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12/01/02
FINDING ζ(2p) FROM A PRODUCT OF SINES
(Final Version Accepted by American Mathematical Monthly)
Thomas J. Osler
Mathematics Department
Rowan University
Glassboro, NJ 08028
[email protected]
∞
1
(valid for Re( z ) > 1 )
z
n =1 n
The zeta function ζ ( z ) given by the series ζ ( z ) = ∑
was first evaluated in closed form by Euler [5] when z is a positive even integer. The
result is
(−1) p +1 22 p −1 B2 p 2 p
1
=
π .
2p
(2 p )!
n =1 n
∞
(1)
ζ (2 p) = ∑
Here the numbers Bn are called Bernoulli’s numbers, and they are all rational. The first
few are
1
1
1
1
B0 = 1, B1 = − , B2 = , B4 = − , B6 = ,... , and
2
6
30
42
B3 = B5 = B7 = ... = 0 .
These can all be calculated recursively by starting with B0 = 1 , and using
n
n
Bn = ∑  Bk ,
k =0  k 
for n = 2,3, 4,". When n = 2, this gives B1 = −1/ 2 . (See Knopp [6], page 183, for an
equivalent formula.) Several additional methods of deriving (1) have been given since
Euler, some of which are found in [1], [3], [4], and [6]. We present a method here that we
were unable to locate in the literature.
We begin our evaluation of ζ (2 p ) by considering the following two lemmas.
2
∞
Lemma 1: If
∑ an is an absolutely convergent series, then the product
n =1
∞
∏ (1 + a z ) is an
n
n =1
entire function of z whose power series expansion about the origin begins as follows:

∞
∞
∏ (1 + a z ) = 1 +  ∑ a
n
n =1
n =1
n

 z +" .

This Lemma is an immediate consequence of [6, p.439, example 1], where
explicit formulas are also given for the coefficients of z 2 , z 3 ," . (Euler also used this
Lemma but he required the coefficients of the higher powers. In this paper, we use only
the constant and linear term.)
The next lemma is the well known factorization of 1 − t p in terms of pth roots of
unity.
Lemma 2: If ω = eπ i / p , where p is a positive integer, the following algebraic identity is
valid for all t:
p −1
∏ (1 − ω t ) = 1 − t
2k
p
.
k =0
z2
When t = 2 2 , Lemma 2 gives
π n
 (ω k z ) 2 
z2 p


1
−
=
1
−
.
∏
π 2n2 
π 2 p n2 p
k =0 


p −1
(2)
∞

z2 
Next use the infinite product representation sin z = z ∏  1 − 2 2  with z replaced by
π n 
n =1 
ω k z and form the finite product of sines
p −1
∏ sin (ω z ) = ω
k =0
k
p ( p −1) / 2
 (ω k z ) 2 
z ∏∏ 1 − 2 2 
π n 
n =1 k = 0 


p
∞
p −1
3
∞

z2 p
= ω p ( p −1) / 2 z p ∏ 1 − 2 p 2 p
π n
n =1 

,

where in the last step we used (2). Now define
p −1
∞

z2 p
g ( z ) = ω p (1− p ) / 2 ∏ sin (ω k z ) = z p ∏ 1 − 2 p 2 p
π n
k =0
n =1 

.

Apply Lemma 1 to get the power series expansion
p −1
g ( z ) = ω p (1− p ) / 2 ∏ sin (ω k z ) = z p −
(3)
k =0
ζ (2 p) 3 p
z +" .
π 2p
To find ζ (2 p ) from (3), we could replace each sine function by its Taylor’s
series and then multiply and equate coefficients of z 3 p . However, to derive (1), it is
easier if we first take the derivative of g ( z ) . From (3) we get
g '( z ) = pz p −1 − 3 p
(4)
ζ (2 p) 3 p −1
+" .
z
π 2p
Another form of expansion (4) can be obtained by logarithmic differentiation of the finite
product defining g ( z ) . We have from (3)
p −1
log g ( z ) = ∑ log sin (ω k z ) ,
k =0
whose derivative is
g '( z ) p −1 k
= ∑ ω cot (ω k z ) ,
g ( z ) k =0
and hence
ζ (2 p)


g '( z ) = g ( z )∑ ω k cot (ω k z ) =  z p − 2 p z 3 p + "  ∑ ω k cot (ω k z ) .
π

 k =0
k =0
p −1
(5)
p −1
To evaluate ζ (2 p ) we equate the coefficient of z 3 p −1 in (4) with that in (5).
Contributions to this coefficient in (5) come from two sources arising from the Laurent
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expansion of the sum of cotangents (from the coefficient of z −1 ), and from the coefficient
of z 2 p −1 . Because
cot z =
1 ∞
22 r B2 r
+ ∑ cr z 2 r −1 , where cr = (−1) r
,
z r =1
(2r )!
we have
∞
1
ω cot (ω z ) = + ω k ∑ cr (ω k z )
z
r =1
k
k
2 r −1
=
1 ∞
+ ∑ crω 2 rk z 2 r −1 .
z r =1
When this is summed over k the total contribution from z −1 is p, while that from z 2 p −1 is
pc p because ω 2 kp = 1 . Equating the coefficient of z 3 p −1 in (4) with that in (5) we find
−3 p
ζ (2 p)
ζ (2 p)
= − p 2 p + pc p .
2p
π
π
22 p −1 B2 p
1
This gives ζ (2 p ) = − c pπ 2 p = (−1) p +1
as required. ,
2
(2 p)!
Acknowledgement
The author wishes to thank the referee for considerably improving the exposition
of this note.
REFERENCES
[1] Apostol, Tom M., Another elementary proof of Euler’s formula for ζ (2n) , American
Mathematical Monthly, 80(1973), pp. 425-431.
[2] Ayoub, Raymond, Euler and the Zeta function, American Mathematical Monthly,
81(1974), pp. 1067-1086.
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[3] Berndt, Bruce C., Elementary evaluation of ζ ( z ) , Mathematics Magazine, 48(1975),
pp. 148-154.
[4] Edwards, H. M., Riemann’s Zeta Function, Academic Press, New York, 1974.
[5] Euler, Leonard Introduction to Analysis of the Infinite, Book I, (Translated by John D.
Blanton) Springer-Verlag, New York, 1988, pp.137-153.
[6] Knopp, Konrad, Theory and Application of Infinite Series, Dover Publications, New
York, 1990. (A translation by R. C. H. Young of the 4th German edition of 1947.)