Recurrence equation and integral representation of Apéry sums

Various methods are used to investigate sums involving a reciprocal central binomial coefficient and a power term. In the first part, new functions are introduced for calculation of sums with a negative exponent in the power term. A recurrence equation for the functions provides an integral representation of the sums using polylogarithm functions. Thus polylogarithms and, in particular, zeta values can be expressed via these functions, too. In the second part, a straightforward recurrence formula is derived for sums having a positive exponent in the power term. Finally, two interesting cases of double sums are presented.


Introduction and main results
The subject of this article are infinite sums, A + m and A − m , involving a reciprocal central binomial coefficient 1/ 2k k and a power term k −m : In the context of these sums-which we also call the Apéry sums-Roger Apéry proved the irrationality of π 2 and ζ(3) [1,9,13]. Furthermore, the tight relationship between these sums and values of the zeta function are the topic of many publications [2,[6][7][8][10][11][12][13][14]. Whereas there exist closed-form expressions for A + 1 , A + 2 , A − 1 , A − 2 , and as well A + 4 , see for instance [10][11][12][13], the sums A + m and A − m for m > 4 can only be expressed by means of polylogarithmic ladders or multiple Clausen values [2,7]. In particular, where φ = ( √ 5 + 1)/2, see e.g. [2,[11][12][13]. In this work two approaches are presented to calculate the sums with a negative exponent in the power term, A + m and A − m for m > 0. First, using a reformulation of the reciprocal central binomial coefficient we define the functions A + m (x) and A − m (x): These functions A + m (x) and A − m (x) together with their derivatives A + m (x) and A − m (x) relate the Apéry sums A + m and A − m with the polylogarithm functions Li n (x), defined as For the proofs of Theorems 1.1 and 1.2 see (11) and Lemma 2.1 in Sect. 2. Thus, integrating (5) the Apéry sums for m > 0 can be calculated as Furthermore, as shown in Sect. 2, the polylogarithm Li m (z), and in particular its value Li m (1) = ζ(m), can be expressed vice versa in terms of A m (x). In Sect. 3 we derive an integral representation of the reciprocal central binomial coefficient, which gives us a direct integral formula of the sums, see (6). In Sect. 4 we calculate the integrals for m = 1, 2, 3.
Due to the exponential convergence of the reciprocal binomial coefficients even the sums with a positive exponent in the power term, A + −m and A − −m for m 0, converge. These sums up to m = 3 are studied in various articles [11,12,15]. In this work a straightforward recurrence formula is presented to calculate them.
For the proof see (26) in Sect. 5. By reiterative application of (7) For the proof see (30) in Sect. 6. Finally, in Sect. 6 the identities, 0 = 6A are presented as vanishing double sums over the products 1/ 2k k 2 j j .

Recurrence equation
We show below that non-alternating sums A + m and alternating sums A − m subordinate, in general, to the same equations, except for a sign. First, we simplify the notation: Reiterative application of the recurrence formula for reciprocal binomial coefficients, Thus the sums A m , (1), for m 1, can be reformulated via coefficients α m jk as Now, for a more detailed investigation of the sums A m the functions A m (x) are introduced, see (3): In the end of this section the coefficients a + 1 j and a − 1 j are calculated. Convergence of the sums A + m (x) and A − m (x) is guaranteed for |x| < 1, respectively |x| < 1/φ, only. However, by the process of analytic continuation the order of summation in (10) can be switched and a comparison with (9) shows The derivatives A m (x) and A m (x) are given by For m = 1 the derivative A 1 (x)-as the analytic continuation of (12)-is calculated as In fact, the coefficients a m j from (10) satisfy the identity Thus, for the functions A m (x) from (10) the following differential equation is obtained: Therefore, the derivatives A m (x) can be calculated recurrently: Since A m−1 (0) = a m1 = 0, by integration A m (x) can be written as Proof (i) For m = 1 both sides are identical and (ii) for m > 1 they are described by the same recurrence equation.
(ii) Applying (15) to the recurrence equation of polylogarithm, (14), again: Due to the symmetry of (15), and the transformation, x → 1 2 − x, the Apéry sums A m = A m (1) are calculated as Integrating (16), the polylogarithm ∓Li m (∓(x − x 2 )) can be described by the function A m (x): Thus, the functions A + m (x), or respectively the coefficients a + m j , enable a formulation for both the sums A + m and the values ζ(m). Next, we want to focus on the coefficients a m j from (9) in the case m = 1. Here, the coefficients a + 1 j of the non-alternating sum A + 1 (x) fulfil the equation With a + 11 = 0 and a + 12 = 1 one gets recurrently, Hence, comparing the sum A + 1 = A + 1 (1) from (10) with (2) gives The coefficients a − 1 j of the alternating sum A − 1 (x) can be written as With a − 11 = 0 and a − 12 = 1 the coefficients a − 1, j+1 are given by the Fibonacci numbers F j : Since Fibonacci numbers grow exponentially with lim j→∞ F j /F

Integral representation
The integrals A m can be obtained directly by the integral representation of the reciprocal central binomial coefficients 1/ 2k k , see e.g. [15]. Looking at the integrals 1/2 0 1 4 − x 2 k−1 dx, partial integration results in Thus, starting with 1/2 0 1 4 − x 2 0 dx = 1 2 , we arrive to the integral by induction: So, the reciprocal central binomial coefficients 1/ 2k k are With (8) and the definition of the polylogarithm function, Li m (z) = ∞ k=1 k −m z k , the sums A m can be calculated as in agreement with (18). A similar result was derived by Taylor [15].
Furthermore, calculating the integral formula for the coefficients 1/ 2k k in (20) one obtains the identity

Integrals for A 1 , A 2 , and A 3
The calculation of the sums A m for m > 0, and in particular A 1 and A 2 , see (2), has been considered in many publications, see e.g. [11][12][13]. In this section we employ the integrals from (18) to calculate A 1 and A 2 and to give an integral formula of A 3 . For m = 1 the integral A 1 from (18) is given by Li 0 (z) = z/(1 − z): In particular, A + 1 and A − 1 are calculated as For m = 2 the integral A 2 from (18) can be calculated via Li 1 (z) = − ln(1 − z): Having the factorizations, the integral ln(a ± x)/x dx = ln x ln a − Li 2 (∓x/a), the dilogarithm Li 2 (0) = 0, and the Landen identities, A + 2 and A − 2 are obtained as The integral A 3 from (18) is calculated by integration of the dilogarithm Li 2 . Integrating by parts one finds Thus the sums A + 3 and A − 3 = 2 5 ζ(3), see (2), are given by the following integrals: leading to another integral form for ζ(3).

Apéry sums for positive powers
The sums for positive powers, A −m for m 0, have also been studied in [11,12,15]. Calculation of A 0 , A −1 , A −2 , and A −3 was done by Lehmer [12]. In this section, a general recurrence formula for A −m is derived. The sums A −m converge, since 1/ 2k k < k 4 k and With (1) the expression 4 A m − 2 A m+1 can be calculated by Thus, we obtain in the case of positive powers, with m 0, that This directly leads to the recurrence equation for the sums A −m : Here, the sums A −m can be calculated recurrently starting with A 1 . For 0 m 4 the sums A + −m and A − −m are given by: , Using the substitution A −m = (4∓1) m+1 A −m , (26) gives Here, w m j are the weights of A − j contributing to A −m .

Lemma 5.1
The numbers A −m are sums of an integer and an odd multiple of 2 A 1 .
Proof First we prove A 0 is a sum of an integer and an odd multiple of 2 A 1 . All weights w m j are integer numbers and the sum of weights, is an odd number. Then, by induction, each number A −m is a sum of an integer and an odd multiple of 2 A 1 .

Further relations on Apéry sums
In the same way as the zeta function ζ(m) = ∞ k=1 k −m and eta function η(m) 1 2 , since all terms for k 2 vanish when m → ∞, and the term for k = 1 establishes the limit. The sum of Apéry sums, S = ∞ m=1 A m − 1 2 , can be written as To calculate S, first we construct the sum of 4 A m − 2 A m+1 − 1, (25), over m: Regrouping (29) leads to Comparing one gets Thus, the sum S + of the non-alternating sums A + m and the sum S − of the alternating sums A − m are given by Again, these equations resemble similar equations for the zeta and eta functions, namely ∞ m=2 (ζ (m) − 1) = 1 and ∞ m=2 (η(m) − 1) = 1 − 2η(1) = 1 − 2 ln 2. Finally, the sums A + 1 and A + 2 from (21) and (23), and also A − 1 and A − 2 of (22) and (24), are related by In addition, the relations between A + 0 and A + 1 , and also between A − 0 and A − 1 , are given by (27) and (28): Thus, combining (31) and (32), the following double sums are shown to be 0: Here, the question remains open whether these equations could be proved in rational terms, i.e., without an explicit calculation of the irrational sums A 1 and A 2 .