Journal of Inequalities in Pure and
Applied Mathematics
ON THE SEQUENCE (p2n − pn−1 pn+1 )n≥2
LAURENŢIU PANAITOPOL
Faculty of Mathematics
14 Academiei St.
RO-70109 Bucharest, Romania.
EMail: pan@al.math.unibuc.ro
volume 3, issue 4, article 53,
2002.
Received 17 December, 2001;
accepted 24 May, 2002.
Communicated by: L. Toth
Abstract
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Abstract
Let pn be the n-th prime number and xn = p2n − pn−1 pn+1 . In this paper, we
study sequences containingPthe terms of the sequence (xn )n≥1 . The main re2
sult asserts that the series ∞
n=1 xn /pn is convergent, without being absolutely
convergent.
On the Sequence
(p2n − pn−1 pn+1 )n≥2
2000 Mathematics Subject Classification: 11A25, 11N05, 11N36
Key words: Prime Numbers, Sequences, Series, Asymptotic Behaviour.
Laurenţiu Panaitopol
Contents
1
2
IntroductionP. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
xn
The Series ∞
n=2 p2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
P∞ n |xn |
3
The Series n=2 p2 logα n . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
n
References
3
5
9
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1.
Introduction
We shall use the following notation:
pn the n-th prime number
xn = p2n − pn−1 pn+1 for n ≥ 2,
dn = pn+1 − pn for n ≥ 1,
pn+1
for n ≥ 1,
qn =
pn
f (x) g(x) if there exist c1 , c2 , M > 0 such that
c1 f (x) < g(x) < c2 f (x) for every x > M .
It will be our aim here to study the sequence (xn )n≥2 defined above.
It was proved in [1] that the sequence (dn )n≥1 is not nonotone. A similar
result holds for the sequence (qn )n≥1 as well. This means that the sequence
(xn )n≥2 has infinitely many positive terms, and infinitely many negative terms,
hence it is not monotone.
In [1], the so-called method of the triple sieve (due to Vigo Brun) was used
to prove that
X q
n
log
log x.
(1.1)
q
n−1
p ≤x
On the Sequence
(p2n − pn−1 pn+1 )n≥2
Laurenţiu Panaitopol
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n
This result plays an essential role in the following paragraph of the present
paper.
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Another useful result is proved in [4]:
(1.2)
the series
n
∞ X
dn
n=1
pn
is convergent.
On the Sequence
(p2n − pn−1 pn+1 )n≥2
Laurenţiu Panaitopol
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2.
The Series
P∞
Theorem 2.1. The series
vergent.
xn
n=2 p2n
P∞ xn
n=2 p2n
is convergent, but it is not absolutely con-
In order to prove this fact, we need the following lemmas.
Lemma 2.2. For x ≥ − 12 , we have
x2
x + |x| ≥ | log(1 + x)| ≥ |x| − .
2
2
On the Sequence
(p2n − pn−1 pn+1 )n≥2
Laurenţiu Panaitopol
Proof. The inequalities are well known for x > 0. When x ∈ − 12 , 0 , they take
on the form
x2
x2 − x ≥ − log(1 + x) ≥ −x − .
2
1 2
Let f, g : − 2 , 0 → R be defined by f (x) = log(1 + x) − x − x2 and g(x) =
log(1 + x) − x + x2 , respectively. We have f 0 (x) = − x(x+2)
≥ 0, and g 0 (x) =
1+x
x(2x+1)
≤ 0. Since f is increasing and f (0) = 0, we get f (x) ≤ 0. On the other
1+x
hand, we have g 0 (x) < 0 and g(0) = 0, so that g(x) ≥ 0.
P
dn −dn−1
Lemma 2.3. The series ∞
is convergent.
n=2
pn
Proof. Denote Sn =
dk −dk−1
,
k=2
pk
Pn
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n
1
dn X d2k−1
+
− .
Sn =
pn k=2 pk pk−1 2
J. Ineq. Pure and Appl. Math. 3(4) Art. 53, 2002
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P
d2k−1
1, it suffices to prove that the series ∞
k=2 pk pk−1 is con
2
k−1
vergent.
and the terms of the series are positive, it
∼ dpk−1
2
P
P∞ dk−1 2
dk−1
follows that the series ∞
and
are simultaneously conk=2 pk pk−1
k=2 pk−1
vergent or not. Now just use (1.2) and the proof ends.
P
x2n
Lemma 2.4. The series ∞
n=2 p4 is convergent.
pn+1
=
pn
2
d
Since pk k−1
pk−1
Since limn→∞
n
Proof. Since xn = dn dn−1 + pn (dn−1 − dn ), it follows that
(2.1)
xn
dn dn−1 dn−1 − dn
=
+
p2n
p2n
pn
Laurenţiu Panaitopol
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hence
(2.2)
On the Sequence
(p2n − pn−1 pn+1 )n≥2
x2n
≤2
p4n
d2n d2n−1 (dn−1 − dn )2
+
p4n
p2n
Contents
.
P
d2n−1
d2n−1
d2n
Since the series ∞
is
convergent
and
∼
, it follows that the se2
2
n=1 pn
p2n
pn−1
P∞ d2n−1
P
max(d2n ,d2n−1 )
ries n=2 p2 is convergent as well. This implies that the series ∞
n=2
p2n
n
is also convergent. Since
d2n d2n−1
max(d2n , d2n−1 )
<
p4n
p2n
we deduce by (2.2) that the series
and
max(d2n , d2n−1 )
(dn−1 − dn )2
<
,
p2n
p2n
x2n
n=2 p4n
P∞
is convergent.
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Lemma 2.5. For x > 0 we have
X qn − qn−1 qn−1 log x.
p ≤x
n
Proof. In view of Lemma 2.2, we have
2 qn − qn−1 qn − qn−1 1 qn − qn−1 2
q
qn − qn−1
n
≥ log
≥
−
+ .
qn−1
qn−1 qn−1 qn−1 2
qn−1
Since
qn − qn−1
=
qn−1
pn +1
pn
−
pn
pn−1
pn
pn−1
=−
xn
,
p2n
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we have
(2.3)
On the Sequence
(p2n − pn−1 pn+1 )n≥2
Contents
1 x2n qn qn − qn−1 qn x2n
·
+ log
>
≥ log
− .
2 p4n qn−1 qn−1 qn−1 p4n
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Now the desired conclusion follows by (1.1) and Lemma 2.4.
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Proof of Theorem 2.1. By the relation (2.1) we have
Sn =
n
X
xk
k=2
p2k
=
n
X
dk dk−1
k=2
p2k
+
n
X
dk−1 − dk
k=2
Since dk dk−1 ≤ max(d2k , d2k−1 ), and since the series
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P∞
n=2
max(d2n ,d2n−1 )
p2n
vergent, see the proof of Lemma 2.4, it follows that the series
P∞
is con-
dn dn−1
n=2
p2n
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P
convergent too. Consequently the sequence (Sn0 )n≥1 , defined by Sn0 = nk=2 dk dpk−1
2
k
00
00
is convergent. Lemma 2.3 implies that the sequence (Sn )n≥1 , defined by Sn =
Pn dk−1 −dk
is convergent as well. It then follows that the sequence (Sn )n≥2
k=2
pk
P
xn
is convergent, that is, the series ∞
n=2 p2n is convergent.
|x |
n−1 On the other hand, Lemma 2.5 and the relation qnq−q
= p2n imply that
n−1
n
X |xn |
log x,
p2n
p ≤x
(2.4)
n
hence the series
P∞
xn
n=2 p2n
is not absolutely convergent.
On the Sequence
(p2n − pn−1 pn+1 )n≥2
Laurenţiu Panaitopol
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3.
|xn |
n=2 p2n logα n
P∞
The Series
P
|xn |
Since the series ∞
n=2 p2n is divergent, it is natural to study what “correction”
does it need to become convergent. In this connection, we prove the following
fact.
P
|xn |
Theorem 3.1. The series ∞
n=2 p2 logα n is convergent if and only if α > 1.
n
Proof. We are going to put to use a technique from [3].
To begin with, wePrecall an inequality due to Abel: Let ak , bk ∈PR, k ∈ 1, n
n
i
such that, if Si =
k=1 bk , then Si ≥ 0 for i ∈ 1, n. Then
i=1 ai bi =
S1 (a1 − a2 ) + S2 (a2 − a3 ) + · · · + Sn−1 (an−1 − an ) + Sn an , which implies the
inequalities
On the Sequence
(p2n − pn−1 pn+1 )n≥2
Laurenţiu Panaitopol
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(3.1)
n
X
ai bi ≥ an Sn provided a1 ≥ · · · ≥ an ,
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i=1
and
(3.2)
n
X
ai bi ≤ an Sn when a1 ≤ · · · ≤ an .
i=1
It follows by (2.4) that there exist c1 and c2 such that 0 < c1 < c2 and
(3.3)
Contents
X |xn |
c1 log x <
< c2 log x for all x ≥ 2.
p2n
p ≤x
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For α > 0 and n ≥ 1, we denote a1 = 1, b1 = 0 and for i ≥ 2 ai =
c0
i
1
logα i
and
|xi |
,
p2i
bi = ·
where c0 > 0 is chosen such that S1 , S2 , . . . , Sn ≥ 0. Such a choice
P
is possible because 2≤i≤x 1i ∼ log x and (3.3) holds.
0
P
P
2
i|
It now follows by (3.1) that ni=2 log1α i ci − |xp2i | ≥ 0, that is, ni=2 p2|xlog
<
α
i
i
i
P
P
n
∞
c0 i=2 i log1 α i . Since the series i=2 i log1 α i is convergent for α > 1, we deduce
P
|xn |
that the series ∞
n=2 p2n logα n is convergent as well.
One can similarly show that there exists c00 > 0 such that
n
X
i=2
Since the series
P
the series ∞
n=2
P∞
n
X 1
|xi |2
00
α > c
α .
2
pi log i
i
log
i
i=2
1
i=2 i logα i is divergent for
|xn |
is in turn divergent.
p2n logα n
α ≤ 1, it follows that in this case
On the Sequence
(p2n − pn−1 pn+1 )n≥2
Laurenţiu Panaitopol
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References
[1] P. ERDŐS AND A. RÉNYI, Some problems and results on consecutive
primes, Simon Stevin, 27 (1950), 115–125.
[2] P. ERDŐS AND P. TURÁN, On some new question on the distribution of
prime numbers, Bull. Amer. Math. Soc., 54 (1948), 371–378.
[3] L. PANAITOPOL, Properties of the series of differences of prime numbers, Publications du Centre de Recherches en Mathématiques Pures. Univ.
Neuchâtel, Sér. I, Fasc., 31 (2000), 21–28.
[4] P. VLAMOS, Inequalities involving the sequence of the differences of the
prime numbers, Publications du Centre de Recherches en Mathématiques
Pures. Univ. Neuchâtel, Sér. I, Fasc., 32 (2001), 32–40.
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(p2n − pn−1 pn+1 )n≥2
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