Bulletin of Mathematical Analysis and Applications
ISSN: 1821-1291, URL: http://www.bmathaa.org
Volume 3 Issue 3(2011), Pages 25-34.
AN EXTENSION OF A RESULT ABOUT THE ORDER OF
CONVERGENCE
(COMMUNICATED BY HAJRUDIN FEJZIC)
DAN ŞTEFAN MARINESCU, MIHAI MONEA
Abstract. In this note, we analize the context in which the calculation of
limits of sequences is performed using the integration of product of Riemann
integrable functions. We will also study the degree of convergence of such
sequences.
1. INTRODUCTION
The integration is one of the most useful concept in introductory calculus. Its
development was started with Newton which considered this process like a reverse
of differentiation. Later, Leibniz and Riemann have seen the integration as ”summation of many quantities”. This approach led to the current definition of Riemann
integral (see e.g. [5], pag. 216).
To evaluate Riemann integral without derivative is not easy. Therefore, in this
article, we will analize the Riemann integral as a limit. First, we recall a classic
result(see e.g. [8], pag. 49).
Lemma 1.1. Let f : [a, b] → R be Riemann integrable. Then
b−a ∑
lim
f
n→∞
n
n
k=1
(
k
a + (b − a)
n
)
∫b
=
f (x)dx.
a
In [8], p. 49, Polya and Szego have formulated the problem of ”the degree of
approximation” (see problem 10) and gave an answer (p. 232) which is presented
in the next lemma.
2000 Mathematics Subject Classification. 26A24, 26A42.
Key words and phrases. Riemann integrable, limit of sequence, degree of convergence.
c
⃝2008
Universiteti i Prishtinës, Prishtinë, Kosovë.
Submitted April 2, 2011. Accepted May 25, 2011.
25
26
D.Ş. MARINESCU, M. MONEA
Lemma 1.2. Let f : [a, b] → R be continuously differentiable. Then


(
) ∫b
n
∑
b−a
k
f (b) − f (a)
lim n 
.
f a + (b − a) − f (x) dx = (b − a)
n→∞
n
n
2
k=1
a
In this paper we want to extend these results for product of two functions. In
order to do this, we recall the next results about the product of two integrable
functions (see e.g. [9], theorem 7.9).
Theorem 1.3. If f, g : [a, b] → R are two Riemann integrable functions, then f g
is Riemann integrable.
Since the transformations x → x−a
b−a change the interval [a, b] into [0, 1], we will
present all our results for the functions defined on the interval [0, 1]. In this context,
by using theorem 1.3, lemmas 1.1 and 1.2 admit the following extensions:
Lemma 1.4. If f, g : [0, 1] → R are two Riemann integrable functions, then
( ) ( ) ∫1
n
k
k
1∑
f
g
= f (x) g (x)dx.
lim
n→∞ n
n
n
k=1
0
Lemma 1.5. If f, g : [0, 1] → R are two continuously differentiable functions, then


( ) ( ) ∫1
n
∑
k
k
f (1) g (1) − f (0) g (0)
1
f
g
− f (x) g (x) dx =
.
lim n 
n→∞
n
n
n
2
k=1
0
We will generalize these lemmas in the next section. The main result is the
proposition 3.1, which will be presented in section 3. This result will give more
information about the convergence’s order of a sequence, more general then the
one from lemma 1.5.
2. SEQUENCES WHOSE LIMITS ARE CALCULATED USING
RIEMANN INTEGRALS
First, we formulate the next lemma to establish a general framework for the
main results of this note.
Lemma 2.1. Let f, g : [0, 1] → R be two Riemann
[
]integrable functions. For any
k
,
n ∈ N∗ and k ∈ {1, 2, ..., n}, we denote Ik = k−1
n
n . We define Ak = sup f (x),
x∈Ik
ak = inf f (x), Bk = sup g (x) and bk = inf g (x) . We consider the sequences
x∈Ik
x∈Ik
x∈Ik
1∑
ak bk
n
n
xn =
k=1
AN EXTENSION OF A RESULT ABOUT THE ORDER OF CONVERGENCE
and
27
1∑
Ak Bk .
n
n
yn =
k=1
Then:
a) lim (yn − xn ) = 0;
n→∞
b) lim xn = lim yn =
n→∞
n→∞
∫1
f (x) g (x)dx.
0
Proof. As the functions are Riemann integrable, we obtain from Darboux’s criten
n
∑
∑
(Ak − ak ) = 0 and lim n1
(Bk − bk ) = 0.
rion, that lim n1
n→∞
n→∞
k=1
k=1
a) First, we assume that the functions f and g are nonnegative. Then
1∑
(Ak Bk − ak bk )
n
n
0 6 yn − xn =
k=1
=
1
n
n
∑
k=1
1
n
6 sup g (x) ·
x∈[0,1]
1∑
(Bk − bk ) ak
n
n
(Ak − ak ) Bk +
n
∑
k=1
1∑
(Bk − bk ).
n
n
(Ak − ak ) + sup f (x) ·
x∈[0,1]
k=1
k=1
But f, g are integrable, so they are bounded. Then, we obtain
)
(
n
n
1∑
1∑
sup g (x) ·
lim
(Ak − ak ) + sup f (x) ·
(Bk − bk ) = 0
n→∞ x∈[0,1]
n
n
x∈[0,1]
k=1
k=1
and lim (yn − xn ) = 0.
n→∞
In the absence of the nonnegative assumption, we note a =
inf
f (x) and,
x∈[0, 1]
respectively b =
inf
x∈[0, 1]
g (x). Consider the functions F : [0, 1] → R, F (x) =
f (x) − a and, G : [0, 1] → R, G (x) = g (x) − b, which are nonnegative functions.
Then :
n
1∑
lim (yn − xn ) = lim
(Ak Bk − ak bk )
n→∞
n→∞ n
k=1
( n
)
n
n
1∑
1∑
1∑
= lim
((Ak − a) (Bk − b) − (ak − a) (bk − b)) +
a (Bk − bk ) +
b (Ak − ak ) = 0,
n→∞
n
n
n
k=1
because
k=1
k=1
1∑
((Ak − a) (Bk − b) − (ak − a) (bk − b)) = 0,
n→∞ n
n
lim
k=1
in accordance with previous calculations for the functions F , respectively G.
b) If f, g are nonnegative, then for any interval I ⊂ [0, 1], the following inequalities are satisfied:
sup (f (x) g (x)) 6 sup f (x) · sup g (x)
x∈I
x∈I
x∈I
and
inf f (x) g (x) > inf f (x) · inf g (x) .
x∈I
x∈I
x∈I
28
D.Ş. MARINESCU, M. MONEA
This implies that
∫1
xn 6
f (x)g (x) dx 6 yn ,
0
for all n ∈ N∗ , as any Riemann sum attached to the function f g lies between xn
and yn . So, we obtain
∫1
0 6 yn −
f (x)g (x) dx 6 yn − xn
0
which means that
∫1
lim yn =
f (x) g (x)dx.
n→∞
0
Similarly
∫1
06
f (x)g (x) dx − xn 6 yn − xn
0
and we obtain
∫1
lim xn =
f (x) g (x)dx.
n→∞
0
If f, g are not necessarily nonnegative, we obtain the same result by applying
the same reasoning for functions F and G, defined at the previous point.
The result from 2.1 help us to prove the next lemma. We should mention that
this result was presented by Bényi and Niţu in [3] but without a solution.
Lemma 2.2. Let f, g : [0, 1] → R be two
integrable functions. For any
[ Riemann
]
k
k ∈ {1, 2, 3, ..., n} and for any αk , βk ∈ k−1
,
,
the
sequence
n
n
1∑
f (αk ) g (βk )
n
n
zn =
k=1
converges to
∫1
f (x)g (x) dx.
0
Proof. We have two cases. First, we assume that the functions f and g are nonnegative. We choose the sequences xn and yn from lemma 2.1. Obviously we have
xn 6 zn 6 yn
and the conclusion follows.
In the absence of the nonnegative assumption, we note a =
inf
f (x) and,
x∈[0, 1]
respectively b =
inf
x∈[0, 1]
g (x). Similar with the proof of lemma 2.1., we consider the
AN EXTENSION OF A RESULT ABOUT THE ORDER OF CONVERGENCE
29
functions F : [0, 1] → R, F (x) = f (x) − a and G : [0, 1] → R, G (x) = g (x) − b,
which are nonnegative functions. Then
1∑
1∑
f (αk ) g (βk ) = lim
(F (αk ) + a) (G (βk ) + b)
n→∞ n
n→∞ n
n
n
k=1
k=1
lim
(
= lim
n→∞
1∑
b∑
a∑
F (αk ) G (βk ) +
F (αk ) +
G (βk ) + ab
n
n
n
n
n
k=1
k=1
∫1
=
n
k=1
∫1
F (x) G (x) dx + b
0
∫1
F (x) dx + a
0
∫1
=
)
G (x) dx + ab
0
∫1
(F (x) + a) (G (x) + b) dx =
0
f (x) g (x) dx
0
which conclude the proof.
A similar result could be found in [10], where Spivak has posed the same problem
(see problem 1, p.263) but for continuous functions . Evidently, our result is more
general.
A consequence of lemma 2.2 is the following result from [3].
Corollary 2.3.(Bényi, Niţu) Let f : [0, 1] → R be a Riemann integrable functions.Then
(
) ( ) ∫1
n
1∑
k−1
k
lim
f
f
= f 2 (x) dx.
n→∞ n
n
n
k=1
0
Proof. We apply 2.2 for g = f and the sequences αn =
k−1
n
and βn =
k
n.
3. THE ORDER OF CONVERGENCE
Lemma 2.2 gives us an answer about the limits of sequences of the type
1∑
f (ak ) g (bk ),
n
n
k=1
but we want to know more details about these limits. We would like to study
their order of convergence and we can deliver a good estimation in the case of the
sequence
) (
)
(
n
k−β
k−α
1∑
g
,
f
n
n
n
k=1
with α, β ∈ [0, 1]. The results are included in next proposition.
30
D.Ş. MARINESCU, M. MONEA
Proposition 3.1. Let f, g : [0, 1] → R be two continuously differentiable functions.
Let define α, β ∈ [0, 1] and the sequence
(
) (
)
∫1
n
1∑
k−α
k−β
an = f (x)g (x) dx −
f
g
,
n
n
n
k=1
0
∗
for all n ∈ N . Then:
1
lim nan = (f (0) g (0) − f (1) g (1)) + α
n→∞
2
∫1
∫1
′
f (x) g (x) dx + β
0
f (x) g ′ (x) dx.
0
Proof. After some algebra we have
(
) (
)
∫1
n
∑
k−α
k−β
nan = n f (x)g (x) dx −
f
g
n
n

= n
k=1
0
∫1
0

( )( ( )
))
( ) ( )
(
n
n
1∑
k
k  ∑
k
k
k−β
f (x) g (x) dx −
+
f
f
g
g
−g
n
n
n
n
n
n
k=1
k=1
(
)( ( )
(
))
n
∑
k−β
k
k−α
+
g
f
−f
.
n
n
n
k=1
But
 1

( ) ( )
∫
n
∑
1
k
k
 = f (0) g (0) − f (1) g (1) .
lim n  f (x)g (x) dx −
f
g
n→∞
n
n
n
2
k=1
0
[
]
k
However, for any k ∈ {1, 2, 3.., n} and any interval k−β
,
, we apply mean value
( k−1 k ) n n
theorem for the function g and we find ck ∈ n , n which verifies the relation
( )
(
)
k
k−β
β
g
−g
= g ′ (ck ) .
n
n
n
So
( )( ( )
( )
(
))
n
n
∑
k
k
k−β
β∑
k
f
g
−g
=
f
g ′ (ck ).
n
n
n
n
n
k=1
k=1
From 2.2, we obtain
lim
n→∞
(
( )
n
β∑
k
f
g ′ (ck )
n
n
k=1
)
∫1
=β
f (x)g ′ (x) dx.
0
]
Similarly, for the function f on the interval n , nk , applying mean value the( k−1 k )
orem, we find bk ∈ n , n so that
(
)
( )
k−α
α
k
−f
= f ′ (bk ) .
f
n
n
n
Then
(
)( ( )
(
))
)
(
n
n
∑
k−β
k
k−α
α∑ ′
k−β
g
f
−f
=
.
f (bk ) g
n
n
n
n
n
k=1
[ k−α
k=1
AN EXTENSION OF A RESULT ABOUT THE ORDER OF CONVERGENCE
31
Applying Lemma 2.2. we find that
(
(
))
∫1
n
α∑ ′
k−β
lim
f (bk ) g
= α f ′ (x) g (x) dx.
n→∞
n
n
k=1
0
Finally, we obtain
1
lim nan = (f (0) g (0) − f (1) g (1)) + α
n→∞
2
∫1
∫1
′
f (x) g (x) dx + β
0
f (x) g ′ (x) dx.
0
If we choose the particular values for the numbers α and β from proposition 3.1,
we obtain some results as:
For α = 0 and β = 1 , we obtain
 1

( ) (
)
∫
n
∑
1
k
k
−
1

lim n  f (x)g (x) dx −
f
g
n→∞
n
n
n
k=1
0
=
1
(f (0) g (0) − f (1) g (1)) +
2
∫1
f (x) g ′ (x) dx;
0
For α =
1
2
and β = 1, we obtain
 1

(
) (
)
∫
n
∑
2k
−
1
1
k
−
1

f
lim n  f (x)g (x) dx −
g
n→∞
n
2n
n
k=1
0
1
1
= (f (0)g(0) − f (1)g(1)) +
2
2
∫1
∫1
′
f (x) g (x) dx +
0
f (x) g ′ (x) dx.
0
Using the formula of integration by parts, we obtain

 1
(
) (
)
∫1
∫
n
∑
2k
−
1
1
1
k
−
1


f
=
lim n
f (x)g (x) dx −
g
f (x) g ′ (x) dx.
n→∞
n
2n
n
2
k=1
0
0
If we choose g(x) = 1 we obtain the next result, which represent the theorem 1
from [4].
Corollary 3.2. (Chiţescu) Let f : [0, 1] → R be a continuously differentiable
functions. Let define α ∈ [0, 1] and the sequence
)
(
∫1
n
1∑
k−α
,
an = f (x)dx −
f
n
n
0
for all n ∈ N∗ . Then:
k=1
)
(
1
[f (1) − f (0)] .
lim nan = α −
n→∞
2
32
D.Ş. MARINESCU, M. MONEA
Proof. We apply 3.1 for g(x) = 1.
4. APPLICATIONS
At the end of this note, we present three applications whose solutions are obtained using the results from the previous paragraph. The first problem reads
Problem 1. Let the sequence
n
∑
k2 − k
an =
,
n3 + kn2
k=1
∗
for any n ∈ N . Show that lim an = ln 2 −
n→∞
1
2
(
and evaluate lim n ln 2 −
n→∞
1
2
)
− an .
Solutions: By using Lema 2.2 for the functions f, g : [0, 1] → R, f (x) =
g (x) = x, we have
k
1 k (k − 1)
1
k2 − k
n
=
·
=
·
n3 + kn2
n n (n + k)
n 1+
so
1 ∑ nk
lim an = lim
n→∞
n→∞ n
1+
n
k=1
But
∫1
x2
dx =
1+x
∫1 (
x−1+
0
1
1+x
)
k
n

= lim n 
∫1
n→∞
·
∫1
x·
k−1
,
n
x
dx,
1+x
0
x2
dx =
2
0
From Prop. 3.1, we obtain
k−1
·
=
n
k
n
x
1+x ,
1
1
1
1
− x + ln (1 + x) = ln 2 − .
0
0
0
2
(
)
1
lim n ln 2 − − an
n→∞
2
1∑
f (x)g (x) dx −
f
n
n
k=1
0
1
= (f (0) g (0) − f (1) g (1)) +
2

( ) (
)
k
k−1 
g
n
n
∫1
f (x) g ′ (x) dx
0
∫1
1
=− +
4
∫1
x
1
dx = − +
1+x
4
0
0
(
1−
1
1+x
)
1
1
1
3
dx = − +x −ln (x + 1) = −ln 2.
0
0
4
4
√
[x]
Problem 2. Let n ∈ N, n > 2 and the function f : [0, n] → R, f (x) = xe− n .
We note Vn , the volume of the solid obtained by rotating the graph of f about x
axis. Find lim Vnn2 .
n→∞
AN EXTENSION OF A RESULT ABOUT THE ORDER OF CONVERGENCE
Solution: As Vn = π
∫n
∫n
f 2 (x) dx = π
0
xe−
2[x]
n
dx, we evaluate first
0
∫n
xe−
33
2[x]
n
dx.
0
We have
∫n
−
xe
2[x]
n
dx =
n ∫k
∑
−
xe
2[x]
n
dx =
k=1k−1
0
n
∑
−
e
2(k−1)
n
k=1
∫k
xdx =
n
∑
2k − 1
2
k=1
k−1
e−
2(k−1)
n
.
Then
Vn
1 ∑ 2k − 1 − 2(k−1)
n
lim 2 = π lim
e
=
n→∞ n
n→∞ n
2n
n
k=1
∫1
xe−2x dx.
0
But
∫1
−2x
xe
xe−2x
dx = −
2
0
∫1
1
1
xe−2x
e−2x dx = −
+
0
2
2
0
1
e−2x
−
0
4
1
1 − 3e−2
,
=
0
4
so
Vn
1 − 3e−2
=
π
.
n→∞ n2
4
lim
Problem 3. (Pr. 11535 from AMM 9/2010) Let f be a continuously differentiable
∫1
function on [0, 1]. Let A = f (1) and let B = x−1/2 f (x) dx. Evaluate

lim n 
0
∫1
f (x) dx −
n→∞
n
∑
(
k=1
0
(k − 1)
k2
−
n2
n2
2
) (
f
(k − 1)
n2
2
)

in terms of A and B.
Solution: We perform the change of variables x = y 2 , so we have
∫1
0
∫1
0
f (x) dx =
( )
( )
∫1
∫1 ( )
2yf y 2 dt and x−1/2 f (x) dx = 2f y 2 dt. Using the notation g (y) = 2f y 2 ,
0
0
the hypothesis becomes g (1) = 2A and
∫1
g (y) dy = B.
0
Then, we compute
 1
(
) (
)
∫
n
2
2
2
∑
k
(k − 1)
(k − 1) 
lim n  f (x) dx −
−
f
n→∞
n2
n2
n2
k=1
0

= lim n 
∫1
n→∞

(
)
n
∑
1
k−1 
2k − 1
yg (y) dy −
g
n
2n
n
k=1
0
1
=
2
∫1
0
yg ′ (y) dy
34
D.Ş. MARINESCU, M. MONEA
from proposition 3.1. But
∫1
∫1
1 ∫1
′
yg (y) dy = yg (y) 0 − g (y) dy = g (1) − g (y) dy = 2A − B.
0
0
0
Finally we obtain that the limit value is A −
B
2.
References
[1] V. Arsinte, Probleme de Calcul Integral, Ed. Univ. Bucureşti, 1995.
[2] A. Bényi, M. Bobeş, Asupra unei probleme de olimpiadă, Gazeta Matematică, CXIII(2008),
no.10. pag. 476 - 481.
[3] A. Bényi, C. Niţu, Asupra unei probleme deschise, Gazeta Matematică, CXV(2010), no.7-8-9,
pag. 338 - 340.
[4] M. Chiriţă, I. Chiţescu, A. Constantinescu, Aplicaţii ale integralei Riemann, Gazeta Matematică, XCII(1988), no.4. pag.145 - 152.
[5] J. B. Dence, Th. P. Dence, Advanced Calculus, Elsevier, London, 2010,
[6] D. Ş. Marinescu, V. Cornea, Asupra unor probleme cu şiruri, Gazeta Matematică, CV(2000),
no.5-6. pag. 202 - 212.
[7] W.J. Kaczor, M.T. Nowak, Problems in Mathematical Analysis III, AMS, 2003.
[8] G. Pólya, G. Szegö, Problems and Theorems in Analysis I, Springer, Berlin, 1998.
[9] J.A. Fridy, Introductory Analysis, Academic Press, 2000.
[10] M. Spivak, Calculus, Cambridge Universitary Press, 1994.
[11] *** - American Mathematical Monthly, 117(2010), no. 9, problem 11535.
Dan Ştefan Marinescu
National College ”Iancu de Hunedoara”,
Romania, Hunedoara, 331078,
Victoriei no.12
E-mail address: marinescuds@yahoo.com
Mihai Monea
National College ”Decebal”,
Romania, Deva, 330018,
1 Decembrie no. 22
E-mail address: mihaimonea@yahoo.com