International Journal of Pure and Applied Mathematics
Volume 91 No. 4 2014, 421-433
ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version)
url: http://www.ijpam.eu
doi: http://dx.doi.org/10.12732/ijpam.v91i4.1
AP
ijpam.eu
ON SUMMATION OF SUBSERIES OF
THE RIEMANN ZETA FUNCTION
Hongwei Chen
Department of Mathematics
Christopher Newport University
Newport News, VA 23606, USA
Abstract: Three classes of subseries of the Riemann zeta function are evaluated in closed form. All the results are expressed in terms of the Riemann zeta
function itself and powers of π. By using the Euler totient function, we prove
that these three classes are unique subseries that exhibit this kind of closed
form.
AMS Subject Classification: 11M06, 11M35, 11B68, 33B10
Key Words: Riemann zeta function, Hurwitz zeta function, Bernoulli polynomials, reflection sums, derivatives of the cotangent.
1. Introduction
Let ζ(s) be the Riemann zeta function defined by
∞
X
1
.
ζ(s) :=
s
n
n=1
The series converges for all s > 1. There are many formulas related the subseries
Received:
October 8, 2013
c 2014 Academic Publications, Ltd.
url: www.acadpubl.eu
422
H. Chen
of the Riemann zeta function
∞
X
1
, (a, b, s ∈ N, gcd(a, b) = 1)
ζ(a, b, s) :=
(ak + b)s
(1)
k=0
to ζ(s) itself, for example, from the well-known expression
ζ(2, 1, s) =
∞
X
k=0
1
= (1 − 2−s ) ζ(s),
(2k + 1)s
to more involved results such as
∞
X
1
1
2 √ 3
−3
ζ(3, 1, 3) =
=
(1
−
3
)ζ(3)
+
3π ,
(3k + 1)3
2
35
ζ(3, 1, 5) =
k=0
∞
X
k=0
1
2 √
1
= (1 − 3−5 )ζ(5) + 7 3 π 5 .
5
(3k + 1)
2
3
(2)
(3)
(4)
Motivated by such intriguing results, it is natural for us to ask
Q1. When does a given series ζ(a, b, s) have a closed form?
Q2. Is the closed form always a linear combination of π s and ζ(s)?
Notice that these series are all subseries of the Riemann zeta function. Appealing to the polylogarithm of Lewin, Lis (z) (see [3, p.189] or [4, p.611]) defined
by
∞
X
zk
.
(5)
Lis (z) :=
ks
k=1
the series multisection formula (see [2, Theorem 10.1, p.110]) yields, for example,
∞
X
1
1
1
1
2
ζ(3, 1, 3) =
Li
(ω)
+
=
Li
(ω
)
,
(6)
Li
(1)
+
3
3
3
(3k + 1)3
3
ω
ω2
k=0
e2πi/3 ,
where ω =
the primitive cubic root. However, due to the limited knowledge of Lis (z) for complex arguments (see [3]), we can’t recover (3) from (6).
Using the contour integral, Zucker and Robertson [7] evaluated some Dirichlet L-series exactly, which are alternating series of (1). The main results are
summarized in the Appendix A of [1]. In this paper, we present another method
to study ζ(a, b, s) in closed form. We will show that the affirmative answers
to Q1 and Q2 only hold for a = 3, 4, 6 and s odd. Our approach is in purely
Eulerian style, and our all manipulation of series and appearance of identities
have the familiar Eulerian flair.
ON SUMMATION OF SUBSERIES OF...
423
2. Proof of (3)
Before we proceed to develop the closed form for ζ(a, b, s), we give an elementary
proof of (3), and will see that the proof of the extensions follows by analogy.
First, inspired by Euler’s infinite product formula
Y
ζ(s) =
(1 − p−s )−1 ,
p prime
we find that
1
1
1
1
1
1
1
ζ(3)(1 − 3−3 ) = 1 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + · · ·
4
5 7 8
10
11
2
1
1
1
1
1
=
1 + 3 + 3 + ··· +
+
+
+ ···
4
7
23 53 83
∞
∞
X
X
1
1
+
.
=
3
(3k + 1)
(3k + 2)3
(7)
k=0
k=0
On the other hand, recall that Hurwitz zeta function (see [4, Chapter 25.11,
p.607]), which is defined by
ζ(s, a) :=
∞
X
k=0
1
.
(k + a)s
First we will need a lemma that converts the series into an integral.
Lemma 1. For s > 1 we have
ζ(s, a) =
1
Γ(s)
Z
∞
0
xs−1 e−ax
dx,
1 − e−x
(8)
where Γ is the Euler gamma function defined by
Z ∞
ts−1 e−t dt.
Γ(s) =
0
Proof. Note that
∞
X
1
=
e−kx .
1 − e−x
k=0
Therefore,
Z
∞
0
X
xs−1 e−ax
dx
=
1 − e−x
∞
k=0
Z
0
∞
xs−1 e−(k+a)x dx.
(9)
424
H. Chen
The change of variable t = (k + a)x yields
Z
0
∞
X
xs−1 e−ax
1
dx =
−x
1−e
(k + a)s
∞
k=0
Z
∞
ts−1 e−t dt = ζ(s, a)Γ(s),
0
which is equivalent to (8).
Next, in terms of the integral representation (8), we establish a lemma
connecting the reflection sum of the Hurwitz zeta function:
Lemma 2. For 0 < µ < 1, n ∈ N we have
(−1)n−1 π
(cot(µπ))(n−1) ,
(n − 1)!
ζ(n, µ) − (−1)n−1 ζ(n, 1 − µ) =
(10)
where f (k) denotes the k-th derivative.
Proof. Applying (8) we see that
ζ(n, µ) − (−1)n−1 ζ(n, 1 − µ)
1
=
(n − 1)!
Z
0
∞
xn−1 (e−µx − (−1)n−1 e−(1−µ)x )
dx. (11)
1 − e−x
Via the parametric differentiation, this can be related to
f (µ) =
Z
∞
0
e−µx − e−(1−µ)x
dx.
1 − e−x
In terms of (9), we have
Z ∞ −(1−µ)x
e
e−µx
dx
−
dx
−x
1
−
e
1 − e−x
0
0
∞
∞
X
X
1
1
=
−
k+µ
k+1−µ
f (µ) =
=
=
Z
∞
k=0
k=0
∞
X
∞
X
1
+
k+µ
k=0
∞
X
k=−∞
k=0
1
.
k+µ
1
−(k + 1) + µ
ON SUMMATION OF SUBSERIES OF...
425
Appealing to Euler’s elegant partial fraction decomposition (see [2, Eq. 6.27,
p.68])
∞
X
1
cot x =
,
kπ + x
k=−∞
we find that
f (µ) = π cot(µπ).
Taking (n − 1)th derivative of f (µ) gives the desired result (10).
Now, returning to (3), notice that
∞
X
k=0
∞
X
1
1
−
(3k + 1)3
(3k + 2)3
k=0
=
1
(ζ(3, 1/3) − ζ(3, 2/3)) .
33
Putting n = 3 in (10) yields
∞
X
1
π
4 √ 3
1
′′
−
=
(cot(µπ))
|
=
3π .
µ=1/3
(3k + 1)3
(3k + 2)3
2! · 33
35
∞
X
(12)
k=0
k=0
In conjunction with (7), this yields (3) as expected. As a byproduct, we also
find the closed form for its reflection sum
∞
X
k=0
2 √ 3
1
1
−3
3π .
(1
−
3
)ξ(3)
−
=
(3k + 2)3
2
35
(13)
In general, observe that if n = 2m + 1, applying (10) yields
∞
X
k=0
=
∞
X
1
1
−
2m+1
(3k + 1)
(3k + 2)2m+1
1
32m+1
k=0
(ζ(2m + 1, 1/3) − ζ(2m + 1, 2/3)) =
π
32m+1 (2m)!
(cot(µπ))(2m) |µ=1/3 .
Coupling with
∞
X
k=0
we obtain
∞
X
1
1
+
= (1 − 3−n )ζ(n),
n
(3k + 1)
(3k + 2)n
k=0
(14)
426
H. Chen
∞
X
k=0
1
(3k + 1)2m+1
=
∞
X
k=0
π
1
(1 − 3−(2m+1) )ζ(2m + 1) +
(cot(µπ))(2m) |µ=1/3 , (15)
2m+1
2
2·3
(2m)!
1
(3k + 2)2m+1
=
π
1
(1 − 3−(2m+1) )ζ(2m + 1) −
(cot(µπ))(2m) |µ=1/3 . (16)
2m+1
2
2·3
(2m)!
√
In particular, setting m = 2 and using that (cot(µπ))(4) |µ=1/3 = 32 3π 4 /3,
we arrive at (4).
However, if n = 2m, (10) yields the same equation as (14). In contrast to
(2) where a = 2, we are unable to find the closed form for
ζ(3, 1, 2m) =
∞
X
k=0
1
(3k + 1)2m
in terms of ζ(2m) and π 2m , even for m = 1.
As we end this section, for rational µ, we show how the values of cot(µπ)(n)
are related to the Bernoulli polynomials — another favorite of Euler. This
lemma will enable us to answer Q2.
Lemma 3. If a, b, n ∈ N, 1 ≤ b < a and (a, b) = 1, then
(cot(µπ))(2n) |µ=b/a = (−1)n−1
(2n−1)
(cot(µπ))
a
2(2aπ)2n X
j
2bjπ
B2n+1
sin
, (17)
2n + 1
a
a
j=1
2n−1
n (2aπ)
|µ=b/a = (−1)
n
a
X
j=1
2bjπ
j
B2n
cos
. (18)
a
a
Here Bn (x) is the Bernoulli polynomial of degree n (see [4, Chapter 24]).
Proof. The approach is based on another way to evaluate ζ(n, µ) ± ζ(n, 1 −
µ). Notice that
ζ(n, b/a) = an ζ(a, b, n).
ON SUMMATION OF SUBSERIES OF...
427
Applying the series multisection formula and (5), we have
ζ(n, b/a) = a
n−1
a
X
ω −jb Lin (ω j ),
j=1
where ω = e2πi/a . Simplification yields
ζ(n, b/a) + ζ(n, 1 − b/a) = 2a
n−1
ζ(n, b/a) − ζ(n, 1 − b/a) = 2a
n−1
a
X
j=1
a
X
j=1
∞
X
cos(2kjπ/a)
kn
!
2bjπ
cos
, (19)
a
k=1
!
∞
X
2bjπ
sin(2kjπ/a)
sin
. (20)
kn
a
k=1
The proposed formulas (17) and (18) now follow from the Fourier series for
Bn (x) (see [4, Eq. 24.8.1 and Eq. 24.8.2, p.592]):
B2n (x) = (−1)n−1
B2n−1 (x) = (−1)n
∞
2(2n)! X cos(2kxπ)
,
(2π)2n
k2n
k=1
∞
X
2(2n − 1)!
(2π)2n−1
k=1
sin(2kxπ)
.
k2n−1
When a = 3, b = 1, in view of the well-known fact that B2m+1 (1) = 0 and
the formulas
Bn (0) = (−1)n Bn (1) = Bn , (see [4, Eq. 24.4.25, p.590]),
Bn (x) = (−1)n Bn (1 − x), (see [4, Eq. 24.4.3, p.589]),
1 1−2m
B2m (1/3) =
(3
− 1)B2m , (see [4, Eq. 24.4.29, p.590]),
2
(21)
(22)
where Bn is the Bernoulli number, applying (17) and (18) we have
2 · 62m B2m+1 (1/3) √ 2m
3π ,
2m + 1
22m−1 (32m − 1)
= (−1)m
B2m π 2m−1 .
2m
(cot(µπ))(2m) |µ=1/3 = (−1)m−1
(cot(µπ))(2m−1) |µ=1/3
(23)
(24)
In conjunction with (14), by (24) we recover another Euler’s famous result:
!
∞
∞
X
X
1
1
1
+
ζ(2m) =
1 − 3−2m
(3k + 1)2m
(3k + 2)2m
k=0
k=0
428
H. Chen
32m
1
(ζ(2m, 1/3) + ζ(2m, 2/3))
2m
2m
3 −1 3
π
(cot(µπ))(2m−1) |µ=1/3
= − 2m
(3 − 1)(2m − 1)!
22m−1
= (−1)m+1
B2m π 2m .
(2m)!
=
Furthermore, (23) and (24) enable us to rewrite (15) and (16) as
∞
X
k=0
1
(3k + 1)2m+1
1
(1 − 3−(2m+1) )ζ(2m + 1)
2
=
+ (−1)m−1
∞
X
k=0
1
(3k + 2)2m+1
√
22m
B2m+1 (1/3) 3 π 2m+1 , (25)
3 · (2m + 1)!
1
(1 − 3−(2m+1) )ζ(2m + 1)
2
=
+ (−1)m
√
22m
B2m+1 (1/3) 3 π 2m+1 .
3 · (2m + 1)!
(26)
These formulas yield the affirmative answer to Q1 and Q2 for a = 3, s = 2m+1.
3. Extensions to a = 4, 6
Observe that finding the closed form of ζ(3, 1, n) is based on a system of two
equations for ζ(3, 1, n) and its reflection sum ζ(3, 2, n). Here, when n is odd,
(10) provides one equation, while the decomposition (14) supplies another.
Along these lines, in this section we establish closed formulas of ζ(a, b, n) for
a = 4, 6 and n odd. First, similar to (7), we have
1
1
1
1
1
1
1
ζ(n)(1 − 2−n ) = 1 + n + n + n + n + n + n + n + · · ·
3
5
7
9
11
13
15
1
1
1
1
1
=
1 + n + n + ··· +
+
+
+
·
·
·
5
9
3n 7n 11n
∞
∞
X
X
1
1
=
+
.
(27)
(4k + 1)n
(4k + 2)n
k=0
k=0
On the other hand, using (10), we find that
∞
X
k=0
X
1
1
(−1)n−1 π
n−1
−
(−1)
=
(cot(µx))(n−1) |µ=1/4 . (28)
(4k + 1)n
(4k + 2)n
4n (n − 1)!
∞
k=0
ON SUMMATION OF SUBSERIES OF...
429
If n = 2m + 1, solving the coupled system of (27) and (28) yields
∞
X
k=0
∞
X
k=0
1
(4k + 1)2m+1
1
(4k + 3)2m+1
=
=
1
(1 − 2−(2m+1) )ζ(2m + 1)
2
π
(cot(µπ))(2m) |µ=1/4 ,
+
2m+1
2·4
(2m)!
1
(1 − 2−(2m+1) )ζ(2m + 1)
2
π
(cot(µπ))(2m) |µ=1/4 .
−
2m+1
2·4
(2m)!
(29)
(30)
By (17) and (21), we have
2 · 82m 2m
π (B2m+1 (1/4) − B2m+1 (3/4))
2m + 1
4 · 82m B2m+1 (1/4) 2m
= (−1)m−1
π .
2m + 1
(cot(µπ))(2m) |µ=1/4 = (−1)m−1
This transforms (29) and (30) into
∞
X
k=0
1
(4k + 1)2m+1
=
1
(1 − 2−(2m+1) )ζ(2m + 1)
2
+ (−1)m−1
∞
X
k=0
1
(4k + 3)2m+1
=
22m+1
B2m+1 (1/4) π 2m+1 , (31)
4 · (2m + 1)!
1
(1 − 2−(2m+1) )ζ(2m + 1)
2
+ (−1)m
22m+1
B2m+1 (1/4) π 2m+1 .
4 · (2m + 1)!
(32)
Therefore, when a = 4, n = 2m + 1, we obtain the positive answers to Q1 and
Q2 again. In particular, letting m = 1, 2, we have
ζ(4, 1, 3) =
ζ(4, 3, 3) =
ζ(4, 1, 5) =
∞
X
k=0
∞
X
k=0
∞
X
k=0
1
1
1
= (1 − 2−3 )ζ3) + 6 π 3 ,
3
(4k + 1)
2
2
(33)
1
1
1
= (1 − 2−3 )ζ(3) − 6 π 3 ,
3
(4k + 3)
2
2
(34)
1
1
5
= (1 − 2−5 )ζ(5) +
π5,
5
(4k + 1)
2
3 · 210
(35)
430
H. Chen
ζ(4, 3, 3) =
∞
X
k=0
1
1
5
= (1 − 2−5 )ζ(3) −
π5.
(4k + 3)5
2
3 · 210
(36)
On the other hand, if n is even, for example, we have
ζ(4, 1, 2) =
∞
X
k=0
1
1
1
= (1 − 2−2 )ζ(2) + G,
2
(4k + 1)
2
2
(37)
where G is the well-known Catalan number, and so Q2 does not valid. Moreover,
we are unable to find the closed form for ζ(4, 1, 2m) for any m > 1.
We now turn to find the closed form for ζ(a, b, n) when a = 6. Applying
(10) yields
∞
X
k=0
X
1
(−1)n−1 π
1
n−1
(cot(µπ))(n−1) |µ=1/6 . (38)
−
(−1)
=
(6k + 1)n
(6k + 5)n
6n (n − 1)!
∞
k=0
On the other hand, stemming from Euler’s infinite product formula again gives
1
1
1
1
1
ζ(n)(1 − 2−n )(1 − 3−n ) = 1 + n + n + n + n + n + · · ·
5
7
11
13
17
1
1
1
1
1
=
1 + n + n + ··· +
+
+
+ ···
7
13
5n 11n 17n
∞
∞
X
X
1
1
=
+
.
(39)
(6k + 1)n
(6k + 5)n
k=0
k=0
When n = 2m + 1, combining (38) and (39) we have
∞
X
k=0
∞
X
k=0
1
(6k + 1)2m+1
1
(6k + 5)2m+1
=
=
1
(1 − 2−(2m+1) )(1 − 3−(2m+1) )ζ(2m + 1)
2
π
(cot(µπ))(2m) |µ=1/6 ,
+
2 · 62m+1 (2m)!
1
(1 − 2−(2m+1) )(1 − 3−(2m+1) )ζ(2m + 1)
2
π
(cot(µπ))(2m) |µ=1/6 .
−
2m+1
2·6
(2m)!
(40)
(41)
Applying (17) and (21) once more we have
√
2m 3
(2m)
m−1 2 · 12
(cot(µπ))
|µ=1/6 = (−1)
(B2m+1 (1/6) + B2m+1 (1/3))π 2m .
2m + 1
ON SUMMATION OF SUBSERIES OF...
431
Setting m = 2 in the Raabe formula (see [4, Eq. 24.4.18, p.590])
Bn (x) =
m−1
1 X
k
B
x
+
n
mn
m
k=0
gives
B2m+1 (1/6) = (1 + 2−2m )B2m+1 (1/3),
and thus
(2m)
(cot(µπ))
m−1 2
|µ=1/6 = (−1)
√
· 122m 3
(2 + 2−2m )B2m+1 (1/3)π 2m .
2m + 1
This converts (40) and (41) into
∞
X
k=0
1
(6k + 1)2m+1
=
1
(1 − 2−(2m+1) )(1 − 3−(2m+1) )ζ(2m + 1)
2
+(−1)m−1
∞
X
k=0
1
(6k + 5)2m+1
=
√
(1 + 22m+1 )
B2m+1 (1/3) 3 π 2m+1 , (42)
6 · (2m + 1)!
1
(1 − 2−(2m+1) )(1 − 3−(2m+1) )ζ(2m + 1)
2
+(−1)m
√
(1 + 22m+1 )
B2m+1 (1/3) 3 π 2m+1 .
6 · (2m + 1)!
(43)
These yield a positive answer to Q1 and Q2 again. In particular, if m = 1, 2,
we have
∞
X
k=0
∞
X
k=0
1
(6k + 1)3
=
1
1 √
(1 − 2−3 )(1 − 3−3 )ζ(3) + 2 3 3 π 3 ,
2
2 ·3
(44)
1
(6k + 1)5
=
1
11 √
(1 − 2−5 )(1 − 3−5 )ζ(5) + 4 6 3 π 5 .
2
2 ·3
(45)
The last one was recently proposed as the Monthly Problem 11715 [5].
4. Uniqueness
In this section, we show that ζ(a, b, n) admitting a closed form in terms of π n
and ζ(n) only occurs for a = 3, 4, 5 and n odd.
432
H. Chen
For Q1 and Q2, (37) offers a counterexample when n is even. Thus, we
assume that n is odd in what follows. The reflection formula (10) and Lemma
3 demonstrate that ζ(a, b, n) − ζ(a, a − b, n) is related to the sine and cosine at
rational multiples of π. To obtain a closed form for ζ(a, b, n) alone, we need to
have another approach to sum ζ(a, b, n) + ζ(a, a − b, n) in terms of ζ(n). This
is equivalent to decomposing ζ(n) as a sum of ζ(a, b, n) and ζ(a, a − b, n). We
show this occurs only for a = 3, 4, 6. To this end, we recall another tool from
Euler. Let φ(a) be the Euler totient function, i.e., φ(a) equals the number of
positive integers less than a that are relatively prime to a. Observe that
∞ X
1
1
+
= c ζ(n)
ζ(a, b, n) + ζ(a, a − b, n) =
(ak + b)n (ak + (a − b))n
k=0
if and only if φ(a) = 2. Thus, the desired result follows from the simple fact:
φ(a) = 2 ⇐⇒ a = 3, 4, 6.
As an illustrative example, by (10) we have
√
7−4 3 3
(46)
ζ(12, 5, 3) − ζ(12, 7, 3) = 3 2 π .
3 ·4
However, since φ(12) = 4, using (39), the decomposition of ζ(3) becomes
(1 − 2−3 )(1 − 3−3 )ζ(3) = ζ(12, 1, 3) + ζ(12, 5, 3) + ζ(12, 7, 3) + ζ(12, 11, 3),
instead of ζ(12, 5, 3) + ζ(12, 7, 3) as expected. So we can’t find ζ(12, 5, 3) in
closed form alone.
Investigating further, the problem of determining the rational numbers b/a
such that the sine and cosine values at bπ/a are radicals was considered by
Gauss ([7]). He associated this problem to constructible regular polygon by
compass and ruler, and proved that a must be of the form
a = 2k p1 p2 · · · pl
where k ∈ N and pi are distinct Fermat primes (a prime of the form 22 + 1). In
conjunction with (10) and Lemma 3, we see that Q2 is true only for a = 3, 4, 6
again.
Based on Euler’s idea and tools, we have found a class of closed forms for
ζ(a, b, n). It is interesting to notice that today we have resources not available to Euler. For example, Mathematica. It is easy to verify our results (3),
(4), (33) — (36), (44) and (45) via Mathematica. However, to evaluate (46),
Mathematica only returns
n
-(PolyGamma[2, 5/12]/3456) + PolyGamma[2, 7/12]/3456
It is hard to see how to reduce this answer to match the result in (46).
ON SUMMATION OF SUBSERIES OF...
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References
[1] J. M. Borwein, I. J. Zucker and J. Boersma, The evaluation of character
Euler double sums, Ramanujan Journal, 15 (2008), 377-405
[2] H. Chen, Excursions in Classical Analysis, Mathematical Association of
America, Washington, DC (2010).
[3] L. Lewin, Polylogarithms and Associated Functions, North Holland, New
York (1981).
[4] F. W. J. Olver et.al (eds), NIST Handbook of Mathematical Functions,
National Institute of Standards and Technology, U.S. Department of Commence & Cambridge University Press, New York (2010).
[5] M. Stofka, Problem 11715, American Mathematical Monthly, 120 (2013),
569.
[6] E. W. Weisstein, Trigonometry Angles, From MathWorld, A Wolfram
Web Resource,
Available at http://mathworld.wolfram.com/TrigonometryAngles.html.
[7] I. J. Zucker and M. M. Robertson, Some properties of Dirichlet L-series,
J. Phys. A: Math. Gen., 9 (1976), 1207-1214.
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