Formulas for Odd Zeta Values and Powers of π

Marc Chamberland and Patrick Lopatto Department of Mathematics and Statistics Grinnell College Grinnell, IA 50112 USA

                  chamberl@math.grinnell.edu

Abstract Plouffe conjectured fast converging series formulas for π^{2n+1} and ζ(2n+1) for small values of n. We find the general pattern for all integer values of n and offer a proof.

1 Introduction

It took nearly one hundred years for the Basel Problem — finding a closed form solution to ∑_{k=1}^{∞} 1/k^{2} — to see a solution. Euler solved this in 1735 and essentially solved the problem where the power of two is replaced with any even power. This formula is now usually written as

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

where ζ(s) is the Riemann zeta function and B_{k} is the k^{th} Bernoulli number uniquely defined by the generating function

\frac{x}{e^{x} − 1} = ∑_{n=0}^{∞} \frac{B_{n}x^{n}}{n!} , |x| < 2π.

and whose first few values are 0, −1/2, 1/6, 0, −1/30, . . .. However, finding a closed form for ζ(2n + 1) has remained an open problem. Only in 1979 did Apéry show that ζ(3) is irrational. His proof involved the snappy acceleration

ζ(3) = \frac{5}{2} ∑_{n=1}^{∞} \frac{(−1)^{n−1}}{n^{3}\binom{2n}{n}} .

This tidy formula does not generalize to the other odd zeta values, but other representations, such as nested sums or integrals, have been well-studied. The hunt for a clean result like Euler’s has largely been abandoned, leaving researchers with the goal of finding formulas which either converge quickly or have an elegant form.

Following his success in discovering a new formula for π, Simon Plouffe[3] postulated several identities which relate either π^{m} or ζ(m) to three infinite series. Letting

S_{n}(r) = ∑_{k=1}^{∞} \frac{1}{k^{n}(e^{πrk} − 1)},

the first few examples are

π = 72S_{1}(1) − 96S_{1}(2) + 24S_{1}(4)
π^{3} = 720S_{3}(1) − 900S_{3}(2) + 180S_{3}(4)
π^{5} = 7056S_{5}(1) − 6993S_{5}(2) + 63S_{5}(4)
π^{7} = \frac{907200}{13}S_{7}(1) − 70875S_{7}(2) + \frac{14175}{13}S_{5}(4)

and

ζ(3) = 28S_{3}(1) − 37S_{3}(2) + 7S_{3}(4)
ζ(5) = 24S_{5}(1) − \frac{259}{10}S_{5}(2) + \frac{1}{10}S_{5}(4)
ζ(7) = \frac{304}{13}S_{7}(1) − \frac{103}{4}S_{7}(2) + \frac{19}{52}S_{7}(4).

Plouffe conjectured these formulas by first assuming, for example, that there exist constants a, b, and c such that

π = aS_{1}(1) + bS_{1}(2) + cS_{1}(4).

By obtaining accurate approximations of each the three series, he wrote some computer code to postulate rational values for a, b, c. Today, such integer relations algorithms have been used to discover many formulas. The widely used PSLQ algorithm, developed by Ferguson and Bailey[2], is implemented in Maple. The following Maple code (using Maple 14) solves the above problem:

> with(IntegerRelations):
> Digits := 100;
> S := r -> sum( 1/k/( exp(Pi*r*k)-1 ), k=1..infinity );
> PSLQ( [ Pi, S(1), S(2), S(4) ] );

The PSLQ command returns the vector [−1, 72, −96, 24], producing the first formula.

While the computer can be used to conjecture the coefficients of a specific power, finding the general sequence of rationals has remained an open problem. This note finds the sequences and offers formal proofs.

2 Exact Formulas

While it does not seem that ζ(2n+1) is a rational multiple of π^{2n+1}, a result in Ramanujan’s notebooks gives a relationship with infinite series which converge quickly.

Theorem 1. (Ramanujan) If α > 0, β > 0, and αβ = π^{2}, then

α^{−n} \{\frac{1}{2}ζ(2n + 1) + S_{2n+1}(2α)\} =

(−β)^{−n} \{\frac{1}{2}ζ(2n + 1) + S_{2n+1}(2β)\} − 4^{n} ∑_{k=0}^{n+1}(−1)^{k} \frac{B_{2k}B_{2n+2−2k}}{(2k)!(2n + 2 − 2k)!}α^{n+1−k}β^{k}.

Using α = β = π in Proposition 1 and defining

F_{n} = ∑_{k=0}^{n+1}(−1)^{k} \frac{B_{2k}B_{2n+2−2k}}{(2k)!(2n + 2 − 2k)!},

we have

(π^{−n} − (−π)^{−n}) (\frac{1}{2}ζ(2n + 1) + S_{2n+1}(2π)) = −4^{n}π^{n+1}F_{n}.

To find formulas for the odd zeta values and powers of π, we will divide these into two classes: ζ(4m − 1) and ζ(4m + 1). Such distinctions can be seen in other studies; see [1, pp. 137–139].

First we find the formulas for π^{4m−1} and ζ(4m − 1). If n is odd, then

\frac{1}{2}ζ(2n + 1) + S_{2n+1}(2π) = \frac{−4^{n}}{2} π^{2n+1}F_{n}  (1)

Using α = π/2 and β = 2π in Proposition 1 and defining

G_{n} = ∑_{k=0}^{n+1}(−4)^{k} \frac{B_{2k}B_{2n+2−2k}}{(2k)!(2n + 2 − 2k)!},

one has

\frac{1}{2}ζ(2n + 1) = \frac{S_{2n+1}(4π) + 4^{n}S_{2n+1}(π) + \frac{4^{n}}{2} π^{2n+1}G_{n}}{−(4^{n} + 1)} .

Combining this with equation (1) yields

\frac{4^{n}S_{2n+1}(π) − (4^{n} + 1)S_{2n+1}(2π) + S_{2n+1}(4π)}{\frac{4^{n}}{2} (4^{n} + 1)F_{n} − \frac{4^{n}}{2} G_{n}} = π^{2n+1}

Substituting n = 2m − 1 and defining

D_{m} = 4^{2m−1}[(4^{2m−1} + 1)F_{2m−1} − G_{2m−1}]/2

produces

π^{4m−1} = \frac{4^{2m−1}}{D_{m}} S_{4m−1}(π) − \frac{4^{2m−1} + 1}{D_{m}}S_{4m−1}(2π) + \frac{1}{D_{m}} S_{4m−1}(4π)

This identity may be used in conjunction with equation (1) to obtain

ζ(4m − 1) = −\frac{F_{2m−1}4^{4m−2}}{D_{m}} S_{4m−1}(π) + \frac{G_{2m−1}4^{2m−1}}{D_{m}} S_{4m−1}(2π) − \frac{F_{2m−1}4^{2m−1}}{D_{m}} S_{4m−1}(4π).

To obtain formulas for the 4m + 1 cases, use α = π/2 and β = 2π in Theorem 1 with n = 2m to obtain

ζ(4m + 1) = \frac{\frac{4^{2m}}{2} π^{4m+1}G_{2m} − S_{4m+1}(4π) + 4^{2m}S_{4m+1}(π)}{\frac{1}{2}(1 − 4^{2m})}.  (2)

Define T_{n}(r) (similar to S_{n}(r)) by

T_{n}(r) = ∑_{k=1}^{∞} \frac{1}{k^{n}(e^{rk} + 1)}

and another finite sum of Bernoulli numbers by

H_{n} = ∑_{k=0}^{n}(−4)^{n+k} \frac{B_{4k}B_{4n+2−4k}}{(4k)!(4n + 2 − 4k)!}.

Vepstas [4] cites an expression credited to Ramanujan:

(1 + (−4)^{m} − 2^{4m+1})ζ(4m + 1) =

2T_{4m+1}(2π) + 2(2^{4m+1} − (−4)^{m})S_{4m+1}(2π) + 2^{4m+1}π^{4m+1}H_{m} + 2^{4m}π^{4m+1}G_{2m}. Vepstas also produces a formula to show the relationship between T_{k} and S_{k}:

T_{k}(x) = S_{k}(s) − 2S_{k}(2x)

Combining the last two equations produces

\frac{1 + (−4)^{m} − 2^{4m+1}}{\frac{1}{2}(1 − 4^{2m})} (\frac{4^{2m}}{2} π^{4m+1}G_{2m} − S_{4m+1}(4π) + 4^{2m}S_{4m+1}(π)) =

2[2^{4m+1} − (−4)^{m} + 1]S_{4m+1}(2π) − 4S_{4m+1}(4π) + 2^{4m+1}π^{4m+1}H_{m} + 2^{4m}π^{4m+1}G_{2m} Letting

K_{m} = \frac{\frac{1}{2}(1 − 4^{2m})}{1 + (−4)^{m} − 2^{4m+1}}

and

E_{m} = \frac{4^{2m}}{2} G_{2m} − 2^{4m+1}K_{m}H_{m} − 2^{4m}K_{m}G_{2m},

one eventually finds

π^{4m+1} = −\frac{4^{2m}}{E_{m}} S_{4m+1}(π) + \frac{2K_{m}[2^{4m+1} − (−4)^{m} + 1]}{E_{m}}S_{4m+1}(2π) + \frac{(1 − 4K_{m})}{E_{m}}S_{4m+1}(4π)

Substituting this into equation (2) produces

ζ(4m + 1) = −\frac{16^{m}(G(2m)16^{m} + 2E_{m})}{(−1 + 16^{m})E_{m}}S_{4m+1}(π)

−\frac{2G_{2m}K_{m}16^{m}(2 · 16^{m} − (−4)^{m} + 1)}{(−1 + 16^{m})E_{m}}S_{4m+1}(2π)

−\frac{G(2m)16^{m} + 4G_{2m}16^{m}K_{m} + 2E_{m}}{(−1 + 16^{m})E_{m}} S_{4m+1}(4π).

References

[1] D. Bailey and J. Borwein. Experimentation in Mathematics. AK Peters, 2004.

[2] H. R. P. Ferguson and D. H. Bailey. A polynomial time, numerically stable integer relation algorithm. RNR Techn. Rept. RNR-91-032, Jul. 14, 1992.

[3] Plouffe, S. Identities inspired by Ramanujan notebooks (part 2),

   http://www.plouffe.fr/simon/, April 2006.

[4] Vepstas, L. On Plouffe’s Ramanujan identities, http://arxiv.org/abs/math/0609775, November 2010.

2010 Mathematics Subject Classification: Primary 11Y60. Keywords: π, Riemann zeta function.

Received January 24 2011; revised version received February 7 2011. Published in Journal of Integer Sequences, February 20 2011.

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