Journal of Inequalities in Pure and
Applied Mathematics
CONSEQUENCES OF A THEOREM OF ERDÖS-PRACHAR
LAURENTIU PANAITOPOL
Faculty of Mathematics,
University of Bucharest,
14 Academiei St.,
RO-70109 Bucharest, Romania
EMail: pan@al.math.unibuc.ro
volume 2, issue 3, article 35,
2001.
Received 27 March, 2001;
accepted 11 June, 2001.
Communicated by: L. Toth
Abstract
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c 2000 Victoria University
ISSN (electronic): 1443-5756
028-01
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Abstract
P
In the present article we study the asymptotic behavior of the sums n≤x pcn+1
− pcnn
n+1
α
P
P
cn+1
cn
n+1
and n≤x pcn+1
− pcnn , and of the series ∞
−
n=1 pn+1
pn , where pn denotes
the n-th prime number while cn stands for the n-th composed number.
2000 Mathematics Subject Classification: 11A25, 11N05.
Key words: Sequences, Series, Prime Numbers, the Sequence of Composed Numbers, Asymptotic Formulae
Consequences of a Theorem of
Erdös-Prachar
Laurenţiu Panaitopol
The author gratefully acknowledges support from Grant #C12/2000 awarded by the
Consiliul Naţional al Cercetării Ştiinţifice din Învăţământul Superior, România.
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Contents
1
2
3
Introduction . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . 3
Properties of the Sequence pnn
.................... 6
n≥1
Properties of the Sequence pcnn
. . . . . . . . . . . . . . . . . . . . 10
References
n≥1
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1.
Introduction
We are going to use the following notation
π(x) the number of prime numbers ≤ x,
C(x) the number of composed numbers ≤ x,
pn the n-th prime number,
cn the n-th composed number; c1 = 4, c2 = 6, . . . ,
log2 n = log(log n).
The present work originates in a result due to Erdös and Prachar [2]: they
proved that there exist c0 , c00 > 0 such that
c0 log2 x >
X pk+1
pk
−
> c00 log2 x for x ≥ 2,
k+1
k
p ≤x
k
that is
(1.1)
X pk+1
pk
−
log2 x.
k+1
k
p ≤x
Consequences of a Theorem of
Erdös-Prachar
Laurenţiu Panaitopol
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In a recent paper [3], Panaitopol proved that
(1.2)
X k+1
k
−
log log x.
p
p
k+1
k
p ≤x
k
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The proofs of these results rely on the following result due to Schnirelmann: if
for x positive and n a positive integer one denotes by M (n, x) the number of
the indices k such that pk ≤ x and pk+1 − pk = n, then
M (n, x) < c000
x X1
,
log2 x d|n d
where c000 is a positive constant.
In the present paper, several well known results will be used:
x
;
log x
(1.3)
π(x) ∼
(1.4)
pn ∼ n log n;
Consequences of a Theorem of
Erdös-Prachar
Laurenţiu Panaitopol
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(1.5)
the series
∞
X
n=3
(1.6)
1
is convergent if and only if α > 1;
n log n(log log n)α
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1
= log log x + O(1);
n log n
2≤n≤x
X
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(1.7)
X 1
= log x + O(1);
n
2≤n≤x
J. Ineq. Pure and Appl. Math. 2(3) Art. 35, 2001
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(1.8)
log p
∼ log x.
p
p prime ≤x
X
We also need the following result of Bojarincev [1]:
(1.9)
cn = n +
n
· un , where lim un = 1.
n→∞
log n
Consequences of a Theorem of
Erdös-Prachar
Laurenţiu Panaitopol
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2.
Properties of the Sequence
n
pn
n≥1
P
n+1
n
The series ∞
n=1 pn+1 − pn is divergent by (1.2). In connection with this fact
we prove the following result.
Theorem 2.1. The series
∞
X
n+1
n
1
−
·
pn+1
pn (log log n)α
n=3
Laurenţiu Panaitopol
is convergent if and only if α > 1.
Let us first prove the following auxiliary result.
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Lemma
2.2. Consider the sequences (an )n≥1 , (xn )n≥1 and (sn )n≥1 , where sn =
Pn
i=1 ai . If the sequence (sn xn )n≥1 is convergent, then one of the series
∞
X
n=1
sn (xn+1 − xn ) and
Consequences of a Theorem of
Erdös-Prachar
∞
X
an x n
n=1
is convergent if and only if the other one is convergent.
P
Proof. If ∞
n=1 sn (xn+1 − xn ) is convergent, then limn→∞ sn (xn+1 − xn ) =
0. But limn→∞ sn xn = k for some k ∈ R, hence limn→∞ sn xn+1 = k and
limn→∞ (sn+1 − sn )xn+1 = 0.
P
On the other hand, if ∞
n=1 an xn is convergent, then limn→∞ an+1 xn+1 = 0,
hence limn→∞ (sn+1 − sn )xn+1 = 0.
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Now let us denote Sn =
each p we have
Pn
i=1
ai xi and σn =
Pn
i=1
si (xi+1 − xi ). Then for
(2.1) Sn+p − Sn = σn+p − σn + sn+p xn+p − sn+1 xn+1 + (sn+1 − sn )xn+1 .
Since we have just seen that in either case we have lim (sn+1 − sn )xn+1 =
n→∞
0, relation (2.1) implies that in either case one of the sequences (Sn )n≥1 and
(σn )n≥1 is Cauchy if and only if the other one is Cauchy. Now, by Cauchy’s
criterion, one of the two series is convergent if and only if the other one is
convergent.
Proof of Theorem 2.1. If α ≤ 0, then the series is divergent by (1.2). Next
1
assume α > 0 and choose an = pn+1
− pnn and xn = (log log
.
n)α
n+1
If we consider the function f : [n, n + 1] → R, f (x) = (log log x)−α , then
Lagrange’s theorem implies
(log log(n + 1))−α − (log log n)−α = −
α(log log θn )−α−1
,
θn log θn
where n < θn < n + 1. Since θn ∼ n, it follows that
(2.2)
xn+1 − xn ∼
−α
.
n log n(log log n)α+1
By (1.2) and (1.4) it follows that sn log log n and (2.2) implies
(2.3)
sn (xn+1 − xn ) −
1
.
n log n(log log n)α
Consequences of a Theorem of
Erdös-Prachar
Laurenţiu Panaitopol
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If α > 1, then we have
log log n
= 0,
n→∞ (log log n)α
lim sn xn = lim
n→∞
while for α = 1 we get limn→∞
α ≥ 1, the above lemma
P∞sn xn = 1. Thus, for P
implies that one of the series n=3 sn (xn+1 −xn ) and ∞
n=3 an xn is convergent
if
and
only
if
the
other
one
is
convergent.
In
view
of
(1.5)
and (2.3), the series
P∞
n=3 an xn is convergent for α > 1 and divergent for α = 1.
Finally, if 0 < α < 1, then
xn
n
1
n+1
n
n+1
−
>
−
pn+1
pn
log log n pn+1
pn
Consequences of a Theorem of
Erdös-Prachar
Laurenţiu Panaitopol
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and the desired conclusion follows.
Contents
Consequence 1. If α > 1, then the series
∞
X
n
n+1
−
pn+1
pn
n=1
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Proof. We have
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n+1
n
−
pn+1
pn
α−1
≤ max
n+1 n
,
pn+1 pn
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α−1
.
J. Ineq. Pure and Appl. Math. 2(3) Art. 35, 2001
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1
For K > 0 and n ≥ 3, we have (log n)
α−1 <
n+1
n
1
0
∼ pn ∼ log n . There exists K such that
pn+1
n+1
n
−
pn+1
pn
α
< K0
and the convergence of the series
K
(log log n)α
and (1.4) implies that
n+1
n
1
−
pn+1
pn (log log n)α
P∞
n+1
n=1 pn+1
−
n
pn
α
follows by Theorem 2.1.
Consequences of a Theorem of
Erdös-Prachar
Laurenţiu Panaitopol
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3.
Properties of the Sequence
Since cn ∼ n (see (1.9)), for the sequence
are similar to those of the sequence pnn
cn
pn
cn
pn
n≥1
we obtain properties which
n≥1
.
n≥1
In connection with (1.2) we have the following fact.
Theorem 3.1. We have
X ck+1
ck
−
log log x
p
p
k+1
k
p ≤x
k
Consequences of a Theorem of
Erdös-Prachar
Laurenţiu Panaitopol
for every x > e.
Proof. If we denote αk = ck+1 − ck − 1, then it follows that αk = 0 if ck + 1 is
a composed number, and αk = 1 if ck + 1 is prime. In the last case ck + 1 = pm .
Setting k = k(m), we deduce by (1.9) that ck − k ∼ logk k and log k ∼ log ck ∼
pm
log pm ∼ log m. It then follows that k(m) − pm ∼ − log
and (1.4) implies that
m
k(m) = pm − mym with lim ym = 1.
(3.1)
m→∞
We have by (1.9)
ck+1
ck
ck + 1 + αk
ck
αk + 1 ck (pk+1 − pk )
−
=
−
=
−
pk+1 pk
pk+1
pk
pk+1
pk pk+1
αk + 1 k(pk+1 − pk )
kuk
=
−
−
pk+1
pk pk+1
(log k)pk pk+1
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=
k+1
k
−
pk+1
pk
+
αk
kuk (pk+1 − pk )
−
.
pk+1
(log k)pk pk+1
We have the inequality
(3.2)
k
k+1
−
−
pk+1
pk
ck+1
≤
−
pk+1
αk
kuk (pk+1 − pk )
−
pk+1
(log k)pk pk+1
k+1
αk
kuk (pk+1 − pk )
ck
k
≤
−
+
+
.
pk
pk+1
pk
pk+1
(log k)pk pk+1
We have
k(pk+1 − pk )
pk+1 − pk
kuk (pk+1 − pk )
∼
=
.
3
(log k)pk pk+1
k 2 log k
k log3 k
P
pn+1 −pn
Panaitopol proves in [4] that for β > 2 the series ∞
n=2 n logβ n is convergent,
P
kuk (pk+1 −pk )
hence the series ∞
k=2 (log k)pk pk+1 is also convergent.
We have furthermore
∞
X
k=1
X0 1
αk
=
,
pk+1
pk+1
Consequences of a Theorem of
Erdös-Prachar
Laurenţiu Panaitopol
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where 0 extends over the values of k such that αk = 1, that is, ck + 1 = pm .
Then by (2.2) and (1.4) we get
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pk+1 ∼ pk ∼ ppm ∼ m log2 m.
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(3.3)
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P
1
Since the series ∞
m=2 m log2 m is convergent, it then follows that the series
P∞ αk
k=1 pk+1 is also convergent. Now (3.2) implies that
n
n
X
X
ck+1
ck
k+1
k
−
=
−
+ O(1)
pk+1 pk
pk+1
pk
k=1
k=1
and the desired conclusion follows.
Analogously to the Erdös-Prachar theorem, we shall prove the following
fact.
Laurenţiu Panaitopol
Theorem 3.2. We have
X pk+1 pk
−
log2 x
ck+1
ck
p ≤x
k
for every x > 1.
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Proof. We have
pk+1 pk
(3.4)
−
=
ck+1
ck
Consequences of a Theorem of
Erdös-Prachar
pk+1
pk
−
k+1
k
+ pk
1
ck+1 − ck
−
k(k + 1)
ck+1 ck
(pk+1 − pk )(k + 1 − ck+1 )
+
.
(k + 1)ck+1
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By (1.9) we get
X (pk+1 − pk )(k + 1 − ck+1 )
(k + 1)ck+1
p ≤x
k
X (pk+1 − pk )uk+1
c
log(k + 1)
pk ≤x k+1


π(x)
X
pk+1 − pk 
= O
k log k
k=2


π(x)
X
1
1
x
.
+
pk
−
= O
π(x) log π(x) k=3
(k − 1) log(k − 1) k log k
=−
By (1.3) we have π(x) log π(x) ∼ x,
1
1
pk log k
1
−
∼ 2 2 ∼
pk
(k − 1) log(k − 1) k log k
k
k log k
Laurenţiu Panaitopol
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and (1.7) implies
(3.5)
Consequences of a Theorem of
Erdös-Prachar
X (pk+1 − pk )(k + 1 − ck+1 )
= O(log x).
(k
+
1)c
k+1
p ≤x
k
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We have also
X
ck+1 − ck
1
pk
−
k(k + 1)
ck+1 ck
pk ≤x
X 00
X0
1
1
pk
≤
+2
,
pk
−
k(k
+
1)
c
(c
+
1)
c
(c
+
2)
k
k
k
k
p ≤x
p ≤x
k
k
P0
where
extends over the values of k such that ck + 1 is a prime number, while
P00
extends over the values of k such that ck + 1 = pm is composed. By (1.9)
we deduce
!
X k log k
X 00 pk (ck − k)(ck + k + 1)
=O
· k 2 log k
4
k(k + 1)ck (ck + 1)
k
pk ≤x
pk ≤x
!
X 1
= O(log x).
=O
k
p ≤x
k
By (1.4) and (1.8) it follows that
X0
pk ≤x
X 0 ck log ck
pk
∼
ck (ck + 2)
ck (ck + 2)
k≤π(x)
X 0 log(ck + 1)
ck + 1
ck ≤x
X log pm
= O(log x).
=
pm
p ≤x
∼
m
Consequences of a Theorem of
Erdös-Prachar
Laurenţiu Panaitopol
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Thus
(3.6)
X
pk ≤x
pk
1
ck+1 − ck
−
= O(log x).
k(k + 1)
ck+1 ck
Now by (1.1), (3.4), (3.5) and (3.6) it follows that
and the proof is completed.
pk+1
pk ≤x ck+1
P
−
pk
ck
log2 x
Consequences of a Theorem of
Erdös-Prachar
Laurenţiu Panaitopol
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References
[1] A.E. BOJARINCEV, Asymptotic expressions for the nth composite number, Ural. Gos. Univ. Mat. Zap., 6 (1967), 21–43 (in Russian).
[2] P. ERDÖS AND K. PRACHAR, Sätze und Probleme über pk /k, Abh. Math.
Sem. Univ. Hamburg, 25 (1961/1962), 251–256.
[3] L. PANAITOPOL, On Erdös-Prachar Theorem, An. Univ. Bucureşti Mat., 2
(1999), 143–148.
[4] L. PANAITOPOL, On the sequence of the differences of consecutive prime
numbers, Gaz. Mat. Ser. A, 79 (1974), 238–242 (in Romanian).
Consequences of a Theorem of
Erdös-Prachar
Laurenţiu Panaitopol
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