}H_{j+1}^{(2)} + 2H_{j+1}^{(3)})}{6 (j + 1)!}

therefore

∑_{n=1}^{∞} \frac{1}{n^{5} ( \begin{matrix}an + j + 1 \\ j + 1 \end{matrix} )} = (j + 1)! [A_{0}ζ (5) + A_{1}ζ (4) + A_{2}ζ (3) + A_{3}ζ (2)]

+ ∑_{r=1}^{j+1} (−1)^{r+1} (\frac{a}{r})^{4} \binom{j + 1}{r} H(1)_{\frac{r}{a}}

and

∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{5} ( \begin{matrix}an + j + 1 \\ j + 1 \end{matrix} )} = (j + 1)! [−\frac{15A_{0}}{16} ζ (5) − \frac{7A_{1}}{8} ζ (4) − \frac{3A_{2}}{4} ζ (3) − \frac{A_{3}}{2} ζ (2)]

+ ∑_{r=1}^{j+1} (−1)^{r} (\frac{a}{r})^{4} \binom{j + 1}{r} (H(1)_{\frac{r}{a}} − H(1)_{\frac{r}{2a}}) .

Remark 9. The very special case of a = 1 and j = 0 allows one to evaluate

A_{s} = lim_{n→0} [\frac{1}{s!} \frac{d^{s}}{dn^{s}} \{\frac{n^{k}}{n^{k} (n + 1)}\}] , s = 0, 1, 2, ...k − 1
= (−1)^{s} ,

and from (11) and (13) we can easily obtain (2) and (1).

A recurrence relation for a degenerate case, j = 0, of Theorem 4 is embodied in the following corollary. Corollary 10. Let the conditions of Theorem 4 hold with j = 0 and put

S_{k+1}^{0} : = S_{k+1}^{0} (a, t) = ∑_{n=1}^{∞} \frac{t^{n}}{n^{k+1} (an + 1)}
= \frac{t}{a + 1} _{k+3}F_{k+2} [ \begin{matrix}\overbrace{1, 1, ....., 1}^{(k+2)−terms}, \frac{a+1}{a} \\ \underbrace{2, 2, ....., 2}_{(k+1)−terms}, \frac{2a+1}{a} \end{matrix} | t] ,

then

S_{k+1}^{0} + aS_{k}^{0} = Li_{k+1} (t) , for k ≥ 1

with solution

S_{k+1}^{0} = (−a)^{k} S_{1}^{0} + ∑_{r=1}^{k} (−a)^{r−1} t Φ (t, k + 2 − r, 1)
= (−a)^{k} S_{1}^{0} + ∑_{r=1}^{k} (−a)^{r−1} Li_{k+2−r} (t)

S_{1}^{0} (a, t) = ∑_{n=1}^{∞} \frac{t^{n}}{n (an + 1)} = \frac{t}{a + 1} _{3}F_{2} [ \begin{matrix}1, 1, \frac{a+1}{a} \\ 2, \frac{2a+1}{a} \end{matrix} | t]

and Φ, Li are the Lerch transcendent and the polylogarithm respectively.

Proof. We notice that

S_{k+1}^{0} + aS_{k}^{0} = ∑_{n=1}^{∞} \frac{t^{n}}{n^{k+1}} = Li_{k+1} (t) ,

and hence the solution follows by iteration.

Related results may be seen in Coffey [10, Lemmas 1 and 2].

Some examples are as follows:

• For t = 1, we know that Li_{k+2−r} (1) = ζ (k + 2 − r) . Hence

S_{k+1}^{0} (a, 1) = (−a)^{k} S_{1}^{0} (a, 1) + ∑_{r=1}^{k} (−a)^{r−1} ζ (k + 2 − r) , for k ≥ 1.

When a = 1, we obtain Coffey’s [8] result, by noting that, from (6), S_{1}^{0} (1, 1) = 1

S_{k+1}^{0} (1, 1) = (−1)^{k} + ∑_{r=1}^{k} (−1)^{r−1} ζ (k + 2 − r) .

When a = \frac{1}{2},

S_{k+1}^{0} (\frac{1}{2}, 1) = \frac{3 (−1)^{k}}{2^{k+1}} + ∑_{r=1}^{k} \frac{(−1)^{r−1}}{2^{r−1}} ζ (k + 2 − r) .

Similarly

S_{k+1}^{0} (8, 1) = (−1)^{k} 2^{3(k+1)} − (−1)^{k} 2^{3k−1}π\sqrt{3 + 2\sqrt{2}} − (−1)^{k} 2^{3k+2} ln (2)

− (−1)^{k} 2^{3k−3/2} ln (3 + 2\sqrt{2}) + ∑_{r=1}^{k} (−2)^{3(r−1)} ζ (k + 2 − r) .

• For t = −1, Li_{k+2−r} (−1) = η (k + 2 − r) = (2^{r−k−1} − 1) ζ (k + 2 − r) , where η (·) is the Dirichlet Eta function. Hence

S_{k+1}^{0} (a, −1) = (−a)^{k} S_{1}^{0} (a, −1) + ∑_{r=1}^{k} (−a)^{r−1} (2^{r−k−1} − 1) ζ (k + 2 − r) , for k ≥ 1.

When a = 1, we obtain Coffey’s [8] result, by noting that

S_{k+1}^{0} (1, −1) = (−1)^{k} (1 − 2 ln 2) + ∑_{r=1}^{k} (−1)^{r−1} (2^{r−k−1} − 1) ζ (k + 2 − r) .

When a = 4,

S_{k+1}^{0} (4, −1) = (−2)^{k} (4 − \sqrt{2} ln (1 + \sqrt{2}) − \frac{π}{2} − ln 2)

+ ∑_{r=1}^{k} (−1)^{r−1} 2^{2(r−1)} (2^{r−k−1} − 1) ζ (k + 2 − r) ,

and similarly

S_{k+1}^{0} (\frac{1}{8}, −1) = − (−1)^{k} \frac{533}{840}2^{−3k} + ∑_{r=1}^{k} \frac{(−1)^{r−1}}{8^{r−1}} (2^{r−k−1} − 1) ζ (k + 2 − r)

= −\frac{8}{9} _{k+3}F_{k+2} [ \begin{matrix}\overbrace{1, 1, ....., 1}^{(k+2)−terms}, 9 \\ \underbrace{2, 2, ....., 2}_{(k+1)−terms}, 10 \end{matrix} | − 1] .

References

[1] H. Alzer, D. Karayannakis and H. M. Srivastava. Series representations for some mathematical constants. J. Math. Anal. Appl. 320 (2006), 145–162.

[2] H. Alzer and S. Koumandos. Series and product representations for some mathematical constants. Period. Math. Hungar.58(1), (2009), 71–82.

[3] N. Batir. Integral representations of some series involving \binom{2k}{k}^{−1}k^{−n} and some related series. Appl. Math. Comp.147 (2004), 645–667.

[4] N. Batir. On the series ∑_{k=1}^{∞} \binom{3k}{k}^{−1}k^{−n}x^{k} . Proceedings of the Indian Academy of Sciences: Mathematical Sciences, 115(4) (2005), 371–381.

[5] J. M. Borwein and R. Girgensohn. Evaluations of binomial series. Aequationes Math., 70, (2005), 25–36.

[6] J. M. Borwein, D. J. Broadhurst and J. Kamnitzer. Central binomial sums, multiple Clausen values and zeta values. Experiment. Math., 10, (2001), 25–34.

[7] J. Choi and D. Cvijovic, Values of the polygamma functions at rational arguments, J. Phys. A: Math. Theor. 40 (2007), 15019–15028.

[8] M. W. Coffey. On one dimensional digamma and polygamma series related to the evaluation of Feynman diagrams. J. Comput. Appl. Maths., 183, (2005), 84–100.

[9] M. W. Coffey. On some series representations of the Hurwitz zeta function. J. Comput. Appl. Maths. 216, (2008), 297- -305.

[10] M. W. Coffey. Some definite logarithmic integrals from Euler sums, and other integration

    results, http://arxiv.org/abs/1001.1366, January 8 2010.

[11] S. R. Finch. Mathematical constants. Encyclopedia of Mathematics and its applications, 94. Cambridge Univ. Press, New York, 2003.

[12] A. P. Isaev. Multi-loop Feynman integrals and conformal quantum mechanics, Nucl. Phys. B662 (3), (2003), 461–465.

[13] M. Yu. Kalmykov, O. Veretin. Single-scale diagrams and multiple binomial sums. Physics Letters, B483, (2000), 315–323.

[14] K. Kölbig, The polygamma function ψ (x) for x = 1/4 and x = 3/4, J. Comput. Appl. Math. 75 (1996), 43–46.

[15] K. Kölbig, The polygamma function and the derivatives of the cotangent function for rational arguments, CERN-IT-Reports CERN-CN., (1996), 96–105.

[16] D. H. Lehmer. Interesting series involving the central binomial coefficient. Amer. Math. Monthly, 92, (1985), 449–457.

[17] L. Lewin. Polylogarithms and associated functions. North Holland, Amsterdam, 1981.

[18] T. Sherman. Summation of Glaisher and Apéry-like series, available at

    http://math.arizona.edu/~rta/001/sherman.travis/series.pdf.

[19] A. Sofo. Computational Techniques for the Summation of Series, Kluwer Academic/Plenum Publishers, 2003.

[20] A. Sofo. General properties involving reciprocals of binomial coefficients, J. Integer Sequences, 9 (2006), Article 06.4.5.

[21] A. Sofo. Integral forms of sums associated with harmonic numbers, Appl. Math. Comput., 207, (2009), 365–372.

[22] A. Sofo. Sums of derivatives of binomial coefficients, Advances in Appl. Math., 42, (2009), 123–134.

[23] A. Sofo. Convexity properties of Reciprocals of binomial coefficients. In T. E. Simos, ed., Numerical Analysis and Applied Mathematics, AIP, Melville, New York, 2007, pp. 703–706.

[24] H. M. Srivastava and J. Choi. Series Associated with the Zeta and Related Functions. Kluwer Academic Publishers, London, 2001.

[25] E. W. Weisstein. Nielsen generalized polylogarithm.

    http://mathworld.wolfram.com/NielsenGeneralizedPolylogarithm.html.
[26] E. W. Weisstein. Binomial sums. http://mathworld.wolfram.com/BinomialSums.html.

[27] I. J. Zucker. On the series ∑_{k=1}^{∞} \binom{2k}{k}^{−1}k^{−n} and related sums. J. Number Theory, 20, (1985), 92–102.

2000 Mathematics Subject Classification: Primary 05A10; Secondary 11B65, 05A19, 33C20. Keywords: double integral, combinatorial identity, harmonic number, polygamma function, recurrence.

Received December 27 2009; revised version received February 16 2010. Published in Journal of Integer Sequences, February 23 2010.

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