Download Reciprocal of the Square Root of a Perfect Power

International Mathematical Forum, Vol. 11, 2016, no. 1, 21 - 26
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/imf.2016.5867
Reciprocal of the Square Root
of a Perfect Power
Rafael Jakimczuk
División Matemática, Universidad Nacional de Luján
Buenos Aires, Argentina
c 2015 Rafael Jakimczuk. This article is distributed under the Creative
Copyright Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
Let us consider the sum
1
√
P
P ≤x
X
where P denotes a perfect power. In a previous article we obtained the
formula
X 1
1
√ = log x + C + o(1)
2
P
P ≤x
where C is a constant. In this article we obtain more precise asymptotic
formulae. Besides we prove that C is positive and
C=γ+
X
P0
1
√
P0
where γ > 0 is the Euler’s constant and P 0 denotes a perfect power not
a square.
Mathematics Subject Classification: 11A99, 11B99
Keywords: Square root of a perfect power, reciprocal, asymptotic formulae
22
1
R. Jakimczuk
Introduction
A natural number of the form mn where m is a positive integer and n ≥ 2 is
called a perfect power. The first few terms of the integer sequence of perfect
powers are
1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128 . . .
A quadratfrei number is a number without square factors, a product of different
primes. The first few terms of the integer sequence of quadratfrei numbers are
1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, . . .
On the other hand, the Möbius function µ(n) is defined as follows: µ(1) = 1,
if n is the product of r different primes, then µ(n) = (−1)r , if the square of a
prime divides n, then µ(n) = 0.
Let N (x) be the number of perfect powers not exceeding x. Jakimczuk [2]
proved the following theorem.
Theorem 1.1 Let ph be the h-th prime with h ≥ 1, where h is an arbitrary
but fixed positive integer. We have
X
N (x) =
−µ(q)x1/q + g(x)x1/ph
(x ≥ 1)
(1)
2≤q≤ph
where q denotes a quadratfrei number and limx→∞ g(x) = 0.
For example. If h = 1 then Theorem 1.1 becomes
√
√
N (x) = x + g(x) x
That is
N (x) ∼
√
x
and if h = 2 then Theorem 1.1 becomes
√
√
√
N (x) = x + 3 x + g(x) 3 x
Let us consider the sum
(2)
1
√
P
P ≤x
X
where P denotes a perfect power. In a previous article [3] we obtained the
formula
X 1
1
√ = log x + C + o(1)
2
P
P ≤x
where C is a constant. In this article we obtain more precise asymptotic
formulae.
We need the following Abel’s identity.
23
Square Root of a Perfect Power
Lemma 1.2 Let a(n) be a sequence of real numbers. Let us consider the
P
sum A(x) = n≤x a(n) and let f (x) be a function with continuous derivative
in the interval [1, x]. Then the following formula holds
X
a(n)f (n) = A(x)f (x) −
Z x
A(t)f 0 (t) dt
(3)
1
n≤x
Proof. See for example [1], page 77 and pages 279-280.
2
Main Results
Theorem 2.1 Let ph be the h-th prime with h ≥ 2, where h is an arbitrary
but fixed positive integer. We have
X 2µ(q) − 1 + 1
1
1
−1+ 1
√ = log x + C +
x 2 q + o x 2 ph
2
P
3≤q≤ph q − 2
P ≤x
X
(4)
where q denotes a quadratfrei number.
Proof. Let us consider the sequence a(n) = 1 if n is a perfect power
√ and a(n) =
0 if n is not a perfect power. On the other hand let f (x) = 1/ x = x−1/2 . We
have
X
a(n)
N (x) =
n≤x
and equation (3) becomes
X
N (x) 1 Z x N (t)
1
N (x) 1 Z x 1
√ =
√
a(n)f (n) =
+
dt = √ +
dt
3/2
x
2
t
x
2 1 t
1
P
n≤x
P ≤x
√
√
1 Z x N (t) − t
N (x) 1
1 Z ∞ N (t) − t
dt = √ + log x +
dt
+
2 1
t3/2
x
2
2 1
t3/2
√
1 Z ∞ N (t) − t
−
dt
(5)
2 x
t3/2
X
Note that the improper integral in (5) is convergent. Use equation (2) and the
comparison criterion for improper integrals.
Now (see (1)) we have
X
1
1
N (x)
−1+ 1
√ =1+
−µ(q)x− 2 + q + o x 2 ph
x
3≤q≤ph
On the other hand (see (1)) we have
√
Z ∞
Z ∞
Z ∞
X
N (t) − t
−3+ 1
− 23 + 1q
dt
=
µ(q)
t
dt
−
g(t)t 2 ph dt
−
3/2
t
x
x
x
3≤q≤ph
(6)
(7)
24
R. Jakimczuk
where
Z ∞
1
3
t− 2 + q dt =
x
and
Z ∞ 3 1
−2+p
t
h
dt =
x
2q − 12 + 1q
x
q−2
(8)
2ph − 21 + p1
h
x
ph − 2
(9)
Let > 0, there exist x0 such that if x ≥ x0 we have
− ≤ g(t) ≤ Therefore
−
Z ∞ 3 1
−2+p
t
h
dt ≤
Z ∞
x
− 23 + p1
g(t)t
h
dt ≤ Z ∞ 3 1
−2+p
x
t
h
dt
x
That is
−
Z ∞
2ph − 21 + p1
2ph − 12 + p1
−3+ 1
h ≤
h
g(t)t 2 ph dt ≤ x
x
ph − 2
ph − 2
x
Hence, since is arbitrarily small we have
Z ∞
− 32 + p1
g(t)t
h
dt = o x
− 12 + p1
(10)
h
x
Substituting (8), (9) and (10) into (7) and then substituting (7) and (6) into
(5) we obtain (4), where
√
1 Z ∞ N (t) − t
dt
C =1+
2 1
t3/2
The theorem is proved.
Example 2.2 If h = 2 then equation (4) becomes
1
1
2
1
√ = log x + C − √
+o √
6
6
2
x
x
P
P ≤x
!
X
(11)
Now, we shall prove that the constant C is positive. Before, we need the
following lemma.
Lemma 2.3 The following asymptotic formula holds
n
X
1
= log n + γ + o(1)
k=1 k
where γ > 0 is called Euler’s constant.
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Square Root of a Perfect Power
Proof. See for example ([4], Chaper VI).
Theorem 2.4 In equation (4) the constant C is positive. Besides we have
C=γ+
X
P0
1
√
P0
where P 0 denotes a perfect power not a square.
Proof. Equation (11) gives
1
1
√ = log x + C + o(1)
2
P
P ≤x
X
Therefore (Lemma 2.3)
n
X
X
1
1
1
√ = log n + C + o(1) =
√ +
√ = log n + γ + o(1)
P
k 2 P 0 <n2 P 0
k=1
P ≤n2
X
1
√
+
P0
P 0 <n2
X
and consequently
X
P 0 <n2
1
√ = (C − γ) + o(1)
P0
That is
lim
n→∞
X
P 0 <n2
X 1
1
√ =
√ =C −γ
P0
P0
P0
The theorem is proved.
To finish, we give a direct proof that the series of positive terms
converges. As usual, b.c denotes the integer-part function.
Theorem 2.5 The series
X
P0
P
P0
√1
P0
1
√
P0
converges.
Proof. Let N1 (x) be the number of perfect powers
P 0 not exceeding x. The
√
number of squares not exceeding√x is clearly b xc, since the inequality k 2 ≤ x
has the solutions k = 1, 2, . . . , b xc. Therefore (see (2))
j√ k
√
N1 (x) = N (x) − x ∼ 3 x
26
R. Jakimczuk
consequently if Pn0 is the n-th perfect power P 0 we have n = N1 (Pn0 ) ∼
3
Pn0 , that is, Pn0 ∼ n3 . Hence
and
q
X
P0
X
X 1
X 1
1
1
q
q
√ =
=
≤A
n3/2
P0
Pn0
f (n)n3
where f (n) → 1 and A > 0. Then, by the comparison criterion, the series
converges. The theorem is proved.
Acknowledgements. The author is very grateful to Universidad Nacional de
Luján.
References
[1] T. M. Apostol, Introduction to Analytic Number Theory, Springer, New
York, 1976. http://dx.doi.org/10.1007/978-1-4757-5579-4
[2] R. Jakimczuk, On the distribution of perfect powers, J. Integer Seq., 14
(2011), Article 11.8.5, 1-9.
[3] R. Jakimczuk, Sums of perfect powers, Int. J. Contemp. Math. Sciences,
8 (2013), 61-67.
[4] W. J. LeVeque, Topics in Number Theory, Addison-Wesley, 1958.
Received: Septemebr 9, 2015; Published: December 14, 2015