andrica_piticari1. 87KB Jun 04 2011 12:08:21 AM
ACTA UNIVERSITATIS APULENSIS
No 10/2005
Proceedings of the International Conference on Theory and Application of
Mathematics and Informatics ICTAMI 2005 - Alba Iulia, Romania
ON A CLASS OF SEQUENCES DEFINED BY USING
RIEMANN INTEGRAL
Dorin Andrica and Mihari Piticari
Abstract.The main result shows that if f : [1, +∞) → R is a continuous
function such that lim xf (x) exists and it is finite, then
x→∞
lim n
n→∞
Z
a
1
f (xn )dx =
for any a > 1. Two applications are given.
Z
+∞
1
f (x)
dx,
x
2000 Mathematical Subject Classification. 26A42, 42A16.
1.Introduction
There are many important classes of sequences defined by using Riemann
integral. We mention here only one which is called the Riemann-Lebesgue
Lemma: Let f : [a, b] → R be a continuous function, where 0 ≤ a < b.
Suppose the function g : [0, ∞) → R to be continuous and T -periodic. Then
Z b
1ZT
f (x)dx.
(1)
g(x)dx
T 0
a
a
For the proof we refer to [4] (in special case a = 0, b = T ) and [5]. In the
paper [1] we proved that a similar relation as (1) holds for all continuous and
bounded functions g : [0, ∞) → R of finite Cesaro mean.
In this note we investigate another class of such sequences, i.e. defined by
Ra
n 1 f (xn )dx, where f : [1, +∞) → R is a continuous function and a > 1 is a
fixed real number.
lim
n→∞
Z
b
f (x)g(nx)dx =
2.The main result
Our main result is the following.
Theorem.Let f : [1, +∞) → R be a continuous function such that x→∞
lim xf (x)
exists and it is finite. Then, the improper integral
381
R∞
1
f (x)
dx
x
is convergent and
D. Andrica, M. Piticari - On a class of sequences defined by using ...
lim n
n→∞
for any a > 1.
Z
a
1
f (xn )dx =
Z
∞
1
f (x)
dx,
x
(2)
Proof. Consider lim xf (x) = l, where l ∈ R. Then, we can find a real
x→∞
number x0 > 1 such that for any x ≥ x0 we have
l−1
f (x)
l+1
≤
≤ 2 .
2
x
x
x
Let us choose a real number m > 0 satisfying the inequality l − 1 + m ≥ 0.
Then, for any x ≥ x0 we have
l−1+m
f (x) m
l+1+m
≤
+ 2 ≤
2
x
x
x
x2
Define the function J : [1, +∞) → R by
0≤
Z
J(t) =
(3)
!
f (x) m
+ 2 dx.
x
x
t
1
The function J is differentiable and we have
J ′ (t) =
f (t) m
+ 2 ≥0
t
t
for any t ≥ x0 . Therefore J is an increasing function on interval [x0 , +∞).
Moreover, by using the last inequality in (3) we get by integration
R x0 f (x)
R t f (x)
m
m
dx
+
dx
+
+
1
x0
x
x
x2
x2
R t dx
R x0 f (x)
m
+ x2 dx + (l + 1 + m) x0 x2
≤ 1
x
R
J(t) =
≤
f (x)
x
x0
1
+
m
x2
dx +
l+1+m
,
x0
for any t ≥ x0 . It follows that lim J(t) is finite. But, we have
t→∞
J(t) =
Z
1
t
1
f (x)
,
dx + m 1 −
x
t
382
D. Andrica, M. Piticari - On a class of sequences defined by using ...
hence
t
Z
lim
t→∞ 1
f (x)
dx = lim J(t) − m,
t→∞
x
which is finite.
For a fixed real number a > 1, denote
J(t) = t
Z
a
1
t
f (x )dxandU (t) =
Z
at
1
f (x)
dx.
x
Because function g : [1, +∞) → R, g(x) = xf (x), is continuous and
lim g(x) is finite, it follows that g is bounded, i.e. we can find M > 0 with
x→∞
the property
|g(x)| ≤ M, x ∈ [1, ∞)
(4)
Changing the variable x by x = ut , we get dx = tut−1 du, hence
U (t) = t
From (4) and (5) we obtain
Z
a
1
f (ut )
du
u
(5)
R
R a f (xt )
a
t
dx =
f
(x
)dx
−
1
1
x
R
R
t)
a
f
(x
a
dx =
= t 1 f (xt ) − x dx ≤ t 1 |f (xt )| x−1
R a x−1 x
Ra t
t x−1
= t 1h x |f (x )| xt+1 dx ≤ tM 1 xt+1i dx =
|J(t) − U (t)| = t
= Mt
Because
1
(a−t+1
1−t
(6)
− 1) − 1t (1 − a−t ) , t > 0
1
1
(a−t+1 − 1) − (1 − a−t ) = 0,
lim
t→∞ 1 − t
t
from (6) it follows that
lim J(t) = lim U (t),
t→∞
i.e. we have
lim
Z
t
t→∞ 1
and the desired result follows.
t→∞
Z a
f (x)
f (xt )dt
dx = lim
t→∞ 1
x
Remark. The relation (2) is a natural reformulation of the first part of
Problem 5.183 in [3] proposed by the second author and S. R˘adulescu.
383
D. Andrica, M. Piticari - On a class of sequences defined by using ...
3.Two applications
Application 1. Let us evaluate
a
dx
,
+k
1
where k > 0, a > 1 are fixed real numbers.
Using the result in Theorem for function f (x) =
lim n
n→∞
lim n
n→∞
Z
a
1
Z
(7)
xn
1
,
x+k
x ≥ 1, we obtain
Z ∞
dx
1
x ∞ 1
dx
= ln(k + 1).
=
=
ln
xn + k
x(x + k)
k x+k 1
k
1
Note that for k = 1 we get
lim n
n→∞
Z
a
dx
= ln 2,
+1
xn
1
i.e. the second part of Problem 5.183 in [3].
Application 2. Let us evaluate
xn−2
dx,
(8)
n→∞
0 x2n + xn + 1
which is a problem proposed by D. Popa to Mathematical Regional Contest
”Grigore Moisil”, 2002 (see [2] for details).
Fix a ∈ (0, 1) and we can write
lim n
n
Z
0
1
Z
1
Z a
Z 1
xn−2
xn−2
xn−2
dx
=
n
dx
+
n
dx.
x2n + xn + 1
0 x2n + xn + 1
a x2n + xn + 1
For the first term in the right side we have
0≤n
Z
0
a
Z a
nan−1
xn−2
n−2
x
dx
=
dx
≤
n
→ 0.
x2n + xn + 1
n−1
0
For the second term we obtain
n
Z
1
a
Z 1
Z 1/a
xn−2
xn dx
tn
dx = n
=n
dt.
x2n + xn + 1
t2n + tn + 1
a x2 (x2n + xn + 1)
1
384
D. Andrica, M. Piticari - On a class of sequences defined by using ...
The function f (t) =
Z
1
∞
t
t2 +t+1
satisfies lim tf (t) = 1 and we have
t→∞
Z ∞
t + 12 ∞
dt
π
2
f (t)
√
√
√
dt = lim
=
arctg
=
3 1
t→∞ 1
t
t2 + t + 1
3
3 3
2
Applying the result in Theorem it follows that
lim n
n→∞
Z
0
1
xn−2
π
dx = √ .
2n
n
x +x +1
3 3
References
[1] Andrica, D., Piticari, M., An extension of the Riemann-Lebesgue lemma
and some applications, Proc. International Conf. on Theory and Applications
of Mathematics and Informatics (ICTAMI 2004), Thessaloniki, Acta Universitatis Apulensis, No.8(2004), 26-39.
[2] Andrica, D., Berinde, V., a.o., Mathematical Regional Contest ”Grigore
Moisil” (Romanian), Cub Press 22, 2006.
[3] B˘atinet¸u-Giurgiu, D.M., a.o., Romanian National Mathematical Olympiads
for High Schools 1954-2003 (Romanian), Editura Enciclopedic˘a, Bucure¸sti,
2004.
[4] Dumitrel, F., Problems in Mathematical Analysis (Romanian), Editura
Scribul, 2002.
[5] Siret¸chi, Gh., Mathematical Analysis II. Advanced Problems in Differential and Integral Calculus, (Romanian), University of Bucharest, 1982.
Dorin Andrica
”Babe¸s-Bolyai” University
Faculty of Mathematics and Computer Science
Cluj-Napoca, Romania
E-mail address: dandrica@math.ubbcluj.ro
Mihai Piticari
”Drago¸s-Vod˘a” National College
Cˆampulung Moldovenesc, Romania
385
No 10/2005
Proceedings of the International Conference on Theory and Application of
Mathematics and Informatics ICTAMI 2005 - Alba Iulia, Romania
ON A CLASS OF SEQUENCES DEFINED BY USING
RIEMANN INTEGRAL
Dorin Andrica and Mihari Piticari
Abstract.The main result shows that if f : [1, +∞) → R is a continuous
function such that lim xf (x) exists and it is finite, then
x→∞
lim n
n→∞
Z
a
1
f (xn )dx =
for any a > 1. Two applications are given.
Z
+∞
1
f (x)
dx,
x
2000 Mathematical Subject Classification. 26A42, 42A16.
1.Introduction
There are many important classes of sequences defined by using Riemann
integral. We mention here only one which is called the Riemann-Lebesgue
Lemma: Let f : [a, b] → R be a continuous function, where 0 ≤ a < b.
Suppose the function g : [0, ∞) → R to be continuous and T -periodic. Then
Z b
1ZT
f (x)dx.
(1)
g(x)dx
T 0
a
a
For the proof we refer to [4] (in special case a = 0, b = T ) and [5]. In the
paper [1] we proved that a similar relation as (1) holds for all continuous and
bounded functions g : [0, ∞) → R of finite Cesaro mean.
In this note we investigate another class of such sequences, i.e. defined by
Ra
n 1 f (xn )dx, where f : [1, +∞) → R is a continuous function and a > 1 is a
fixed real number.
lim
n→∞
Z
b
f (x)g(nx)dx =
2.The main result
Our main result is the following.
Theorem.Let f : [1, +∞) → R be a continuous function such that x→∞
lim xf (x)
exists and it is finite. Then, the improper integral
381
R∞
1
f (x)
dx
x
is convergent and
D. Andrica, M. Piticari - On a class of sequences defined by using ...
lim n
n→∞
for any a > 1.
Z
a
1
f (xn )dx =
Z
∞
1
f (x)
dx,
x
(2)
Proof. Consider lim xf (x) = l, where l ∈ R. Then, we can find a real
x→∞
number x0 > 1 such that for any x ≥ x0 we have
l−1
f (x)
l+1
≤
≤ 2 .
2
x
x
x
Let us choose a real number m > 0 satisfying the inequality l − 1 + m ≥ 0.
Then, for any x ≥ x0 we have
l−1+m
f (x) m
l+1+m
≤
+ 2 ≤
2
x
x
x
x2
Define the function J : [1, +∞) → R by
0≤
Z
J(t) =
(3)
!
f (x) m
+ 2 dx.
x
x
t
1
The function J is differentiable and we have
J ′ (t) =
f (t) m
+ 2 ≥0
t
t
for any t ≥ x0 . Therefore J is an increasing function on interval [x0 , +∞).
Moreover, by using the last inequality in (3) we get by integration
R x0 f (x)
R t f (x)
m
m
dx
+
dx
+
+
1
x0
x
x
x2
x2
R t dx
R x0 f (x)
m
+ x2 dx + (l + 1 + m) x0 x2
≤ 1
x
R
J(t) =
≤
f (x)
x
x0
1
+
m
x2
dx +
l+1+m
,
x0
for any t ≥ x0 . It follows that lim J(t) is finite. But, we have
t→∞
J(t) =
Z
1
t
1
f (x)
,
dx + m 1 −
x
t
382
D. Andrica, M. Piticari - On a class of sequences defined by using ...
hence
t
Z
lim
t→∞ 1
f (x)
dx = lim J(t) − m,
t→∞
x
which is finite.
For a fixed real number a > 1, denote
J(t) = t
Z
a
1
t
f (x )dxandU (t) =
Z
at
1
f (x)
dx.
x
Because function g : [1, +∞) → R, g(x) = xf (x), is continuous and
lim g(x) is finite, it follows that g is bounded, i.e. we can find M > 0 with
x→∞
the property
|g(x)| ≤ M, x ∈ [1, ∞)
(4)
Changing the variable x by x = ut , we get dx = tut−1 du, hence
U (t) = t
From (4) and (5) we obtain
Z
a
1
f (ut )
du
u
(5)
R
R a f (xt )
a
t
dx =
f
(x
)dx
−
1
1
x
R
R
t)
a
f
(x
a
dx =
= t 1 f (xt ) − x dx ≤ t 1 |f (xt )| x−1
R a x−1 x
Ra t
t x−1
= t 1h x |f (x )| xt+1 dx ≤ tM 1 xt+1i dx =
|J(t) − U (t)| = t
= Mt
Because
1
(a−t+1
1−t
(6)
− 1) − 1t (1 − a−t ) , t > 0
1
1
(a−t+1 − 1) − (1 − a−t ) = 0,
lim
t→∞ 1 − t
t
from (6) it follows that
lim J(t) = lim U (t),
t→∞
i.e. we have
lim
Z
t
t→∞ 1
and the desired result follows.
t→∞
Z a
f (x)
f (xt )dt
dx = lim
t→∞ 1
x
Remark. The relation (2) is a natural reformulation of the first part of
Problem 5.183 in [3] proposed by the second author and S. R˘adulescu.
383
D. Andrica, M. Piticari - On a class of sequences defined by using ...
3.Two applications
Application 1. Let us evaluate
a
dx
,
+k
1
where k > 0, a > 1 are fixed real numbers.
Using the result in Theorem for function f (x) =
lim n
n→∞
lim n
n→∞
Z
a
1
Z
(7)
xn
1
,
x+k
x ≥ 1, we obtain
Z ∞
dx
1
x ∞ 1
dx
= ln(k + 1).
=
=
ln
xn + k
x(x + k)
k x+k 1
k
1
Note that for k = 1 we get
lim n
n→∞
Z
a
dx
= ln 2,
+1
xn
1
i.e. the second part of Problem 5.183 in [3].
Application 2. Let us evaluate
xn−2
dx,
(8)
n→∞
0 x2n + xn + 1
which is a problem proposed by D. Popa to Mathematical Regional Contest
”Grigore Moisil”, 2002 (see [2] for details).
Fix a ∈ (0, 1) and we can write
lim n
n
Z
0
1
Z
1
Z a
Z 1
xn−2
xn−2
xn−2
dx
=
n
dx
+
n
dx.
x2n + xn + 1
0 x2n + xn + 1
a x2n + xn + 1
For the first term in the right side we have
0≤n
Z
0
a
Z a
nan−1
xn−2
n−2
x
dx
=
dx
≤
n
→ 0.
x2n + xn + 1
n−1
0
For the second term we obtain
n
Z
1
a
Z 1
Z 1/a
xn−2
xn dx
tn
dx = n
=n
dt.
x2n + xn + 1
t2n + tn + 1
a x2 (x2n + xn + 1)
1
384
D. Andrica, M. Piticari - On a class of sequences defined by using ...
The function f (t) =
Z
1
∞
t
t2 +t+1
satisfies lim tf (t) = 1 and we have
t→∞
Z ∞
t + 12 ∞
dt
π
2
f (t)
√
√
√
dt = lim
=
arctg
=
3 1
t→∞ 1
t
t2 + t + 1
3
3 3
2
Applying the result in Theorem it follows that
lim n
n→∞
Z
0
1
xn−2
π
dx = √ .
2n
n
x +x +1
3 3
References
[1] Andrica, D., Piticari, M., An extension of the Riemann-Lebesgue lemma
and some applications, Proc. International Conf. on Theory and Applications
of Mathematics and Informatics (ICTAMI 2004), Thessaloniki, Acta Universitatis Apulensis, No.8(2004), 26-39.
[2] Andrica, D., Berinde, V., a.o., Mathematical Regional Contest ”Grigore
Moisil” (Romanian), Cub Press 22, 2006.
[3] B˘atinet¸u-Giurgiu, D.M., a.o., Romanian National Mathematical Olympiads
for High Schools 1954-2003 (Romanian), Editura Enciclopedic˘a, Bucure¸sti,
2004.
[4] Dumitrel, F., Problems in Mathematical Analysis (Romanian), Editura
Scribul, 2002.
[5] Siret¸chi, Gh., Mathematical Analysis II. Advanced Problems in Differential and Integral Calculus, (Romanian), University of Bucharest, 1982.
Dorin Andrica
”Babe¸s-Bolyai” University
Faculty of Mathematics and Computer Science
Cluj-Napoca, Romania
E-mail address: dandrica@math.ubbcluj.ro
Mihai Piticari
”Drago¸s-Vod˘a” National College
Cˆampulung Moldovenesc, Romania
385