International Mathematical Forum, Vol. 9, 2014, no. 34, 1673 - 1676
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/imf.2014.49166
On a Sum in Number Theory where
Appear the Integer-Part Function
Rafael Jakimczuk
División Matemática, Universidad Nacional de Luján
Buenos Aires, Argentina
c 2014 Rafael Jakimczuk. This is an open access article distributed under
Copyright the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
Abstract
Let us consider a strictly increasing sequence of positive integers, we
denote a a positive integer in this sequence. Suppose that this sequence
has positive density ρ. We prove the following formula
X
a≤x
k
a
ρ
x
=
ζ(k + 1)xk+1 + o xk+1
a
k+1
where k is an arbitrary but fixed positive integer and ζ(s) is the Riemann
zeta-function.
Mathematics Subject Classification: 11A99, 11B99
Keywords: Sequences of integers with positive density, integer-part function, k-th powers-free integers
1
Main Theorems
We shall need the following theorem (see ([1], chapter XXII) )
Theorem 1.1 Let cn (n ≥ 1) a sequence of real numbers. Let us consider
the function
X
A(x) =
cn
n≤x
1674
Rafael Jakimczuk
Suppose that f (x) has a continuous derivative f 0 (x) on the interval [1, ∞], then
the following formula holds
X
cn f (n) = A(x)f (x) −
n≤x
Z x
A(t)f 0 (t) dt
1
Now, we prove the main theorem.
Theorem 1.2 Let us consider a strictly increasing sequence of positive integers, we denote a a positive integer in this sequence. Let A(x) be the number
P
of positive integers in this sequence not exceeding x. That is A(x) = a≤x 1.
Suppose that A(x) = ρx + o(x), where ρ is a positive real number, that is, ρ is
the positive density of these integers. Then
ak
X
ρ
x
=
ζ(k + 1)xk+1 + o xk+1
a
k+1
a≤x
(1)
where k is an arbitrary but fixed positive integer and ζ(s) is the Riemann zetafunction.
Proof. We use theorem 1.1 con f (x) = xk . Therefore
X
ak
a≤x
k
= A(x)x −
Z x
k−1
A(t)kt
k
dt = (ρx + o(x)) x −
Z x
1
1
= ρx
k+1
+o x
k+1
− kρ
Z x
k
t dt +
Z x
1
= ρx
=
k+1
+o x
k+1
ρ
xk+1 + o xk+1
k+1
x
j+1
<a≤
x
j
k
o t
dt
1
k
xk+1 + o
−ρ
k+1
Note that if
(see (2))
(ρt + o(t)) ktk−1 dt
Z x
k
t dt
1
(2)
then
j k
x
a
= j, where j is a positive integer. Hence

X
ak
x
<a≤ xj
j+1
x
=j
a

X k
a
X
x
<a≤ xj
j+1
ak = j 
−
a≤ xj
X
ak 

x
a≤ j+1
!
=
ρ
1
1
k+1
k+1
x
+
o
x
j k+1 −
k+1 j
(j + 1)k+1
(3)
Now, we have (see (3))
X
a≤x
ak
h
X

X
x
x

=
ak
+

a
a
j=1
a≤ x
h+1

X
x
<a≤ xj
j+1
ak
X
x 
x
ak
=
a
a
a≤ x
h+1
1675
On a sum in number theory
h
X
1
1
ρ
j k+1 −
xk+1
+
k+1
j
(j + 1)k+1
j=1
h
X
ρ
1
+
xk+1
k+1
k+1
i=1 i
ak
X
=
−
+o x
k+1
X
=
a
k
x
a≤ h+1
x
a
h
1
ρ
k+1
xk+1
+
o
x
k+1
h + 1 (h + 1)k


∞
X
x
ρ
1 
ρ
xk+1 ζ(k + 1) −
xk+1 
+
k+1
a
k+1
k+1
i=h+1 i
x
a≤ h+1
−
!
!!
h
ρ
1
ρ
xk+1
xk+1 o(1)
+
k
k+1
h + 1 (h + 1)
k+1
(4)
On the other hand we have (see (2) and note that (2) is also true if k = 0)
0≤
k
X
a
x
a≤ h+1
X
X
x
x
ρ xk
≤
ak ≤ x
ak−1 = x
+
o
xk
k
a
a
k (h + 1)
a≤ x
a≤ x
h+1
k+1
!
h+1
!
ρ
k+1
1
ρ k+1 x
+ o xk+1 =
xk+1
+ o(1)
=
k
k + 1 k (h + 1)
k+1
k (h + 1)k
ρ
≤
xk+1 3
(5)
k+1
We have choose h such that
1
< ,
(h + 1)k
∞
X
1
i=h+1
ik+1
<
(6)
and have choose x0 such that if x ≥ x0 then |o(1)| < in equations (5) and
(4).
Equation (4) gives
j k
x
a
ρ
k+1
x
k+1
k
a≤x a
P
j k
k x
x a
a≤ h+1
a
ρ
k+1
x
k+1
P
− ζ(k + 1) =
h
1
−
+ o(1)
h + 1 (h + 1)k

−
∞
X
i=h+1

1 
ik+1
(7)
Therefore, see (7), (5) and (6), we have if x ≥ x0
j k
P
k x
a≤x a a
ρ
k+1 xk+1
− ζ(k
+ 1)
≤ 6
Consequently, since > 0 is arbitrarily small, we have
P
lim
x→∞
j k
x
a
ρ
k+1
x
k+1
k
a≤x a
= ζ(k + 1)
(8)
1676
Rafael Jakimczuk
That is, equation (1). The theorem is proved.
A positive integer n is (k +1)-th-power-free if all prime in their prime factorization have exponent not exceeding k, where k is a positive integer. In particular,
if k = 1 we obtain the square-free integers. The (k + 1)-th-power-free inte1
(see for example [2]). In particular the
gers have positive density ρ = ζ(k+1)
1
. In this note qk denotes a
square-free integers have positive density ρ = ζ(2)
(k + 1)-th-power-free integer. On the other hand, n denotes a positive integer.
We have the following corollary.
Corollary 1.3 The following asymptotic formula holds
$
X
qk ≤x
qkk
X
x
xk+1
∼
nk ∼
qk
k+1
n≤x
%
In particular, if k = 1 we have the following formula for the square-free integers
q1
$ %
X
X
x2
x
∼
n∼
q1
q1
2
n≤x
q1 ≤x
Proof. The proof is an immediate consequence of theorem 1.2 and the formula
P
k+1
(see (2)) n≤x nk ∼ xk+1 , since the positive integers have density ρ = 1. The
corollary is proved.
ACKNOWLEDGEMENTS. The author is very grateful to Universidad
Nacional de Luján.
References
[1] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers,
Oxford, 1960.
[2] R. Jakimczuk, A simple proof that the square-free numbers have density
6/π 2 , Gulf Journal of Mathematics 1 (2013), Issue 2, 55-58.
Received: September 15, 2014; Published: November 12, 2014