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Journal of Integer Sequences, Vol. 16 (2013),
Article 13.6.3
23 11
An Alternating Sum Involving the Reciprocal
of Certain Multiplicative Functions
Olivier Bordellès
2, allée de la Combe
43000 Aiguilhe
France
borde43@wanadoo.fr
Benoit Cloitre
19, rue Louise Michel
92300 Levallois-Perret
France
benoit7848c@yahoo.fr
To the memory of Patrick Sargos
Abstract
We establish an asymptotic formula for an alternating sum of the reciprocal of
a class of multiplicative functions. The proof is straightforward and uses classical
convolution techniques. Numerous examples are given.
1
Introduction and main results
In 1900, E. Landau [2] proved that
ζ(2)ζ(3)
1
=
ϕ(n)
ζ(6)
n≤x
X
log x + γ −
1
X
p
log p
2
p −p+1
!
+O
log x
x
where ϕ is the classic Euler totient function and γ is Euler’s constant, but it seems that in
the usual literature there is not any similar result for the alternating sum
X (−1)n
ϕ(n)
n≤x
.
It is fairly easy to show that there exists an absolute constant c > 0 such that this sum
is > c log x + O(1), and therefore the question of an asymptotic formula arises naturally.
It should be mentioned that Landau’s result could eventually provide such an asymptotic
formula via the simple identity
X (−1)n
n≤x
ϕ(n)
=
X
1
1
−2
ϕ(n)
ϕ(n)
n≤x
n≤x
X
(n,2)=1
and the use of known estimates as in [3, Theorem 1], but this method does not seem to be
easily generalized to a larger class of multiplicative functions than that of [3]. In this article,
we will follow a slightly different approach working with the set of non-zero complex-valued
multiplicative functions f satisfying the following assumptions: there exist constants λ1 > 0
and 0 ≤ λ2 < 2 such that, for each prime power pα , we have
1
1
1
1
≤ λ1 λα−1
− ≤ λ1 and p2α−1 −
(α ≥ 2) .
(1)
p2 2
f (p) p
f (pα ) pf (pα−1 ) It will be convenient to set
λ :=
λ1 (2 + λ2 )
.
2 − λ2
(2)
For any prime number p, define
Sf (p) :=
∞
X
α=1
1
f (pα )
and sf (p) :=
∞
X
α=1
α
f (pα )
(3)
!
(4)
and set
Pf :=
Y
p
1
1−
p
1+
∞
X
α=1
1
f (pα )
where the product is absolutely convergent.
We are now in a position to state our main result.
Theorem 1. Let f be a non-zero multiplicative function satisfying (1). Then, for x ≥ e, we
have
X (−1)n
Sf (2) − 1
(log x)λ+1
=
Pf log x + Cf + O
f
(n)
S
(2)
+
1
x
f
n≤x
2
where
Sf (2) − 1
Cf :=
Pf
Sf (2) + 1
γ+
X
p
log p
1
sf (p)
−
p − 1 1 + Sf (p)
2sf (2) log 2
−
Sf (2)2 − 1
!
and Sf (p), sf (p) and Pf are given in (3) and (4).
Corollary 2. Let f be a non-zero multiplicative function satisfying (1) and assume that, for
any prime p and any integer α ≥ 1, f (pα ) = pα−1 f (p). Then, for x ≥ e, we have
X (−1)n
n≤x
f (n)
!
X (f (p) − p) log p 8f (2) log 2
2 − f (2) Y
1
1
=
log x + γ +
−
1− +
2 + f (2) p
p f (p)
(p
−
1)f
(p)
+
p
4 − f (2)2
p
(log x)λ1 +1
.
+O
x
Remark 3. If Sf (2) = 1 in Theorem 1, i.e. f (2) = 2 in Corollary 2, then the constant Cf
reduces to
√
Cf = −Pf sf (2) log 2.
Corollary 4. For x ≥ e, the following estimates hold.
(i)
X (−1)n
n≤x
ζ(2)ζ(3)
=
ϕ(n)
3ζ(6)
log x + γ −
X
p
log p
8 log 2
−
2
p −p+1
3
!
+O
(log x)3
x
.
(ii) Let χ be a non-principal Dirichlet character modulo q > 2 and ϕ(n, χ) be the twisted
Euler function attached to χ given in (6). Then
X (−1)n
χ(2)L(1, χ) Y
χ(p) χ(p)
=
1−
+ 2
log x + γ
ϕ(n,
χ)
4
−
χ(2)
p
p
p
n≤x
!
X
8(2 − χ(2)) log 2
(log x)3
χ(p) log p
−
−
+O
.
p2 − pχ(p) + χ(p)
χ(2)(4 − χ(2))
x
p
(iii) Let Ψ be the Dedekind arithmetic function defined in (5). Then
X (−1)n
n≤x
Ψ(n)
X log p
1Y
24 log 2
1
= −
+
1−
log x + γ +
2
5 p
p(p + 1)
p +p−1
5
p
(log x)2
+O
.
x
3
!
(iv) Let γ(n) :=
Q
p|n
X (−1)n γ(n)
ϕ(n)2
n≤x
p be squarefree kernel of n. Then
5 Y
p2 + p − 1
log x + γ
=
1+
11 p
p(p + 1)(p − 1)2
X (p3 − 2p2 − p + 1) log p
64 log 2
−
−
2
4
3
(p − 1)(p − p + 2p − 1)
55
p
!
+O
(log x)7
x
.
(v) Let σ be the sum-of-divisors function. Then
X (−1)n
n≤x
σ(n)
!
∞
Sσ (2) − 1 Y
1 (p − 1)2 X 1
=
log x + γ
1− +
α−1
Sσ (2) + 1 p
p
p
p
α=2
!
X
(log x)4
sσ (p)
1
+ κ log 2 + O
−
+
log p
p − 1 1 + Sσ (p)
x
p
.
.
where κ = 3.6. Note that the leading constant is = −0.16 468.
(vi) Let σ ∗ be the sum-of-unitary-divisors function. Then
X (−1)n
n≤x
σ ∗ (n)
!
∞
Sσ∗ (2) − 1 Y
1 p−1X 1
log x + γ
=
1− +
Sσ∗ (2) + 1 p
p
p α=1 pα + 1
!
X
∗
1
(log x)4
sσ (p)
+
log p
+ ν log 2 + O
−
p − 1 1 + Sσ∗ (p)
x
p
.
.
where ν = 8.04. Note that the leading constant is = −0.10 259.
2
Notation
The celebrated Möbius function is as always denoted by µ, Id(n) = n, ϕ is the Euler totient
function, Ψ is the Dedekind function defined by
Y
1
(5)
Ψ(n) = n
1+
p
p|n
and, for each fixed non-principal Dirichlet character χ modulo q > 2, the χ-twisted Euler
function ϕ(n, χ) recently introduced in [1] is defined by
Y
χ(p)
.
(6)
ϕ(n, χ) = n
1−
p
p|n
4
For any two arithmetic functions u and v, u ⋆ v is the usual Dirichlet convolution product of
u and v defined by
X
(u ⋆ v)(n) =
u(d)v(n/d).
d|n
The Eratosthenes transform of u is the arithmetic function u ⋆ µ. Finally, f is a non-zero
multiplicative function satisfying (1) and g is the Eratosthenes transform of Id f −1 . Note
that the assumption (1) can be written as
and pα−1 |g (pα )| ≤ λ1 λα−1
2
p|g(p)| ≤ λ1
3
(α ≥ 2) .
(7)
Tools
Lemma 5. Let δ > 0. The Dirichlet series of the arithmetic function g(n) is absolutely
convergent in the half-plane σ > δ.
Proof. Let s = δ + it ∈ C with δ > 0. The function g is multiplicative and using (7) we get
for all z ≥ e
!
∞ ∞ XX
X |g(p)| X
g (pα ) g (pα ) +
psα psα =
δ
p
α=2
p≤z α=1
p≤z
α−1 !
∞ X 1
X 1 X
λ2
≤ λ1
+
pδ+1 p≤z pδ α=2 pδ+1
p≤z
X 1
λ2
+ δ δ+1
= λ1
δ+1
p
p (p
− λ2 )
p≤z
X 1
2λ2
1
+
≤ λ1
pδ+1 pδ (2 − λ2 ) pδ+1
p≤z
X 1
≤ λ
pδ+1
p≤z
where we used the inequality
1
1
2
≤
p θ − λ2
2 − λ2 p θ
(θ ≥ 1) .
This implies the asserted result.
Lemma 6. For all real numbers z ≥ e and a ∈ {0, 1}
X
(i) :
|g(n)| ≪ (log z)λ .
n≤z
(ii) :
X |g(n)|(log n)a
n>z
n
5
≪ z −1 (log z)λ+a .
Proof.
(i) As in the proof of Lemma 5, we get for all z ≥ e
)
(
∞
X
XX
X
|g(p)| +
|g (pα )|
|g(n)| ≤ exp
p≤z α=2
p≤z
n≤z
(
X1
≤ exp λ1
p≤z
(
p
+
α−1
∞ XX
λ2
p≤z α=2
p
2λ2 X 1
+
≤ exp λ1
p 2 − λ2 p≤z p
p≤z
!
X1
≪ (log z)λ
= exp λ
p
p≤z
X1
!)
!)
as asserted.
(ii) Follows from Lemma 5, (i) and partial summation. We leave the details to the reader.
The proof is complete.
Lemma 7. For any real number x ≥ 1 and any integer 1 ≤ d ≤ 2x, we have
X1
ρ(d)
ρ(d)x
1
=
+γ +O
log
n
d
d
x
n≤x
d|2n
where
ρ(d) :=
(
2,
1,
if d is even;
if d is odd.
(8)
Proof. Let S(x, d) be the sum of the left-hand side. If x < d ≤ 2x, then
S(x, d) =
2
2
< .
d
x
If d ≤ x is odd, then by Gauss’s theorem we have
X1
1
S(x, d) =
=
n
d
n≤x
d|n
X
1≤k≤x/d
1
1
=
k
d
as required.
6
x
log + γ + O
d
d
x
2x
3
< d ≤ x, then xd −
X
X1
+
S(x, d) =
n
n≤x
n≤x
Now assume that d is even. If
1
d
< 1 and therefore
1
n
n≡d/2 (mod d)
d|n
=
1
2
X
1≤k≤x/d
1 1
+
k d
X
0≤k≤x/d−1/2
1
k + 1/2
x
2
d
1
log + γ + O
+
=
d
d
x
d
1
x
1
=
log + γ + O
d
d
x
implies d2 < x3 . On the other hand, since 1 ≤ xd < 23 , we have
since the assumption d > 2x
3
2
1
4x
2x
γ
1
x
log 6 + γ
3 (log 6 + γ)
log
log
+ − log ≤
+γ <
<
d
d
d d
d
d
d
d
2x
and thus in this case we also get
2
S(x, d) =
d
If 1 ≤ d ≤
2x
,
3
2x
log
+γ
d
1
.
+O
x
since
X
1≤k≤x/d−1/2
x
1
= log + γ − 2 + log 4 + O
k + 1/2
d
d
x
2x
3
implies xd − 21 ≥ 2x
, then
3d
1
x
2
d
S(x, d) =
log + γ + O
+
d
d
x
d
x
d
1
log + γ − 2 + log 4 + O
+
d
d
x
2
2x
1
=
+γ +O
log
d
d
x
where we used the fact that d ≤
completing the proof.
4
Sums of reciprocals
Lemma 8. Let f satisfying the conditions (1). Then
∞
X
X 1
g(d)
(log x)λ+1
= log x
+ Kf + O
f
(n)
d
x
n≤x
d=1
7
where
Kf :=
∞
X
g(d)
d
d=1
and
(γ − log d)
∞
X
n≤2x
n even
log x X g(d)ρ(d)
1
=
+ Lf + O
f (n)
2 d=1
d
(9)
(log x)λ+1
x
where
∞
1 X g(d)ρ(d)
Lf :=
2 d=1
d
ρ(d)
γ + log
.
d
(10)
and where ρ(d) is given in (8) and λ is defined in (2).
Proof. The proof of the first estimate follows the classical lines. We have
X 1
X (g ⋆ 1)(n) X 1 X
=
=
g(d)
f
(n)
n
n
n≤x
n≤x
n≤x
d|n
X g(d) X 1
=
d
k
d≤x
k≤x/d
X g(d)
x
d
=
log + γ + O
d
d
x
d≤x
= (log x + γ)
X g(d)
d≤x
d
−
X g(d) log d
d≤x
d
+O
1X
|g(d)|
x d≤x
!
and by Lemma 6 we get
∞
∞
X
X 1
g(d)ρ(d) X g(d) log(d)
(log x)λ+1
= (log x + γ)
−
+O
f
(n)
d
d
x
n≤x
d=1
d=1
as asserted.
The second estimate is similar, with the additional use of Lemma 7 which gives
X
n≤2x
n even
X 1
X (g ⋆ 1)(2n)
1
=
=
f (n)
f (2n) n≤x
2n
n≤x
=
X1
1X1X
1X
g(d)
g(d) =
2 n≤x n
2 d≤2x
n
n≤x
d|2n
8
d|2n
1X
ρ(d)x
1
ρ(d)
=
log
+γ +O
g(d)
2 d≤2x
d
d
x
=
X g(d)ρ(d) 1 X g(d)ρ(d)
ρ(d)
1
(log x + γ)
+
log
2
d
2 d≤2x
d
d
d≤2x
!
1 X
|g(d)|
+O
x d≤2x
∞
∞
X
X
1
g(d)ρ(d) 1
g(d)ρ(d)
=
(log x + γ)
− (log x + γ)
2
d
2
d
d=1
d>2x
∞
ρ(d) 1 X g(d)ρ(d)
ρ(d)
1 X g(d)ρ(d)
log
−
log
+
2 d=1
d
d
2 d>2x
d
d
!
1 X
+O
|g(d)|
x d≤2x
and by Lemma 6 we get
X
n≤2x
n even
∞
∞
X
1
1
ρ(d)
g(d)ρ(d) 1 X g(d)ρ(d)
(log x)λ+1
= (log x + γ)
+
log
+O
f (n)
2
d
2
d
d
x
d=1
d=1
completing the proof.
5
5.1
Proof of Theorem 1
First step: Asymptotic formula
From Lemma 8 we get
X (−1)n
n≤x
f (n)
= 2
X
n≤x
n even
X 1
1
−
f (n) n≤x f (n)
∞
∞
X
g(d)
x X g(d)ρ(d)
− log x
+ 2Lf − Kf
2 d=1
d
d
d=1
λ+1
(log x)
+O
x
= log
9
where Kf et Lf are given in (9) et (10), so that setting
Cf := 2Lf − Kf − log 2
=
=
=
∞
X
g(d)
d=1
∞
X
d=1
d even
∞
X
d=1
=
d
d=1
d
ρ(d)
(ρ(d) − 1) (γ − log d) + ρ(d) log
2
∞
X
g(d)
ρ(d)
g(d)ρ(d)
(γ − log d) +
log
d
d
2
d=1
d odd
∞
X
g(d)
g(2d)
(γ − log 2d) − log 2
2d
d
d=1
∞
1 X g(2d)
2
∞
X
g(d)
d=1
d
d odd
∞
X
(γ − log d) − log 2
d=1
g(d)
d
we obtain
X (−1)n
n≤x
f (n)
∞
X
g(d) (ρ(d) − 1)
(log x)λ+1
= log x
+ Cf + O
d
x
d=1
∞
(log x)λ+1
log x X g(2d)
+ Cf + O
=
2 d=1 d
x
completing the proof.
5.2
Second step: Series expansions
The unique decomposition d = 2α m with α ∈ Z+ and m ≥ 1 odd provides
∞
X
g(d)
d=1
d
∞
∞
X
g(2α ) X g(m)
=
2α m=1 m
α=0
m odd
!
∞
X
g(2α ) Y
=
1+
2α
α=1
p≥3
∞
1X 1 Y
1−
=
2 α=0 f (2α ) p≥3
= Pf
10
1+
∞
X
g(pα )
α=1
1
p
1+
pα
∞
X
α=1
!
1
f (pα )
!
and
∞
X
g(2d)
d=1
d
=
∞
∞
X
g (2α+1 ) X g(m)
2α
m
m=1
α=0
m odd
g(2) + 2
=
Similarly
∞
X
g(2d) log d
d=1
d
!
∞
X
g (pα )
Y
!
1+
α
2
pα
α=1
α=2
p≥3
!
!
∞
∞
Y
X
X
1
1
1
1−
−1
1+
α)
f
(2
p
f (pα )
α=1
α=1
p≥3
=
=
∞
X
g (2α )
2 (Sf (2) − 1)
Pf .
Sf (2) + 1
∞
∞
X
g (2α+1 ) X g(m) log(2α m)
=
2α
m
m=1
α=0
=
∞
X
α=0
α+1
∞
X
g (2 ) 
α log 2
2α
= log 2
∞
X
αg (2α+1 )
2α
α=0
= log 2
m=1
m odd
∞
X
m=1
∞
X
α=1
f (2α )
1
−1
f (2α )
p≥3
!
g(m)
+
m
∞
X

g(m) log m 

m
m=1
m odd
∞
α+1
X
∞
g(m)
g (2 ) X g(m) log m
+
m
2α
m
m=1
α=0
! m odd
∞
X α−2 Y
α=1
+
m odd

1
1−
p
∞
X
m=1
m odd
1+
∞
X
α=1
m odd
1
f (pα )
!
g(m) log m
m
∞
X
2 log 2 (sf (2) − 2Sf (2))
g(m) log m
Pf + (Sf (2) − 1)
.
=
Sf (2) + 1
m
m=1
m odd
Now since for s ∈ C such that Re s > δ > 0, we have
!
∞
∞
X
X
1
g(m) Y
1
=
1− s
1+
:= G(s)
s
α(s−1) f (pα )
m
p
p
m=1
α=1
p≥3
m odd
we infer
∞
X
g(m) log m
= −G′ (1).
m
m=1
m odd
11
Using the logarithmic derivative, we get
G′f (s) X
=
log p
G(s)
p≥3
and therefore
!
P∞
α
α(s−1)
α
1
α=1 p
f (p )
P∞
−
s
p − 1 1 + α=1 pα(s−1)1 f (pα )
∞
X
X
g(m) log m
2Pf
sf (p)
1
=−
−
log p
m
Sf (2) + 1 p≥3
p − 1 1 + Sf (p)
m=1
m odd
thus completing the proof of Theorem 1.
6
Acknowledgments
We express our gratitude to the referee for his careful reading of the manuscript and the
many valuable suggestions and corrections he made.
References
[1] J. Kaczorowski and K. Wiertelak, On the sum of the twisted Euler function, Internat. J.
Number Theory 8 (2012), 1741–1761.
[2] E. Landau, Über die Zahlentheoretische Function ϕ(n) und ihre Beziehung zum
Goldbachschen Satz, Nachrichten der Koniglichten Gesellschaft der Wissenschaften zu
Göttingen, Mathematisch-Physikalische Klasse, 1900, 177–186.
[3] K. Wooldridge, Mean value theorems for arithmetic functions similar to Euler’s phifunction, Proc. Amer. Math. Soc. 58 (1976), 73–78.
2010 Mathematics Subject Classification: Primary 11A25; Secondary 11N37.
Keywords: alternating sum, average order over a polynomial sequence.
(Concerned with sequences A001615, A058026, A065463, A082695, A211117, and A211178.)
Received February 9 2013; revised versions received March 3 2013; June 12 2013. Published
in Journal of Integer Sequences, June 16 2013.
Return to Journal of Integer Sequences home page.
12