Note di Matematica 29, n. 1, 2009, 73–77.
Irrationality for a class of rational series
Eduardo Pascali
Department of Mathematics ”‘Ennio De Giorgi”’, University of Salento, P.O.B.123, 73100, Lecce, Italy eduardo.pascali@unile.it
Received: 10/09/2007; accepted: 22/04/2008.
Abstract. The aim of this paper is to prove the irrationality of the sum of some classes of series of positive rational numbers.
Keywords:Irrationality, infinite series, infinite sequences MSC 2000 classification:primary 11J72
1 Irrationality of rational series
Various criteria for the irrationality of the sum of infinite series of positive rationals are known, see the references. We prove another criterion, which is strictly related to the proof in [6] and to the results in [3], [4], [8].
The first result is an easy step for the final criterion and clarify the role of the hypotheses.
1 Proposition. Given a
1> 1, a
2> 1 integers numbers, let (a
n) be the sequence defined by
a
n+1= (a
1a
2· · · a
n)
γ1, n > 2, with 1 < γ
1∈ N.
Let (b
n) be a sequence of positive integers such that:
∃k
1, γ
2∈]0, +∞[ b
n≤ k
1a
γn−12∀n > 2.
If γ
2∈ [0, 1] and γ
1(1 − γ
2) ≥ 1, then the series
j bj
aj
is convergent to an irrational number. Moreover for every non decreasing sequence (c
j) of positive integers the series
j bj
ajcj
is convergent to an irrational number.
Proof. We remark that (a
n) is an increasing sequence and that by induc- tion it can be checked that a
n+2≥ 2
(2γ1)nfor all n. Moreover, the following inequalities hold:
0 < b
ja
j≤ k
1(a
j−1)
γ2a
j= k
11
(a
1. . . a
j−2)
γ1(1−γ2)(a
j−1)
γ1≤k
11
a
γj−11≤ k
11
2
γ1(2γ1)j−3. Note Mat. 29 (2009), n. 1, 73-78
ISSN 1123-2536, e-ISSN 1590-0932 DOI 10.1285/i15900932v29n1p73
http://siba-ese.unisalento.it, © 2009 Università del Salento
__________________________________________________________________
As a consequence, the series
j bj
aj
is convergent. Arguing as in [6], assume that the sum is a rational number α =
pq, where p, q ∈ N and 1 ≤ q, and observe that:
A
n= &
p − q
n j=1b
ja
j) *
nj=1
a
j= q & *
nj=1
a
j)
+∞j=n+1
b
ja
jis a positive integer and A
n≥ 1. Now:
A
n=q & *
nj=1
a
j)
+∞j=n+1
b
ja
j= qa
γ11 n+1
+∞j=n+1
b
ja
j= q
+∞
j=n+1
b
ja
γ11 n+1
a
j≤q
+∞j=n+1
b
ja
γ11 j
a
j≤ k
1q
+∞
j=n+1
a
γ21 j−1
a
γ1−1γ1 j
≤ k
1q
+∞j=n+1
1
(a
1. . . a
j−2)
γ1−γ1γ2−1a
γj−11−1≤k
1q
+∞
j=n+1
1 [2
γ1−1]
(2γ1)j−3.
Then from the convergence of the series
+∞j=n+1 1
[2γ1−1](2γ1)j−3
, we have A
n< 1 for n large and a contradiction follows. The last statement easily follows from the inequalities:
b
jc
ja
j≤ b
ja
jand, since (c
j) is non a decreasing sequence:
+∞
j=n+1
b
jc
ja
j*
n j=1(c
ja
j) ≤
+∞
j=n+1
c
n+1b
jc
ja
j*
n j=1a
j≤
+∞
j=n+1
b
ja
j*
n j=1a
j.
QED
Now we can state our main result.
2 Theorem. Given a
1> 1, a
2> 1, and let (a
n) be a sequence of positive integers such that:
∃k
1, k
2, α
1, α
2∈]0, +∞[ k
1(a
1a
2· · · a
n)
α1≤ a
n+1≤ k
2(a
1a
2· · · a
n)
α2(1) for all n > 3. Let (b
n) be a sequence of positive integers such that
∃k
3, γ ∈ [0, +∞[ b
n≤ k
3a
γn−1∀n > 3. (2) If the following conditions hold:
α
1− γα
2≥ 0 (3)
(α
1+ α
2) α
1α
2> 1; k
α21 2
k
1α11 2
α21((α1+α2)α1α2−1)< 1 (4)
α
1> 1 (5)
k
1k
α1α2 2
≥ 1, (6)
then the series
j bj
aj
converges to an irrational number; moreover if (c
j) is an increasing sequences of positive integers then also
j bj
cjaj
converges to an irrational number.
Proof. By (1) and (6) we have:
a
n+1≥ k
1k
α1α2 2
a
(α1+α2)α1α2
n
≥ a
n.
Hence the sequence (a
n) is non decreasing. Moreover from (1) we deduce:
a
n+3≥ k
n
j=1(α1α2(α1+α2))j 1
k
α1α2
n−1
j=1(α1α2(α1+α2))j 2
2
2α1((α1+α2)α1α2)n≡ D
nand the following inequalities hold:
b
n+4a
n+4≤ k
3a
γn+3a
n+4≤ k
3k
γ2k
11
(a
1a
2. . . a
n+2)
α1−γα2a
αn+31≤ k
3k
2γk
11 a
αn+31≤ k
3k
γ
2
k
1D
−1n. Consider now the series
n
D
−1n, and remark that:
D
−1n+1D
−1n=
⎡
⎣k
α21 2
k
α111 2
2α1((α1+α2)α1α2)⎤
⎦
[α1α2(α1+α2]n
Then by (4), the series
n
D
−1nis convergent; hence also
j bj
aj
is convergent.
Assume that the sum is a rational number
pq, where p, q ∈ N and 1 ≤ q, as in the previous proposition. Observe that
A
n= &
p − q
n j=1b
ja
j) *
nj=1
a
j= q & *
nj=1
a
j)
+∞j=n+1
b
ja
jis a positive integer and A
n≥ 1. But we remark that:
1 ≤A
n≤ q k
α11 1
+∞
j=n+1
b
ja
α11 j
a
j≤ k
3q k
α11 1
+∞
j=n+1
a
γj−1a
α1−1α1 j
≤ k
3q k
α11 1
+∞j=n+1
k
2γ(a
1. . . a
j−2)
γα2k
α1−1α1
1
(a
1. . . a
j−1)
α1−1≤ qk
3k
γ2k
1 +∞j=n+1
1
(a
1. . . a
j−2)
α1−α2γa
αj−11−1≤ qk
3k
2γk
1+∞
j=n+1
1 a
αj−11−1≤ qk
3k
γ2k
1 +∞j=n+1
[ k
α1α2
j−5
i=0(α1α2(α1+α2))i 2
k
j−4
i=0(α1α2(α1+α2))i 1
1 2
2α1(α1α2(α1+α2))j−4]
α1−1= qk
3k
γ2k
1 +∞j=n+1
[ k
α21
j−5
i=0(α1α2(α1+α2))i 2
k
α1j−4
i=0(α1α2(α1+α2))i 1
1 2
2α12(α1α2(α1+α2))j−4]
α1−1α1= qk
3k
γ2k
1 +∞j=n+1
(D
−1j−4)
α1−1α1.
Since the series
j
(D
j−1)
α1−1α1is convergent, we have lim
nA
n= 0, a contradic- tion follows and the irrationality of the sum is proved.
The last statement can be proved as in Proposition 1.
QEDLet us show some examples to which our result can be applied. If we choose α
1= 2 and α
2= 4, we can select 0 < γ ≤
12and 0 < k
2≤ k
12.
Theorem 2 ensures the following abstract scheme for the irrationality of the sum of a rational series.
3 Theorem. Consider a pair ((a
n), ψ), where (a
n) is a strictly increasing sequence of positive integers and ψ : [0, +∞[→]0, +∞[ is an increasing function.
Assume that
*
n j=1a
j≤ ψ(a
n+1). (7)
and take h : [1, + ∞[→ [0, +∞[ non increasing such that
+∞1
h(ξ)dξ < +∞. (8)
Then for every sequence (b
j) of positive integers for which:
b
ja
jψ(a
j) ≤ h(a
j), (9)
the series
j bj
aj
is convergent and the sum is an irrational number.
Proof. Since b
ja
j≤ b
ja
jψ(a
j) 1
ψ(a
j) ≤ h(a
j)
ψ(a
j) ≤ h(a
j) ψ(1) , for n > a
1we have
n j=1b
ja
j≤
n j=1h(a
j) ψ(1) ≤ 1
ψ(1)
n j=1h(a
j)(a
j+1− a
j) a
j+1− a
j≤ 1 ψ(1)
n j=1h(a
j)(a
j+1− a
j) ≤ 1 ψ(1)
n a1h(ξ)dξ.
Then the series
j bj
aj
is convergent. Arguing as in the proof of Theorem 2, we have:
1 ≤ A
n=q(
*
n j=1a
j)
+∞
j=n+1
b
ja
j≤ qψ(a
n+1)
+∞
j=n+1
b
ja
j≤ q
+∞j=n+1
b
jψ(a
j) a
j≤
+∞j=n+1
h(a
j) ≤
+∞j=n+1
h(a
j)(a
j+1− a
j) ≤ q
+∞an
h(ξ)dξ.
From (8), A
n→ 0 and we get a contradiction.
QEDReferences
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