1 Introduction and main results

The subject of this article are infinite sums, \(A^{+}_m\) and \(A^{-}_m\), involving a reciprocal central binomial coefficient \(1/{2k\atopwithdelims ()k}\) and a power term \(k^{-m}\):

(1)

In the context of these sums—which we also call the Apéry sums—Roger Apéry proved the irrationality of \(\pi ^2\) and \(\zeta (3)\) [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,

(2)

where \(\phi =(\sqrt{5}+1)/2\), 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 \(A^{-}_m(x)\):

(3)

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 \(\mathrm{Li}_{\,n}(x)\), defined as

$$\begin{aligned} \mathrm{Li}_{\,n}(x) = \sum _{k=1}^\infty \frac{x^k}{k^n}, \quad \mathrm{Li}'_{\,n}(x) = \frac{1}{x}\, \mathrm{Li}_{\,n-1}(x). \end{aligned}$$
(4)

Theorem 1.1

\( A^{+}_m = A^{+}_m(1) \), \( A^{-}_m = A^{-}_m(1) \).

Theorem 1.2

(5)

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

$$\begin{aligned} A^{+}_m = \int _0^1 \frac{\mathrm{Li}_{\,m-1}(x-x^2)}{1-x}\, \mathrm{d}x , \quad A^{-}_m = \int _0^1 \frac{\mathrm{Li}_{\,m-1}(x^2-x)}{x-1}\, \mathrm{d}x . \end{aligned}$$
(6)

Furthermore, as shown in Sect. 2, the polylogarithm \(\mathrm{Li}_{\,m}(z)\), and in particular its value \(\mathrm{Li}_{\,m}(1)=\zeta (m)\), 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 \(m\geqslant 0\), 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.

Theorem 1.3

$$\begin{aligned} {\begin{matrix} 3A^{+}_{-m}&{} = 1 + 2 A_{1-m} + \sum _{j=0}^{m-1}{m\atopwithdelims ()j} A_{-j} , \\ 5A^{-}_{-m}&{}= 1 + 2 A_{1-m} - \sum _{j=0}^{m-1} {m\atopwithdelims ()j} A_{-j} . \end{matrix}} \end{aligned}$$
(7)

For the proof see (26) in Sect. 5. By reiterative application of (7), the sums \(A^{+}_{-m}\) and \(A^{-}_{-m}\) are calculated as linear combinations of 1 and \(A^{+}_1\), respectively 1 and \(A^{-}_1\), where \(A^{+}_1=\sqrt{3}\pi /9\) and \(A^{-}_1=2\ln \phi /{\sqrt{5}}\), see (2).

Theorem 1.4

The numbers \(3^mA^{+}_{-m}\) and \(5^mA^{-}_{-m}\) are sums of an integer and odd—and therefore non-vanishing—multiple of \(2A^{+}_1\), respectively \(2A^{-}_1\).

For the proof see Lemma 5.1 in Sect. 5. Since \(\pi \) and (due to the Gelfond–Schneider theorem [3,4,5]) \(\ln \phi \) are transcendental, \(A^{+}_1\) and \(A^{-}_1\) are transcendental, too. Then, due to Theorem 1.4, \(A^{+}_{-m}\) and \(A^{-}_{-m}\) for all \(m\geqslant 0\) are transcendental numbers. For instance, this is different from \(A^{-}_3\) which is proven only to be irrational [1].

Although there is up to now—with the exception of \(A^{+}_4\)—no closed-form expression for \(A^{+}_m\) and \(A^{-}_m\) for all \(m\geqslant 3\), the sums of \(A^{+}_m-{1}/{2}\) and \(A^{-}_m-{1}/{2}\) over m are calculated as follows.

Theorem 1.5

(8)

For the proof see (30) in Sect. 6. Finally, in Sect. 6 the identities, \(0= 6A^{+}_0A^{+}_2-4A^{+}_1A^{+}_2-3A^{+}_1A^{+}_1\) and \(0=10A^{-}_0A^{-}_2-4A^{-}_1A^{-}_2-5A^{-}_1A^{-}_1\), are presented as vanishing double sums over the products \(1/\bigl [{2k\atopwithdelims ()k}{2j\atopwithdelims ()j}\bigr ]\).

2 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: , , and . Reiterative application of the recurrence formula for reciprocal binomial coefficients,

$$\begin{aligned} {n\atopwithdelims ()m}^{-1}=\frac{m}{m-1}\left[ {n-1\atopwithdelims ()m-1}^{-1}\!\! -\, {n-1\atopwithdelims ()m}^{-1} \right] , \end{aligned}$$

to the reciprocal central binomial coefficient \(1/{2k\atopwithdelims ()k}\) gives

Thus the sums \(A_m\), (1), for \(m\geqslant 1\), can be reformulated via coefficients \(\alpha _{mjk}\) as

(9)

Now, for a more detailed investigation of the sums \(A_m\) the functions \(A_m(x)\) are introduced, see (3):

(10)

In the end of this section the coefficients \(a^{+}_{1j}\) and \(a^{-}_{1j}\) are calculated. Convergence of the sums \(A^{+}_m(x)\) and \(A^{-}_m(x)\) is guaranteed for \(|x|<1\), respectively \(|x|<1/\phi \), only. However, by the process of analytic continuation the order of summation in (10) can be switched and a comparison with (9) shows

$$\begin{aligned} A_m(x) = \sum _{j=1}^\infty \sum _{k=1}^{j-1} \,\frac{(-x)^j\!}{j}\, \alpha _{mjk} = \sum _{k=1}^\infty \sum _{j=k+1}^{2k}\! \frac{(-x)^j}{j}\, \alpha _{mjk}, \quad A_m(1)=A_m. \end{aligned}$$
(11)

The derivatives \(A'_m(x)\) and \(A''_m(x)\) are given by

(12)

For \(m=1\) the derivative \(A'_1(x)\)—as the analytic continuation of (12)—is calculated as

(13)

In fact, the coefficients \(a_{mj}\) 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:

$$\begin{aligned} \frac{2x-1}{x}\, A'_{m-1}(x) = (x-1) A''_m(x) + A'_m(x) = \bigl [ (x-1) A'_m(x) \bigr ]'. \end{aligned}$$
(14)

Since \(A'_{m-1}(0)=a_{m1}=0\), by integration \(A'_m(x)\) can be written as

Lemma 2.1

The function \((x-1)A'_m(x)\) and the polylogarithm \({{\mp }} \mathrm{Li}_{\,m-1}({\mp }(x^2-x))\) are identical:

$$\begin{aligned} (x-1) A'_m(x) = {\mp } \mathrm{Li}_{\,m-1}({\mp }(x^2-x)) . \end{aligned}$$
(15)

Proof

(i) For \(m=1\) both sides are identical and (ii) for \(m>1\) they are described by the same recurrence equation.

(i) In the case \(m=1\), it follows from (13) and the definition \(\mathrm{Li}_0(z)=z/(1-z)\):

(ii) Applying (15) to the recurrence equation of polylogarithm, \(\mathrm{Li}_{\,m-1}'(z) =\mathrm{Li}_{\,m-2}(z)/z\) with \(z={\mp }(x^2-x)\), results in (14), again:

$$\begin{aligned} \begin{aligned} \bigl [ {\mp } \mathrm{Li}_{m-1}({\mp }(x^2-x)) \bigr ]'&= {\mp } (2x-1)\, \frac{\mathrm{Li}_{m-2}({\mp }(x^2-x))}{x^2-x} \\&= \bigl [ (x-1) A'_m(x) \bigr ]' = \frac{2x-1}{x} \,A'_{m-1}(x). \end{aligned} \end{aligned}$$
(16)

\(\square \)

Due to the symmetry of (15),

$$\begin{aligned} x A'_m(1-x) = (1-x) A'_m(x) , \end{aligned}$$
(17)

and the transformation, \(x\mapsto {\frac{1}{2}}-x\), the Apéry sums \(A_m=A_m(1)\) are calculated as

$$\begin{aligned} A_m = \int _0^1\! A'_m(x)\, \mathrm{d}x = \int _0^1 \frac{\mathrm{Li}_{m-1}({\mp }(x^2-x))}{{\mp }(x-1)}\, \mathrm{d}x = \int _0^{\frac{1}{2}} \frac{\mathrm{Li}_{m-1} \bigl ( \pm \bigl ( {\frac{1}{4}}-x^2 \bigr ) \bigr ) }{\pm \bigl ( {\frac{1}{4}}-x^2\bigr )}\, \mathrm{d}x.\nonumber \\ \end{aligned}$$
(18)

Integrating (16), the polylogarithm \({\mp }\mathrm{Li}_m({\mp }(x-x^2))\) can be described by the function \(A_m(x)\):

$$\begin{aligned} \begin{aligned} {\mp }\mathrm{Li}_m({\mp }(x^2-x))&= \int _0^x \frac{2t-1}{t}\, A'_m(t)\, \mathrm{d}t\\&= \int _0^x \!\bigl (A'_m(t) - A'_m(1-t) \bigr )\, \mathrm{d}t = A_m(x) + A_m(1-x) - A_m(1) , \end{aligned} \end{aligned}$$

in which \(\mathrm{Li}_m(0)=A_m(0)=0\) and the symmetry of (17) are used.

The values of the zeta function are given by \(\zeta (m)=\mathrm{Li}_{\,m}(1)\). Using the definition of the coefficients \(a^{+}_{mj}\) from (9), and writing with \(\varphi -\varphi ^2=1\) and \(1-\varphi =\varphi ^{-1}\), one obtains

Thus, the functions \(A^{+}_m(x)\), or respectively the coefficients \(a^{+}_{mj}\), enable a formulation for both the sums \(A^{+}_m\) and the values \(\zeta (m)\).

Next, we want to focus on the coefficients \(a_{mj}\) from (9) in the case \(m=1\). Here, the coefficients \(a^{+}_{1j}\) of the non-alternating sum \(A^{+}_1(x)\) fulfil the equation

With \(a^{+}_{11}=0\) and \(a^{+}_{12}=1\) one gets recurrently,

$$\begin{aligned} a^{+}_{1,3j+1} = 0, \quad a^{+}_{1,3j+2} = 1, \quad a^{+}_{1,3j+3} =-1, \; \ j \geqslant 0 . \end{aligned}$$

Hence, comparing the sum \(A^{+}_1=A^{+}_1(1)\) from (10) with (2) gives

The coefficients \(a^{-}_{1j}\) of the alternating sum \(A^{-}_1(x)\) can be written as

$$\begin{aligned} a^{-}_{1j} = \sum _{k=1}^{j-1}{k-1 \atopwithdelims ()j-k-1} = a^{-}_{1,j-1} + a^{-}_{1,j-2} . \end{aligned}$$

With \(a^{-}_{11}=0\) and \(a^{-}_{12}=1\) the coefficients \(a^{-}_{1,j+1}\) are given by the Fibonacci numbers \(F_j\):

(19)

Since Fibonacci numbers grow exponentially with \(\lim _{j\rightarrow \infty }F_j/F_{j-1}=\phi \), at \(x=1\) the sum \(A^{-}_1(1)=\sum _{j=1}^\infty (-1)^ja^{-}_{1j}/j\) does not converge.

Having \(|a^{+}_{1j}|\leqslant 1\) for all j, the convergence radius of \(A^{+}_1(x)\), (10), is \(R^+_1=1\). The convergence radius of \(A^{-}_1(x)\) is given by \(R^-_1=\lim _{j\rightarrow \infty }|a^{-}_{1,j-1}/a^{-}_{1j}|=1/\phi \). Furthermore, with \(|a^{+}_{mj}|\leqslant |a^{+}_{m-1,j}|\) and \(|a^{-}_{mj}|\leqslant |a^{-}_{m-1,j}|\) it follows \(R^+_m\geqslant 1\) and \(R^-_m\geqslant 1/\phi \).

3 Integral representation

The integrals \(A_m\) can be obtained directly by the integral representation of the reciprocal central binomial coefficients \(1/{2k\atopwithdelims ()k}\), see e.g. [15]. Looking at the integrals \(\int _0^{1/2}\bigl ({\frac{1}{4}}-x^2\bigr ){}^{k-1} \mathrm{d}x\), partial integration results in

$$\begin{aligned} \int _0^{\frac{1}{2}} \biggl (\frac{1}{4}-x^2\biggr )^{k-1} \mathrm{d}x = \frac{4k+2}{k} \int _0^{\frac{1}{2}}\biggl (\frac{1}{4}-x^2\biggr )^k \mathrm{d}x. \end{aligned}$$

Thus, starting with \(\int _0^{1/2} \bigl ({\frac{1}{4}}-x^2\bigr )^0 \mathrm{d}x = \frac{1}{2}\), we arrive to the integral by induction:

So, the reciprocal central binomial coefficients \(1/{2k\atopwithdelims ()k}\) are

(20)

With (8) and the definition of the polylogarithm function, \(\mathrm{Li}_{\,m}(z)=\sum _{k=1}^\infty k^{-m}z^k\), the sums \(A_m\) can be calculated as

$$\begin{aligned} A_m = \sum _{k=1}^\infty \, \frac{(\pm 1)^{k-1}}{k^m{2k\atopwithdelims ()k}} = \int _0^{\frac{1}{2}} \sum _{k=1}^\infty \, \frac{\bigl (\pm \bigl ({\frac{1}{4}}-x^2\bigr )\bigr ){}^{k-1}}{k^{m-1}}\, \mathrm{d}x = \int _0^{\frac{1}{2}} \frac{\mathrm{Li}_{\,m-1}\bigl (\pm \bigl ({\frac{1}{4}}-x^2\bigr )\bigr )}{\pm \bigl ({\frac{1}{4}}-x^2\bigr )}\, \mathrm{d}x , \end{aligned}$$

in agreement with (18). A similar result was derived by Taylor [15].

Furthermore, calculating the integral formula for the coefficients \(1/{2k\atopwithdelims ()k}\) in (20) one obtains the identity

4 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 \(\mathrm{Li}_0(z)=z/(1-z)\):

In particular, \(A^{+}_1\) and \(A^{-}_1\) are calculated as

(21)
(22)

For \(m=2\) the integral \(A_2\) from (18) can be calculated via :

$$\begin{aligned} A_2 = \pm \int _0^1 \frac{\mathrm{Li}_1({\mp }(x^2-x))}{x}\, \mathrm{d}x = {\mp } \int _0^1 \frac{\ln (1\pm (x^2-x))}{x} \, \mathrm{d}x . \end{aligned}$$

Having the factorizations,

the integral , the dilogarithm \(\mathrm{Li}_2(0)=0\), and the Landen identities,

\(A^{+}_2\) and \(A^{-}_2\) are obtained as

(23)
(24)

The integral \(A_3\) from (18) is calculated by integration of the dilogarithm \(\mathrm{Li}_2\). Integrating by parts one finds

Thus the sums \(A^{+}_3\) and \(A^{-}_3={\frac{2}{5}}\zeta (3)\), see (2), are given by the following integrals:

leading to another integral form for \(\zeta (3)\).

5 Apéry sums for positive powers

The sums for positive powers, \(A_{-m}\) for \(m\geqslant 0\), 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 \(1/{2k\atopwithdelims ()k}<{\frac{k}{4^k}}\) and

$$\begin{aligned} |A_{-m}| = \Biggl |\sum _{k=1}^\infty \,\frac{(\pm 1)^k k^m}{{2k\atopwithdelims ()k}}\Biggr | < \sum _{k=1}^\infty \,\frac{k^{m+1}}{4^k} = \mathrm{Li}_{-(m+1)}\biggl (\frac{1}{4}\biggr ). \end{aligned}$$

With (1) the expression \(4 A_m - 2 A_{m+1}\) can be calculated by

(25)

Thus, we obtain in the case of positive powers, with \(m\geqslant 0\), that

$$\begin{aligned} 4 A_{-m} - 2 A_{1-m} = 1 \pm \sum _{k=1}^\infty \, (k+1)^m \,\frac{(\pm 1)^{k-1}}{{2k\atopwithdelims ()k}} = 1 \pm \sum _{j=0}^m {m\atopwithdelims ()j} A_{-j} . \end{aligned}$$

This directly leads to the recurrence equation for the sums \(A_{-m}\):

(26)

Here, the sums \(A_{-m}\) can be calculated recurrently starting with \(A_1\). For \(0 \leqslant m \leqslant 4\) the sums \(A^{+}_{-m}\) and \(A^{-}_{-m}\) are given by:

(27)
(28)

Using the substitution \({\widetilde{A}}_{-m} = (4 {\mp } 1)^{m+1} A_{-m}\), (26) gives

$$\begin{aligned} {\widetilde{A}}_0&= 1 + 2 A_1 , \quad {\widetilde{A}}_{-m} = (4 {\mp } 1)^m + \sum _{j=0}^{m-1} w_{mj} {\widetilde{A}}_{-j}, \\ w_{mj}&= 2\delta _{m-1,j} \pm {m\atopwithdelims ()j} \,(4 {\mp } 1)^{m-1-j}. \end{aligned}$$

Here, \(w_{mj}\) are the weights of \({\widetilde{A}}_{-j}\) contributing to \({\widetilde{A}}_{-m}\).

Lemma 5.1

The numbers \({\widetilde{A}}_{-m}\) are sums of an integer and an odd multiple of \(2 A_1\).

Proof

First we prove \({\widetilde{A}}_0\) is a sum of an integer and an odd multiple of \(2 A_1\). All weights \(w_{mj}\) are integer numbers and the sum of weights,

is an odd number. Then, by induction, each number \({\widetilde{A}}_{-m}\) is a sum of an integer and an odd multiple of \(2 A_1\). \(\square \)

6 Further relations on Apéry sums

In the same way as the zeta function \(\zeta (m)=\sum _{k=1}^\infty k^{-m}\) and eta function \(\eta (m)=\sum _{k=1}^\infty (-1)^{k-1}k^{-m}\) converge to 1 for \(m\rightarrow \infty \), the sums \(A_m=\sum _{k=1}^\infty (\pm 1)^{k-1}k^{-m}/{2k\atopwithdelims ()k}\) converge to \(\frac{1}{2}\), since all terms for \(k\geqslant 2\) vanish when \(m\rightarrow \infty \), and the term for \(k=1\) establishes the limit. The sum of Apéry sums, \(S=\sum _{m=1}^\infty \bigl (A_m-{\frac{1}{2}}\bigr )\), can be written as

To calculate S, first we construct the sum of \(4A_m-2A_{m+1}-1\), (25), over m:

$$\begin{aligned} \begin{aligned} \sum _{m=1}^\infty \,( 4 A_m - 2 A_{m+1} - 1 )&= \pm \sum _{m=1}^\infty \sum _{k=1}^\infty \frac{(\pm 1)^{k-1}}{(k+1)^m {2k\atopwithdelims ()k}}\\&= \pm \sum _{k=1}^\infty \frac{(\pm 1)^{k-1}}{k{2k\atopwithdelims ()k}} = \pm A_1. \end{aligned} \end{aligned}$$
(29)

Regrouping (29) leads to

$$\begin{aligned} \sum _{m=1}^\infty ( 4 A_m - 2 A_{m+1} - 1 ) = 4\sum _{m=1}^\infty \,\biggl (A_m-\frac{1}{2}\biggr ) - 2\sum _{m=2}^\infty \,\biggl (A_m-\frac{1}{2}\biggr ) = 2S + 2A_1 - 1 . \end{aligned}$$

Comparing one gets

$$\begin{aligned} 2S + (2{\mp }1) A_1 - 1 = 0 . \end{aligned}$$
(30)

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 \(\sum _{m=2}^\infty (\zeta (m)-1)=1\) and \(\sum _{m=2}^\infty (\eta (m)-1)=1-2\eta (1)=1-2\ln 2\).

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

(31)

In addition, the relations between \(A^{+}_0\) and \(A^{+}_1\), and also between \(A^{-}_0\) and \(A^{-}_1\), are given by (27) and (28):

$$\begin{aligned} 3A^{+}_0 - 2A^{+}_1 = 1, \quad 5A^{-}_0 - 2A^{-}_1 = 1. \end{aligned}$$
(32)

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 \(A_2\).