A Generalized Apéry Series

M. L. Glasser Department of Physics Clarkson University Postdam, NY 13699-5820 USA

                      laryg@clarkson.edu

Abstract

The inverse central binomial series

S_{k}(z) = ∑_{n=1}^{∞} \frac{n^{k}z^{n}}{\binom{2n}{n}} ,

popularized by Apéry and Lehmer, is evaluated for positive integers k along with the asymptotic behavior for large k. We show that the value z = 2, as commented on by D. H. Lehmer, provides a unique relation to π.

1 Introduction

Since the appearance of S_{−3}(1) in Apéry’s famous proof [1] in 1979 that ζ(3) is irrational, an extensive literature has been devoted to the series

S_{k}(z) = ∑_{n=1}^{∞} \frac{n^{k}z^{n}}{\binom{2n}{n}}  (1)

For example, in 1985 Lehmer [2] presented a number of special cases which could be obtained from the Taylor series for f(x) = x^{−1/2}(1 − x)^{−1/2} sin^{−1} x using only elementary calculus. In passing, he noted that when k is a positive integer, S_{k}(2) had the form a_{k} − b_{k}π and that the rational number a_{k}/b_{k} “is a close approximation to π. This remark was recently taken up by Dyson et al. [3], who proved that |a_{k}/b_{k} − π| = O(Q^{−k}) as k → ∞ where Q = \sqrt{1 + (2π/ ln 2)^{2}}. Lehmer also showed that for positive integer k

S_{k}(z) = \frac{2^{k+z^{5/2}}z^{1/2}}{(4 − z)^{k+3/2}} (A_{k}(z/4) sin^{−1}(\sqrt{z/4}) + \sqrt{z(4 − z)}B_{k}(z/4))  (2)

where A_{k} and B_{k} are recursively defined polynomials. It was apparently not until 2005 that (2) was evaluated explicitly, for z = 1, by J. Borwein and P. Girgensohn [4] who showed

S_{k}(1) = \frac{1}{2}(−1)^{k+1} ∑_{j=1}^{k+1}(−1)^{j}j!S(j + 1, j)3^{−j}\binom{2j}{j} (∑_{i=1}^{j−1} \frac{3^{i}}{(2i + 1)\binom{2i}{i}} + \frac{2}{3\sqrt{3}}π) . (3)

where the Stirling numbers of the second kind are defined by

S(k, j) = \frac{(−1)^{j}}{j!} ∑_{m=0}^{j}(−1)^{m}m^{k}\binom{j}{m}.  (4)

The aim of the present note is to extend (3) to complex z and thus to continue (1) analytically beyond its circle of convergence |z| = 4.

2 Calculation

We begin with the observation that (m\binom{2m}{m})^{−1} = B(m, m + 1), where B denotes Euler’s beta integral. Hence,

S_{k}(z) = ∫_{0}^{1} \frac{dt}{t} ∑_{m=1}^{∞} m^{k+1}(zt(1 − t))^{m}.  (5)

Next, equation (21) of Girgensohn and Borwein [4],

∑_{m=1}^{∞} m^{p}X^{m} = ∑_{n=1}^{p} ∑_{m=1}^{n}(−1)^{m+n}\binom{n}{m}m^{p}X^{n}(1 − X)^{−n−1},  (6)

gives

S_{k}(z) = ∑_{n=1}^{k+1} ∑_{m=1}^{n}(−1)^{m+n}\binom{n}{m}m^{k+1} ∫_{0}^{1} \frac{dt}{t} \frac{(zt(1 − t))^{n}}{(1 − zt(1 − t))^{n+1}} .  (7)

In the appendix it is shown that

∫_{0}^{1} \frac{dt}{t} \frac{(zt(1 − t))^{n}}{(1 − zt(1 − t))^{n+1}} = \frac{\sqrt{π}Γ(n)}{Γ(n + 1/2)}X^{n} _{2}F_{1}(−1/2, n; n + 1/2; −X)  (8)

where X = z/(4 − z), so

S_{k}(z) = ∑_{n=1}^{k+1} n!B(n, 1/2)S(k + 1, n)X^{n} _{2}F_{1}(−1/2, n; n + 1/2; −X).  (9)

By induction, starting with the tabulated value for n = 1 and using Gauss’ contiguity relations we find (some details are given in the appendix)

2F_{1}(−1/2, n; n + 1/2; −X) =

(\frac{1}{2})_{n} (\frac{1}{n!} + \frac{1}{\sqrt{π}Γ(n)} ∑_{k=0}^{n−1} \frac{(−1)^{k}Γ(k + 1/2)}{(k + 1)!}\binom{n − 1}{k} (\frac{X + 1}{X})^{k+1} ×

[\sqrt{X} sin^{−1} \sqrt{\frac{X}{X + 1}} − \frac{1}{2} ∑_{l=1}^{k} \frac{(l − 1)!}{(1/2)_{l}} (\frac{X}{X + 1})^{l}]) .  (10)

(We have used the ascending factorial notation (a)_{n} = Γ(a + n)/Γ(a)). Therefore we have the principal result

S_{k}(z) = ∑_{n=1}^{k+1} n! (\frac{z}{4 − z} )^{n} S(k + 1, n)×

(\frac{1}{n} + ∑_{p=0}^{n−1}(−1)^{p} \frac{(1/2)_{p}}{(p + 1)!}\binom{n − 1}{p} (\frac{4}{z})^{p+1} (\sqrt{\frac{z}{4 − z}} sin^{−1} \frac{\sqrt{z}}{2} − \frac{1}{2} ∑_{l=1}^{p} \frac{Γ(l)}{(1/2)_{l}} (\frac{z}{4})^{l})) (11)

Equation (11) is rather condensed; in unpacking it, sums with upper limit less than the lower limit are to be interpreted as 0. It is clear from (11) that for rational z

∑_{m=1}^{∞} \frac{n^{k}z^{n}}{\binom{2n}{n}} = R_{1}(z, k) + R_{2}(z, k)\sqrt{\frac{z}{4 − z}} sin^{−1} \frac{\sqrt{z}}{2} ,  (12)

where R_{j} is a rational number.

One sees from (11) that S_{k}(z) is analytic on the two-sheeted Riemann surface formed by two planes cut and rejoined along the real half-line x > 4. The numbers in (12) have the explicit expressions

R_{1}(z, k) =  (13)

∑_{n=1}^{k+1} n!S(k + 1, n) (\frac{z}{4 − z} )^{n} (\frac{1}{n} − \frac{1}{2} ∑_{p=1}^{n−1} ∑_{l=1}^{p} \frac{(−1)^{p}(1/2)_{p}}{(p + 1)!(1/2)_{l}} \binom{n − 1}{p}Γ(l) (\frac{4}{z})^{p−l+1}) ,

R_{2}(z, k) = ∑_{n=1}^{k+1} n!S(k + 1, n) ∑_{p=0}^{n−1} \frac{(−1)^{p}}{(p + 1)!}\binom{n − 1}{p} (\frac{4}{z})^{p+1} .  (14)

3 Asymptotics

It is convenient to work in terms of the exponential generating function

G(z, t) := ∑_{k=0}^{∞} S_{k}(z)\frac{t^{k}}{k!} = S_{0}(ze^{t}) = \frac{z}{4 − ze^{t}} + \frac{4\sqrt{z}e^{t/2}}{(4 − ze^{t})^{3/2}} sin^{−1} \frac{\sqrt{z}e^{t/2}}{2}  (15)

To find the generating functions ρ_{j}(z, t) := ∑ R_{j}(z, k)t^{k}/k!, it would be simplest to start with a series D_{k}(z) = R_{1}(z, k) − R_{2}(z, k)\sqrt{\frac{z}{4−z}} sin^{−1} \sqrt{z}/2, work out its generating function D(z, t) and by taking the sum and difference identify ρ_{1} and ρ_{2}. However, this series has not been found and there is nothing to guarantee its existence in tractable form. Therefore, the ρ_{j} were evaluated directly from (13) and (14). The details are omitted as the results

ρ_{1}(z, t) = \frac{ze^{t}}{4 − ze^{t}} +
\frac{8}{π} \sqrt{\frac{ze^{t}}{(4 − ze^{t})^{3/2}}} (sin^{−1} \frac{\sqrt{z}e^{t/2}}{2} cos^{−1} \frac{\sqrt{z}}{2} − cos^{−1} \frac{\sqrt{z}e^{t/2}}{2} sin^{−1} \frac{\sqrt{z}}{2} ) , (16)

ρ_{2}(z, t) = 4\sqrt{\frac{(4 − z)e^{t}}{(4 − ze^{t})^{3}}}  (17)

are easily verified. In the case z = 2, (15) and (16) are identical to Dyson’s formulas [3, 5] obtained empirically.

In view of the prominent role that the ratio R_{1}(z, t)/R_{2}(z, t) plays in Dyson et al. [3] for z = 2 it is interesting to examine it for general z. From (17) we have

R_{2}(z, k) = \frac{2k!\sqrt{4 − z}}{πi} ∮ \frac{ds}{s^{k+1}} \frac{e^{s/2}}{(4 − ze^{s})^{3/2}} .  (18)

The non-zero singularity closest to s = 0 is s_{0} = ln(4/z) and it dominates the asymptotic behavior. Ignoring the other singularities, distorting the contour to a small circle about s_{0} and translating back to the origin by t = s − s_{0}, we have

R_{2}(z, k) ∼ −\frac{k!\sqrt{4 − z}}{zs_{0}^{k+1}} ∮ \frac{dt}{2πi} \frac{e^{t/2}}{(1 − e^{t}))^{3/2}} .  (19)

The exact value of the integral in (19) is −(2/π)\sqrt{e/(e − 1)}, and so

R_{2}(z, k) ∼ \frac{k!}{(ln(4/z))^{k+1}} \frac{2}{π} \sqrt{\frac{e(4 − z)}{z(e − 1)}}.  (20)

In the same way we obtain

R_{1}(z, k) ∼ \frac{k!}{(ln(4/z))^{k+1}} (\sqrt{2} + \frac{2}{π} (\sqrt{\frac{e}{e − 1}} − \sqrt{2}) cos^{−1} \frac{\sqrt{z}}{2} − \frac{2^{3/2}}{π} sin^{−1} \frac{\sqrt{z}}{2} ) . (21)

4 Discussion

From (20) and (21) we find

lim_{k→∞} (\frac{R_{1}(z, k)}{R_{2}(z, k)} − \sqrt{\frac{z}{4 − z}} sin^{−1} \frac{\sqrt{z}}{2} ) = \sqrt{\frac{z}{4 − z}} (cos^{−1} \frac{\sqrt{z}}{2} − sin^{−1} \frac{\sqrt{z}}{2} ) .  (22)

It thus appears that Lehmer’s choice, z = 2, is the unique permissible case for which the limit (22) vanishes. (Also the Lehmer limit, as defined by Dyson et al. [3], relates to π/4 here rather than π). Finally, for negative integer indices, since

2S_{−k}(z) = _{k+1}F_{k}(1, …, 1; \frac{3}{2}, 2, …, 2; \frac{1}{4}z),  (23)

the fact that S_{−k}(z) can be obtained from S_{0}(z) by successive integrations with respect to z and the explicit evaluations by Lehmer [2], Borwein and Girgensohn [4] and others [6, 7, 8, 9, 10] it should be possible to obtain explicit values for sundry generalized hypergeometric functions.

5 Appendix: Derivation of Equations (8) and (10)

Let us consider, for any integrable function F,

I = ∫_{0}^{1} \frac{dt}{t}F (t(1 − t))

Let u = t(1 − t), so u(0) = u(1) = 0; u(1/2) = 1/4. Then there are two expressions for t:

t_{+} = \frac{1}{2}(1 + \sqrt{1 − 4t}) for \frac{1}{2} ≤ t ≤ 1, with \frac{dt_{+}}{t_{+}} = (1 − \frac{1}{\sqrt{1 − 4u}}) \frac{du}{u}

and

t_{−} = \frac{1}{2}(1 − \sqrt{1 − 4t}) for 0 ≤ t ≤ \frac{1}{2}, with \frac{dt_{+}}{t_{+}} = (1 + \frac{1}{\sqrt{1 − 4u}}) \frac{du}{u}.

Consequently,

I = ∫_{0}^{1/2} \frac{dt_{−}}{t_{−}} F (u) + ∫_{1/2}^{1} \frac{dt_{+}}{t_{+}} F (u) = 2 ∫_{0}^{1/4} \frac{du}{u\sqrt{1 − 4u}}F (u)
= 2 ∫_{0}^{1} \frac{dx}{x\sqrt{1 − x}}F (\frac{1}{4}x) = 2 ∫_{0}^{1} \frac{dt}{(1 − t)\sqrt{t}}F (\frac{1 − t}{4})

and, with t = x^{2},

I = 4 ∫_{0}^{1} \frac{dx}{1 − x^{2}} F (\frac{1 − x^{2}}{4} ) .

Therefore,

L = ∫_{0}^{1} \frac{dt}{t} \frac{(zt(1 − t))^{α}}{(1 − zt(1 − t))^{β}} = 2 (\frac{z}{4})^{α−β} ∫_{0}^{1} dx\frac{(1 − x^{2})^{α−1}}{(a^{2} + x^{2})^{β}} ,

where a^{2} = 1/X = (4 − z)/z.

From standard references

∫_{0}^{1} dx cos(xy)(1 − x^{2})^{α−1} = \sqrt{\frac{πy}{8}} (\frac{2}{y})^{α} Γ(α)J_{α−1/2}(y),

∫_{0}^{∞} dx cos(xy)(a^{2} + x^{2})^{−β} = \frac{\sqrt{π}}{Γ(β)} (\frac{y}{2a})^{β−1/2} K_{β−1/2}(ay)

so, by the Parseval relation for the Fourier transform

∫_{0}^{1} dx\frac{(1 − x^{2})^{α−1}}{(a^{2} + x^{2})^{β}} = \frac{2^{α−β}}{a^{β−1/2}} \frac{Γ(α)}{Γ(β)} ∫_{0}^{∞} dyy^{β−α}J_{α−1/2}(y)K_{β−1/2}(ay).

This is a tabulated Hankel Transform and yields

L = \sqrt{π} (\frac{4}{z})^{β−α} X^{β} \frac{Γ(α)}{Γ(α + 1)} _{2}F_{1} (\frac{1}{2}, β; α + \frac{1}{2}; −X) .

Consequently

∫_{0}^{1} \frac{dx}{x} \frac{(zt(1 − t))^{n}}{(1 − zt(1 − t))^{n+1}} = \sqrt{π} (\frac{4}{z}) X^{n+1} _{2}F_{1} (\frac{1}{2}, n + 1; n + \frac{1}{2}; −X)

However, since _{2}F_{1}(a, b; c; z) = (1 − z)^{c−a−b} _{2}F_{1}(c − a, c − b; c; z),

2F_{1} (\frac{1}{2}, n + 1; n + \frac{1}{2}; −X) = (1 + X)^{−1} _{2}F_{1} (−\frac{1}{2}; n; n + \frac{1}{2}; −X)

Next, we note that [11, p . 590]

2F_{1}(−1/2, 1; 3/2; z) = \frac{1}{2} (1 + (1 − z)\begin{matrix}tanh & −1 & \sqrt{z} \\ √ & \overline{z} \end{matrix} ) .

With z → −z, noting that −i tanh^{−1} iw = sin^{−1} \sqrt{\frac{w}{1+w}} one has

2F_{1}(−1/2, 1; 3/2; −z) = \frac{1}{2}(1 + (1 + z)\begin{matrix}sin & −1 & \sqrt{\frac{z}{1+z}} \\ √ & \overline{z} \end{matrix}).  (24)

We next apply Gauss’ differentiation formula

\frac{d}{dz} ((1 + z)^{k} _{2}F_{1}(−1/2, k; k + 1/2; −z)) =

\frac{2k(k + 1)}{2k + 1}(1 + z) _{2}F_{1}(−1/2, k + 1; k + 3/2; −z).  (25)

Iteration of (25) starting with (24), after a great deal of tedious algebra, aided by Mathematica, results in (10).

6 Acknowledgements

The author is grateful to Profs. N. E. Frankel for suggesting this problem and F. J. Dyson for enlightening correspondence.

References

[1] R. Apéry, Irrationalité de ζ(2) et ζ(3), Asterique 61 (1979), 11–13.

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

[3] F. J. Dyson, N. E. Frankel, and M. L. Glasser, Lehmer’s second interesting series,

   preprint. Available at http//:arxiv.org/abs/1009.4274.

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

[5] F. J. Dyson, personal correspondence.

[6] I. J. Zucker, On the series ∑ \binom{2n}{n}^{−1}n^{−k} and related series, J. Number Theory 20 (1981), 92–102.

[7] N. Batir, On the series ∑ \binom{2k}{k}^{−1}k^{−a}x^{k}, Proc. Indian Acad. Sci. 115 (2005), 371–382.

[8] J. -H. Yang and F. -Z. Zhao, Certain sums involving inverses of binomial coefficients and some integrals, J. Integer Sequences 10 (2007), 1–11.

[9] R. Sprugnoli, Sums of the reciprocals of the central binomial coefficients, INTEGERS 6 (2006), 1–18.

[10] R. J. Mathar, Corrigenda to “Interesting series involving the central binomial coefficient by D. H. Lehmer, preprint. Available at http//:arxiv.org/abs/0905.0215v1.

[11] A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, Tables of Integrals and Series Vol. 3, Gordon and Breach, 1998.

2000 Mathematics Subject Classification: Primary 11B65; Secondary 33B05. Keywords: Binomial coefficient, infinite series, generating function.

(Concerned with sequences \underline{A008277} and \underline{A145557}.)

Received October 8 2011; revised version received April 2 2012. Published in Journal of Integer Sequences, April 6 2012.

Return to Journal of Integer Sequences home page.