Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 4, Article 53, 2002
ON THE SEQUENCE (p2n − pn−1 pn+1 )n≥2
LAURENŢIU PANAITOPOL
FACULTY OF M ATHEMATICS
14 ACADEMIEI S T.
RO-70109 B UCHAREST, ROMANIA
pan@al.math.unibuc.ro
Received 17 December, 2001; accepted 24 May, 2002
Communicated by L. Toth
A BSTRACT. Let pn be the n-th prime number and xn = p2n − pn−1 pn+1 . In this paper, we study
sequences
containing the terms of the sequence (xn )n≥1 . The main result asserts that the series
P∞
2
x
/p
n
n is convergent, without being absolutely convergent.
n=1
Key words and phrases: Prime Numbers, Sequences, Series, Asymptotic Behaviour.
2000 Mathematics Subject Classification. 11A25, 11N05, 11N36.
1. I NTRODUCTION
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
qn =
for n ≥ 1,
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 x.
log
(1.1)
qn−1 pn ≤x
ISSN (electronic): 1443-5756
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090-01
2
L AUREN ŢIU PANAITOPOL
This result plays an essential role in the following paragraph of the present paper.
Another useful result is proved in [4]:
n
∞ X
dn
is convergent.
(1.2)
the series
pn
n=1
2. T HE S ERIES
P∞
xn
n=2 p2n
P
xn
Theorem 2.1. The series ∞
n=2 p2n is convergent, but it is not absolutely convergent.
In order to prove this fact, we need the following lemmas.
Lemma 2.2. For x ≥ − 12 , we have
x2
.
2
Proof. The inequalities are well known for x > 0. When x ∈ − 21 , 0 , they take on the form
x2 + |x| ≥ | log(1 + x)| ≥ |x| −
x2
.
2
2
Let f, g : − 21 , 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) = x(2x+1)
≤ 0. Since f is increasing
1+x
1+x
and f (0) = 0, we get f (x) ≤ 0. On the other 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 n=2 pn is convergent.
P
k−1
Proof. Denote Sn = nk=2 dk −d
, so that
pk
x2 − x ≥ − log(1 + x) ≥ −x −
n
1
dn X d2k−1
+
− .
Sn =
pn k=2 pk pk−1 2
P∞ d2k−1
Since limn→∞ pn+1
=
1,
it
suffices
to
prove
that
the
series
k=2 pk pk−1 is convergent. Since
pn
2
2
P∞ d2k−1
dk−1
dk−1
∼
and
the
terms
of
the
series
are
positive,
it
follows
that
the
series
k=2 pk pk−1
pk pk−1
pk−1
P∞ dk−1 2
and k=2 pk−1
are simultaneously convergent or not. Now just use (1.2) and the proof
ends.
P∞ x2n
Lemma 2.4. The series n=2 p4 is convergent.
n
Proof. Since xn = dn dn−1 + pn (dn−1 − dn ), it follows that
xn
dn dn−1 dn−1 − dn
(2.1)
=
+
2
pn
p2n
pn
hence
2 2
dn dn−1 (dn−1 − dn )2
x2n
(2.2)
≤2
+
.
p4n
p4n
p2n
P
P∞ d2n−1
d2n−1
d2n−1
d2n
Since the series ∞
is
convergent
and
∼
,
it
follows
that
the
series
2
2
2
n=1 pn
n=2 p2n is
pn
pn−1
P∞ max(d2n ,d2n−1 )
convergent as well. This implies that the series n=2
is also convergent. Since
p2
n
d2n d2n−1
p4n
<
max(d2n , d2n−1 )
p2n
J. Inequal. Pure and Appl. Math., 3(4) Art. 53, 2002
and
max(d2n , d2n−1 )
(dn−1 − dn )2
<
,
p2n
p2n
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O N THE S EQUENCE (p2n − pn−1 pn+1 )n≥2
we deduce by (2.2) that the series
x2n
n=2 p4n
P∞
3
is convergent.
Lemma 2.5. For x > 0 we have
X qn − qn−1 qn−1 log x.
pn ≤x
Proof. In view of Lemma 2.2, we have
2 qn − qn−1 qn − qn−1 1 qn − qn−1 2
qn − qn−1
q
n
≥ log
≥
−
+ .
qn−1
qn−1 qn−1 qn−1 2
qn−1
Since
pn +1
pn
− pn−1
qn − qn−1
xn
pn
=
=− 2,
pn
qn−1
pn
pn−1
we have
1 x2n qn qn − qn−1 qn x2n
(2.3)
·
+ log
>
≥ log
− .
2 p4n qn−1 qn−1 qn−1 p4n
Now the desired conclusion follows by (1.1) and Lemma 2.4.
Proof of Theorem 2.1. By the relation (2.1) we have
n
n
n
X
xk X dk dk−1 X dk−1 − dk
Sn =
=
+
.
2
2
p
p
p
k
k
k
k=2
k=2
k=2
P
max(d2n ,d2n−1 )
Since dk dk−1 ≤ max(d2k , d2k−1 ), and since the series ∞
is convergent, see the
n=2
p2n
P∞ dn dn−1
proof of Lemma 2.4, it follows that the series n=2 p2 is convergent too. Consequently the
n
P
sequence (Sn0 )n≥1 , defined by Sn0 = nk=2 dk dpk−1
is
convergent.
Lemma 2.3 implies that the
2
k
P
sequence (Sn00 )n≥1 , defined by Sn00 = nk=2 dk−1pk−dk is convergent as well. It then follows that the
P
xn
sequence (Sn )n≥2 is convergent, that is, the series ∞
is convergent.
n=2 p2n
|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,
2
p
n
p ≤x
(2.4)
n
hence the series
P∞
xn
n=2 p2n
is not absolutely convergent.
3. T HE S ERIES
P∞
|xn |
n=2 p2n logα n
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 P
begin with, we recall an inequality due P
to Abel: Let ak , bk ∈ R, k ∈ 1, n such that, if
Si = ik=1 bk , then Si ≥ 0 for i ∈ 1, n. Then ni=1 ai bi = S1 (a1 − a2 ) + S2 (a2 − a3 ) + · · · +
Sn−1 (an−1 − an ) + Sn an , which implies the inequalities
n
X
(3.1)
ai bi ≥ an Sn provided a1 ≥ · · · ≥ an ,
i=1
J. Inequal. Pure and Appl. Math., 3(4) Art. 53, 2002
http://jipam.vu.edu.au/
4
L AUREN ŢIU PANAITOPOL
and
n
X
(3.2)
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
X |xn |
(3.3)
c1 log x <
< c2 log x for all x ≥ 2.
2
p
n
p ≤x
n
0
For α > 0 and n ≥ 1, we denote a1 = 1, b1 = 0 and for i ≥ 2 ai = log1α i and bi = ci · |xp2i | , where
P i
c0 > 0 is chosen such that S1 , S2 , . . . , Sn ≥ 0. Such a choice is possible because 2≤i≤x 1i ∼
log x and (3.3) holds.
0
P
P
P
2
i|
It now follows by (3.1) that ni=2 log1α i ci − |xp2i | ≥ 0, that is, ni=2 p2|xlog
< c0 ni=2 i log1 α i .
α
i
i
i
P
P
|xn |
1
Since the series ∞
is
convergent
for
α
>
1,
we
deduce
that
the
series ∞
α
i=2 i log i
n=2 p2n logα n is
convergent as well.
One can similarly show that there exists c00 > 0 such that
n
n
X
X
|xi |2
1
00
>
c
α
α .
2
p
log
i
i
log
i
i
i=2
i=2
P
1
Since the series ∞
i=2 i logα i is divergent for α ≤ 1, it follows that in this case the series
P∞
|xn |
n=2 p2 logα n is in turn divergent.
n
R EFERENCES
[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.
J. Inequal. Pure and Appl. Math., 3(4) Art. 53, 2002
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