Huaning, Liu

Wenpeng, Zhang

Osaka City University Repository

Huaning, L. and Wenpeng, Z.
Osaka J. Math.
41 (2004), 745–758
ON CERTAIN HARDY SUMS AND THEIR 2m-TH POWER MEAN
LIU HUANING and ZHANG WENPENG
(Received April 4, 2003)
1. Introduction
For a positive integer and an arbitrary integer , the classical Dedekind sums
( ) is defined by
( ) =
∑
=1
(( ))(( ))
where
(( )) =
 − [ ]−
1
2
if is not an integer;
0 if is an integer
The sum ( ) plays an important role in the transformation theory of the Dedekind
η function; See the Chapter 3 of [1]. There is an extensive literature about the Dedekind
sums. H. Rademacher [8] wrote an introductory book on the subject.
Perhaps the most famous property of the Dedekind sums is the reciprocity formula
( ) + ( ) =
2 + 2 + 1
12
− 1
4
for positive coprime integers and . Some three term versions of this formula were
discovered by H. Rademacher [8], R.R. Hall, M.N. Huxley [5] and J. Pommersheim [7].
J.B. Conrey, E. Fransen, R. Klein and C. Scott [4] studied the mean value of
Dedekind sums and proved the following proposition.
Proposition 1. Suppose that is a given positive integer and is any suffi-
ciently large integer. Then
∑′
=1
2 ( ) = ( )
(
12
)2
+
((
9/5 + 2 −1+1/( +1)
)
log3
)
where
∑′ denotes the summation over all such that ( ) = 1, and ( ) is defined
This work is supported by N.S.F. (10271093) and P.N.S.F. of P. R. China
746 L. HUANING AND Z. WENPENG
by the Dirichlet series
∞∑
=1
( )
= 2 · ζ
2(2 )
ζ(4 ) ·
ζ( + 4 − 1)
ζ2( + 2 ) · ζ( )
where ζ( ) is the Riemann zeta-function.
In [3], J. Chaohua improved the error terms in Proposition 1. H. Walum [10]
showed that for prime ,
∑
χ mod
χ(−1)=−1
| (1 χ)|4 = π
4( − 1)
2
∑
=1
| ( )|2
In the spirit of [4] and [10], the second author [11] used an estimate for character
sums to prove the following:
Proposition 2. Suppose that is any sufficiently large prime number and is
any positive integer. Then for = , we have
∑′
=1
| ( )|2 = 5
144
· (
2 − 1)2
( 3 − 1) ·
2 +
(
exp
(
3 log
log log
))
where exp( ) = and the constant implied in the -symbol is absolute.
Also some interesting relations between Dedekind sums and Hurwitz zeta-function
were established (see references [12], [13], [14] and [16]).
B.C. Berndt [2] gave an analogous transformation formula for the logarithm
of the classical theta function
θ( ) =
+∞∑
=−∞
exp(π 2 ) Im > 0
and showed that for =
( )
in the theta group
log θ( ) = log θ( ) + 1
2
log( + )− 1
4
π +
1
4
π ( )
where
( ) =
−1∑
=1
(−1) +1+[ / ]
The sums ( ) (and certain related ones) are sometimes called Hardy sums. They
are closely connected with Dedekind sums [9]. Some arithmetical properties of ( )
CERTAIN HARDY SUMS 747
can be found in B.C. Berndt [2] and R. Sitaramachandra Rao [9]. In [15], the second
author studied the 2 -th power mean of ( ), and proved the following:
Proposition 3. Let be an odd prime and be a positive integer, then
−1∑
=1
| ( )|2 = 2 ζ
2(2 ) (1− 1/4 )
ζ(4 ) (1 + 1/4 ) +
(
2 −1 exp
(
6 ln
ln ln
))
In this paper, we use the important works of J.B. Conrey et al. [4] and
J. Chaohua [3] to study the 2 -th power mean of ( ), and give a sharp asymp-
totic formula for
∑′
=1
2 ( ). That is, we shall prove the following theorem.
Theorem. For any fixed integer ≥ 2 and any sufficiently large integer , we
have the asymptotic formula
∑′
=1
2 ( ) = ( ) 2 + ( 2 −1)
where ( ) is defined by the Dirichlet series
∞∑
=1
( )
=
2
(
2 +4 − 2) (22 − 1)(
2 +2 − 1)2 (22 + 1) · ζ
2(2 )
ζ(4 )
ζ( + 4 − 1)
ζ2( + 2 ) ζ( )
+
2
(
22 − 1)(
2 +2 − 1) · ζ( )ζ(2 )ζ( + 2 )
2. Some lemmas
To prove the Theorem, we need following lemmas. First we have
Lemma 1. For any given positive integer and any integer with ( ) = 1
and any > 1, there exist a positive integer ≤ and an integer with ( ) = 1
such that ∣∣∣∣ − ∣∣∣∣ < 1
Proof. This is a well-known result; See Theorem 36 of [6].
Lemma 2. Let , , , , and be positive integers with − = 1 and
( ) = 1. Let ( )( )
=
( )
748 L. HUANING AND Z. WENPENG
then we have
( ) + ( ) − ( ) =
2 + 2 + 2
12
− 1
4
Proof. This is equation (26) of [5].
Lemma 3. Let and denote relatively prime integers with > 0, then
( ) =
{
4 ( ) − 8 ( + 2 ) if + is odd;
0 if + is even
Proof. This formula is an immediate consequence of (5.9) and (5.10) in [9].
Lemma 4. For any positive integer , we have
∑′
=1
| ( )| ≪ log2
Proof. This is Lemma 6 of [4].
Lemma 5. Let and be positive integers with ( ) = ( ) = 1, and
set = − . If + is odd and 1 ≤ | | ≤ / , then we have
( ) =
− +
(
| ( )| +
∣∣∣ ( +2 )∣∣∣ + | |) if + is an even number;(
+ | |) if + is an odd number
Proof. Suppose that + is even. Since ( ) = 1, and must be odd num-
bers.
First We consider the case that < 0. Since ( ) = 1, there exist positive integers
and such that
− = 1 1 ≤ <
Let = 2 − 2 . Then we have(
2 −2
−
)( )
=
(
−
)
CERTAIN HARDY SUMS 749
and  +2 +
2
(
−
)
=
(
+
2
)
The fact that < and ≥ − / yields
= 2
(
−
)
= 2
(
− + 1
)
= 2
(
+
1
)
=
2
(
+
1
)
> 0
On the other hand, since ( ) = 1 and is odd, we get ( − ) = 1. Then
by Lemma 2, ( +
2
)
+ ( − )− ( + 2 ) = −4
2 + 2 + 2
24
− 1
4
That is,
( + 2 ) = 6 +
(∣∣∣ ( +2 )∣∣∣ + | |)
From Lemma 8 of [4] we also have
( ) =
12
+
(| ( )| + | |)
Therefore by Lemma 3 we immediately have
( ) = − +
(
| ( )| +
∣∣∣ ( +2 )∣∣∣ + | |) if < 0
For > 0, we can find positive integers and satisfying
− = −1 1 ≤ <
Let = 2 − 2 . Then we have(
2 −2
−
)( )
=
( )
and  2+
2
+
( ) = ( 2
+
)
Similarly we can get ( ) = 1 and
= 2
(
−
)
= 2
(
1
+ −
)
= 2
(
1 −
)
=
2
(
1 −
)
> 0
750 L. HUANING AND Z. WENPENG
Then by Lemma 2,
( +
2
)
+ ( )− (2 + ) =
(( + )/2)2 + 2 + ( + )2
12
(( + )/2) · · ( + ) − 14
Noting that
( +
2
)
+
( +
2
)
=
(( + )/2)2 + 2 + 1
12
(( + )/2) · − 14
and
(2 + ) + ( + 2 ) = ( + )
2 + (2 )2 + 1
12 ( + ) · 2 −
1
4
we have
( + 2 ) = 6 +
(∣∣∣ ( +2 )∣∣∣ + | |)
So from Lemma 8 of [4] and Lemma 3 we immediately have
( ) = − +
(
| ( )| +
∣∣∣ ( +2 )∣∣∣ + | |) for > 0
This proves that
( ) = − +
(
| ( )| +
∣∣∣ ( +2 )∣∣∣ + | |)
if + is an even number
On the other hand, if + is an odd number, using the similar methods we can
get
( + 2 ) =
24
+
(
+ | |)
so we have
( ) = 4 ( ) − 8 ( + 2 ) = ( + | |)
This completes the proof of Lemma 5.
Lemma 6. For any real > 1, we have the identities
∞∑
=1
2|
µ( )
=
1
(1− 2 )ζ( )
∞∑
=1
2∤
µ( )
=
2
(2 − 1)ζ( ) ;
CERTAIN HARDY SUMS 751
∞∑
=1
2∤
∑
|
µ( )( / )1−2
+2 =
2 +2 − 21−2
2 +2 − 1 ·
ζ( + 4 − 1)
ζ( + 2 )
and
∞∑
=1
2|
∑
|
µ( ) ( / )1−2
+2 =
21−2 − 1
2 +2 − 1 ·
ζ( + 4 − 1)
ζ( + 2 )
Proof. Using elementary methods we can easily deduce these identities.
3. Proof of Theorem
We suppose that ≥ 2 and a sufficiently large number are given. We set
=
[
1/2
]
= 2
For integers and with 1 ≤ ≤ , let ( ) be an open interval given by
( ) =
(
− 1 + 1
)
When / 6= ´/ ´ and ´ ≤ , one has∣∣∣∣ − ´
´
∣∣∣∣ ≥ 1
´
≥
(
1
+
1
´
)
Thus the intervals ( ) are pairwise disjoint.
If 1 ≤ ≤ , ( ) = 1 and + is odd, then by Lemma 1, / falls into
an interval ( ) with 1 ≤ ≤ , 0 ≤ ≤ and ( ) = 1.
Let = − . It is easy to see that 6= 0 and
| | =
∣∣∣∣ − ∣∣∣∣ ≤ ≤
If / falls into an interval ( ) with 1 ≤ ≤ , 0 ≤ ≤ , ( ) = 1
and + is an odd number, then by Lemma 5, we have
( ) = ( + | |)≪ + ≪ 1/2
Thus, ∑∗ 2 ( ) ≪ +1 ≪ 2 −1
752 L. HUANING AND Z. WENPENG
where the asterisk indicates summation over those integers , 1 ≤ ≤ , ( ) = 1
and + is odd, for which / falls into an interval ( ) with 1 ≤ ≤ ,
0 ≤ ≤ , ( ) = 1 and + is an odd number.
If / falls into an interval ( ) with ≤ ≤ , 0 ≤ ≤ , ( ) = 1
and + is an even number, then by Lemma 5, we have
( ) = − +
(
| ( )| +
∣∣∣ ( +2 )∣∣∣ + | |)
≪ + + ≪ + + ≪ 1/2
Thus, ∑∗ 2 ( ) ≪ +1 ≪ 2 −1
where the asterisk indicates summation over those integers 1 ≤ ≤ ( ) = 1
and + is odd, for which / falls into an interval ( ) with ≤ ≤ 0 ≤
≤ , ( ) = 1 and + is an even number.
Therefore
∑′
=1
2 ( ) =
∑′
=1
2∤ +
2 ( ) =
∑
=1
2∤
∑′
=1
2∤
∑∗
/ ∈ ( )
2 ( ) + ( 2 −1)
where the asterisk means that 1 ≤ ≤ , ( ) = 1 and + is odd.
Lemma 5 produces
( ) = − +
(
| ( )| +
∣∣∣ ( +2 )∣∣∣ + | |)
if and are odd numbers
Using the estimate
( + + )2 = 2 +
(
| |2 −1 (| | + | |)) + ( 2 + 2 )
we obtain
2 ( ) =
( )2
+
((
| |
)2 −1 (
| ( )| +
∣∣∣ ( +2 )∣∣∣ + | |)
)
+
((
| ( )| +
∣∣∣ ( +2 )∣∣∣)2 + 2
)
CERTAIN HARDY SUMS 753
Therefore
∑
=1
2∤
∑′
=1
2∤
∑∗
/ ∈ ( )
2 ( ) ≡ 1 + ( 2) + ( 3)
where
1 =
∑
=1
2∤
∑′
=1
2∤
∑∗
/ ∈ ( )
( )2
2 =
∑
=1
2∤
∑′
=1
2∤
∑∗
/ ∈ ( )
(
| |
)2 −1 (
| ( )| +
∣∣∣ ( +2 )∣∣∣ + | |)
3 =
∑
=1
2∤
∑′
=1
2∤
∑∗
/ ∈ ( )
((
| ( )| +
∣∣∣ ( +2 )∣∣∣)2 + 2
)
Noting that for the fixed , , and , the equation = − has at most one
solution . By Lemma 4, we have
2 ≪ 2 −1
∑
=1
2∤
∑′
=1
2∤
∑∗
/ ∈ ( )
1
2 −1 ·
1
2 −2
(
| ( )| +
∣∣∣ ( +2 )∣∣∣ + 1)
≪ 2 −1
∑
=1
2∤
1
2 −1
∑′
=1
(
| ( )| +
∣∣∣ ( +2 )∣∣∣ + 1)∑6= 0
1
2
≪ 2 −1
∑
=1
2∤
1
2 −1 · · log2( + 1) ≪ 2 −1
∑
=1
2∤
log2( + 1)
2 ≪ 2 −1
Moreover,
3 ≪
∑
=1
2∤
∑′
=1
2∤
∑∗
/ ∈ ( )
(
2 +
( )2 )
≪
∑′
=1
1 ≪ +1 ≪ 2 −1
Combining these estimates, we obtain
∑′
=1
2 ( ) = 1 + ( 2 −1)
754 L. HUANING AND Z. WENPENG
where
1 =
2
∑
=1
2∤
1
2
∑′
=1
2∤
∑∗
/ ∈ ( )
1
2
It remains to obtain an asymptotic formula for 1. Noting that if 1 ≤ ≤ ,
then / /∈ ( ) if and only if | | ≥ / . Hence
2
∑
=1
2∤
1
2
∑′
=1
2∤
∑∗
/ /∈ ( )
1
2 ≤ 2
∑
=1
2∤
1
2
∑′
=1
2∤
∑
| |≥ /
1
2
≪ 2
( )2 −1∑
=1
2∤
1
2 −1
≪ 2 −1 ≪ +1/2 ≪ 2 −1
Thus
1 =
2
∑
=1
2∤
1
2
∑′
=1
2∤
∑′
=1
2∤ +
1
( − )2 +
( 2 −1)
Using the estimate
∑
≥ +1
1
( − )2 ≤
∫ ∞
( − )2 =
∫ ∞
( − ) 2
≪ 12 −1
we get
2
∑
=1
2∤
1
2
∑′
=1
2∤
∑
≥ +1
1
( − )2 ≪
2
∑
=1
2∤
1
2
∑′
=1
2∤
1
2 −1 ≪
Since∑
≤0
1
( − )2 ≤
1
2 +
∑
≥1
1
( + )2 ≤
1
2 +
∫ ∞
0 ( + )2
=
1
2 +
∫ ∞
2 ≪
1
2 +
1
2 −1 ≪
1
2 −1
CERTAIN HARDY SUMS 755
we have
2
∑
=1
2∤
1
2
∑′
=1
2∤
∑
≤0
1
( − )2 ≪
2
∑
=1
2∤
1
2
∑′
=1
2∤
1
2 −1 ≪
Therefore
1 =
2
∑
=1
2∤
1
2
∑′
=1
2∤
∞∑
=−∞
( )=1
2∤ +
1
( − )2 + (
2 −1)
Since
2
∑
>
2∤
1
2
∑′
=1
2∤
∞∑
=−∞
( )=1
2∤ +
1
( − )2 ≪
2
∑
>
2∤
1
2
∑′
=1
2∤
∞∑
=−∞
1
2
≪ 2
∑
>
2∤
1
2
∑′
=1
2∤
1 ≪ 2
∑
>
2∤
1
2 −1
≪
2
2 −2 ≪ +1 ≪ 2 −1
we have
1 =
2
∞∑
=1
2∤
1
2
∑′
=1
2∤
∞∑
=−∞
( )=1
2∤ +
1
( − )2 +
( 2 −1)
Therefore
∑′
=1
2 ( ) = ( ) 2 + ( 2 −1)
where
( ) =
∞∑
=1
2∤
1
2
∑′
=1
2∤
∞∑
=−∞
( )=1
2∤ +
1
( − )2
756 L. HUANING AND Z. WENPENG
Let
( ) =
∞∑
=1
( )
=
∞∑
=1
2∤
1
2
∑′
=1
2∤
∞∑
=1
1 ∞∑
=−∞
( )=1
2∤ +
1
( − )2
=
∞∑
=1
2∤
1
2
∑′
=1
2∤
∞∑
=1
2|
1 ∞∑
=−∞
( )=1
2∤
1
( − )2 +
∞∑
=1
2∤
1
2
∑′
=1
2∤
∞∑
=1
2∤
1 ∞∑
=−∞
( )=1
2|
1
( − )2
≡ 1( ) + 2( )
We proceed to find an expression for 1( ). We remove the coprimality conditions by
use of the Mo¨bius relation
∑
|
µ( ) =
{
1 if = 1;
0 if 6= 1
After rearranging the sums, we have
1( ) =
∞∑
=1
2∤
∞∑
=1
2∤
µ( )
+2
µ( )
4
∞∑
=1
2|
1 ∞∑
=1
2∤
1
2
∑
=1
2∤
∞∑
=−∞
2∤
1
( − )2
Let = ( ), then the inner double sum is
=
1
2
∑
=1
2∤
∞∑
=−∞
2∤
≡− ( / ) mod /
1
2
=
1
2

∑
=1
2∤
∞∑
=1
2∤
≡− ( / ) mod /
1
2 +
∑
=1
2∤
∞∑
=1
2∤
≡−( − )( / ) mod /
1
2

=
1
2
∑
=0
∞∑
=1
2∤
≡− ( / ) mod /
1
2 =
1
2

∑
=1
∞∑
=1
2∤
≡− ( / ) mod /
1
2 +
∞∑
=1
2∤
≡0 mod /
1
2

=
1
2
[ (
22 − 1) ζ(2 )
22
+
2
2
(
22 − 1) ζ(2 )
22
]
=
(
22 − 1) ζ(2 )
22
[
1−2 +
1
2
]
CERTAIN HARDY SUMS 757
Thus by Lemma 6,
1( ) = 2
+4(
2 +2 − 1) (22 + 1) ζ(2 )ζ( + 2 )ζ(4 )
 ∞∑
=1
2|
∞∑
=1
2∤
1−2
2 +
∞∑
=1
2|
∞∑
=1
2∤
1
4

By Lemma 6 we have
∞∑
=1
2|
∞∑
=1
2∤
1−2
2 =
∞∑
=1
2|
∞∑
=1
2∤
1
2
∑
|( )
∑
|
µ( )
( )1−2
=
∞∑
=1
2∤
∑
|
µ( )( / )1−2
+2
∞∑
=1
2|
∞∑
=1
2∤
1
2
=
(
2 +2 − 21−2 ) (22 − 1)
2 +2
(
2 +2 − 1) · ζ( + 4 − 1)ζ( )ζ(2 )ζ( + 2 )
Therefore
1( ) =
(
2 +4 − 2) (22 − 1)(
2 +2 − 1)2 (22 + 1) · ζ
2(2 )
ζ(4 )
ζ( + 4 − 1)
ζ2( + 2 ) ζ( )
+
(
22 − 1)(
2 +2 − 1) · ζ( )ζ(2 )ζ( + 2 )
Using the same methods we can prove
2( ) =
(2 − 1) (2 +4 − 2) (22 − 1)(
2 +2 − 1)2 (22 + 1) · ζ
2(2 )
ζ(4 )
ζ( + 4 − 1)
ζ2( + 2 ) ζ( )
+
(2 − 1) (22 − 1)(
2 +2 − 1) · ζ( )ζ(2 )ζ( + 2 )
So we have
( ) = 2
(
2 +4 − 2) (22 − 1)(
2 +2 − 1)2 (22 + 1) · ζ
2(2 )
ζ(4 )
ζ( + 4 − 1)
ζ2( + 2 ) ζ( )
+
2
(
22 − 1)(
2 +2 − 1) · ζ( )ζ(2 )ζ( + 2 )
This completes the proof of Theorem.
ACKNOWLEDGEMENTS. The authors express their gratitude to the referee for his
very helpful and detailed comments in improving this paper.
758 L. HUANING AND Z. WENPENG
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Liu Huaning
Department of Mathematics
Northwest University
Xi’an, Shaanxi, P. R. China
e-mail: hnliu@etang.com
Zhang Wenpeng
Department of Mathematics
Northwest University
Xi’an, Shaanxi, P. R. China
e-mail: wpzhang@nwu.edu.cn


On certain Hardy sums and their 2m-th power mean

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On certain Hardy sums and their 2m-th power mean

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