1. Introduction The Riemann zeta function can be defined by ... - RGMIA

LOWER AND UPPER BOUNDS OF ζ(3) QIU-MING LUO, ZONG-LI WEI, AND FENG QI

Abstract. In this short note, using refinements of Jordan’s inequality and an integral expression of ζ(3), the lower and upper bounds of ζ(3) are obtained, and some related results are improved.

1. Introduction The Riemann zeta function can be defined by the integral Z ∞ x−1 1 u ζ(x) = du, Γ(x) 0 eu − 1

(1)

where x > 1. If x is an integer n, we obtain the most common form of the function ζ(n), which is given by ∞ X 1 . kn

ζ(n) =

(2)

k=1

For n = 1, the zeta function reduces to the harmonic series, which is divergent. The Riemann zeta function can also be defined in terms of multiple integrals by Z 1 Z 1 Qn dx i=1 Qn i . ζ(n) = ··· (3) 1 − xi 0 0 i=1 | {z } n

An additional identity is lim ζ(s) −

s→1

1 = γ, s−1

(4)

where γ is the Euler-Mascheroni constant. 2000 Mathematics Subject Classification. 11R40, 42A16. Key words and phrases. Lower and upper bounds, Riemann zeta function, Jordan inequality. The third author was supported in part by NSF (#10001016) of China, SF for the Prominent Youth of Henan Province, SF of Henan Innovation Talents at Universities, NSF of Henan Province (#004051800), SF for Pure Research of Natural Science of the Education Department of Henan Province (#1999110004), Doctor Fund of Jiaozuo Institute of Technology, China. This paper was typeset using AMS-LATEX. 1

2

Q.-M. LUO, Z.-L. WEI, AND F. QI

The Euler product formula can also be written as "∞ #−1 Y −x ζ(x) = (1 − i ) .

(5)

i=2

The function ζ(n) was proved to be transcendental for all even n. For n = 2k we have: 2n−1 |Bn | π n , (6) n! where Bn is the Bernoulli number. Another intimate connection with the Bernoulli ζ(n) =

numbers is provided by Bn = (−1)n+1 nζ(1 − n).

(7)

The Riemann zeta function is related to the gamma function and it has important applications in mathematics, especially in the Analytic Number Theory. The Riemann zeta function may be computed analytically for even n using either Contour integration or Parseval’s theorem with the appropriate Fourier series. No analytic form for ζ(n) is known for odd n = 2k + 1, but ζ(2k + 1) can be expressed as the sum limit ζ(2k + 1) =

 π 2k+1 2

1

lim

t→∞ t2k+1

∞ h X

cot

i=1



i i2k+1 . 2t + 1

(8)

The values for the first few integral arguments are 1 ζ(0) = − , 2

ζ(1) = ∞,

ζ(3) = 1.2020569032 · · · ,

ζ(4) =

π4 , 90

In [9, p. 81], using the Jordan inequality 1