General Mathematics Vol. 12, No. 3 (2004), 19–30
Section of Euler summability method
Ioan T
. incu
Abstract
In this paper, we determine sections of Euler summability method,
using random variables which follow Bernoulli0 s law and the central
limit theorem.
2000 Mathematical Subject Classification: 65B15, 65C20
Keywords: Euler summability method
1
Introduction
The paper aims to obtain some regular transformation of the sequences of
real numbers, see (12), (13), (19), (22), (24), (28), (30), (32), which are
sections of Euler, s summability method.
Let A = kan,k kn,k∈N be a real elements matrix.
A sequence (sn )n∈N is said to be A - summable to the value s ∈ R if
∞
X
each of the series σn =
an,k · sk , n = 0, 1, ... is convergent and if σn → s
for n → ∞.
k=0
19
20
Ioan T
. incu
The method A is called regular if each convergent sequence is A -
summable to its limit.
Theorem 1. (Toeplitz) (see [2]) The summation method A is regular if
and only if
(1)
lim an,k = 0 , for every k natural,
n→∞
(2)
lim
n→∞
∞
X
(3)
∞
X
an,k = 1,
k=0
|an,k | ≤ M , for every n natural,
k=0
M being a constant independent of n.
Remark 1.1. If the elements of matrix A are positive and
every n natural, then conditions (2), (3) are verified.
∞
X
an,k = 1, for
k=0
Euler summability method is obtained for the matrix A = kan,k k n ∈ N ,
k=0,n
where
an,k
 
n
=   αk (1 − α)n−k ,
k
α ∈ (0, 1).
In the followings we will use the next notations:
M (f ) represents the expectation of a random variable f
D2 (f ) represents the variance (dispersion) of a random variable f .
[x] represents the whole part of x
We remind that the whole part verifies:
i) [m + x] = m + [x], for m ∈ Z, x ∈ R
ii) [−x] = −1 − [x], for x > 0.
Now, we recall the well-known Lyapunov central limit theorem and a
result on the distribution functions of a random variables.
Section of Euler summability method
21
Theorem 2. (Lyapunov) (see [1], [3]) Let (fn )n∈N a sequence of independent random variables. Let us suppose that Mk = M (fk ), Dk2 = D2 (fk ),
p
Hk = 3 M (|fk − Mk |3 ) exists for every k natural.
Note with
q
q
Sn =
D12
+ ... +
Dn2 , Kn
3
=
H13 + ... + Hn3 ,
βn − M (βn )
D(βn )
and with Fn,βn∗ (x) the distribution function of variable βn∗ . Thus, if
βn = f1 + ... + fn , βn∗ =
Kn
=0
n→∞ Sn
lim
(4)
we have
1
lim Fn,βn∗ (x) = √ ·
n→∞
2π
(5)
1
Function Φ(x) = √
2π
bution function.
Zx
e−t
2 /2
dt , for every x real.
−∞
Zx
e−t
2 /2
dt represent the standard normal distri-
−∞
Theorem 2 is also true in the case when the independent random variables have the same distribution.
Theorem 3. If the random variables η1 and η2 verify the condition
η2 = aη1 + b with a, b real, then fη2 (t) = fη1 (at)eibt where fη2 (t) and fη1 (t)
denote the characteristic functions of the variables η2 and η1 .
If the random variables η1 and η2 verify the condition η2 = a · η1 + b with
a, b real, a > 0, then the distribution functions of these random variables
verify
(6)
µ
Fη2 (x) = Fη1
x−b
a
¶
.
22
2
Ioan T
. incu
Principal results

If the independent random variables f1 , ..., fn have the distribution 

1 0
,
p q
q = 1 − p, p ∈
(0, 1), then therandom variable βn = f1 + ... + fn has the
k
 



distribution  n
and the distribution function

   pk · q n−k 
k
k=0,n

 
n
 0, x<0
X n
  pk · q n−k θ(t − k), where θ(x) =
Fn,βn (t) =
represents
 1, x≥0
k
k=0
Heaviside0 s function.
From Theorem 2 and (6), we get
(7)
 
n
  pk q n−k · θ(np + t√npq − k) =
lim Fn,βn∗ (t) = lim
n→∞
n→∞
k
k=0
n
X
= Φ(t) , for all t ≥ 0
or
(8)

√
[np+t npq]
lim
n→∞
X
k=0

n
  pk · q n−k = Φ(t) ,
k
for all t ≥ 0.
Next, (8) is generalized.
 independent random variables f1 , ..., fn with the distributions
 Let the
n
X
1 0
, qk = 1 − pk , pk ∈ (0, 1), k = 1, n, and lim

pk = ∞.
n→∞
pk qk
k=1


k
The random variable βn = f1 +...+fn has the distribution  Pn (0) 
,
k!
k=0,n
where Pn (x) = (p1 x + q1 )...(pn x + qn ).
Section of Euler summability method
23
From Theorem 2 and (6) we get
lim Fn,βn∗ (t) =
(9)
n→∞
= lim
n→∞
n
X
(k)
Pn (0)
k!
k=0
or
"s
t
n
P
i=1
(10)
lim
n→∞
v

u n
n
X
uX
pi qi +
pi − k  = Φ(t) , for all t ≥ 0
θ tt
i=1
pi qi +
X
n
P
i=1
i=1
#
pi
k=0
1 (k)
P (0) = Φ(t) for all t real positive.
k! n
Remark 2.1. As the function Φ(t) is bounded for x ≥ 0, we can consider
Fn,βn∗ (t)
= 1 , for all t real positive.
n→∞ Φ(t)
(11)
lim
Using Theorem 1, (8) and (10) we can build the following regular summation methods of the sequence of real numbers (sn )n∈N :
√
[np+t npq]
X
Tn(1) (t) =
(12)
(1)
cn,k (t) · sk ,
k=0
 
· r ¸
1 n k
nq
(1)
n−k
p · (1 − p) , p ∈ (0, 1), t ∈ 0,
where cn,k (t) =
,
Φ(t) k
p
"s
t
n
P
i=1
Tn(2) (t)
(13)
=
pi qi +
X
n
P
i=1
#
pi
(2)
cn,k (t) · sk ,
k=0
n
Y
1
P (k) (0)
where
=
·
, P (x) =
(pi x + qi ), pi ∈ (0, 1), qi = 1 − pi
Φ(t)
k!
i=1


n
X
 n−
pi 


n
X


i=1
, lim
v
0,
pk = ∞.
for i = 1, n and t ∈ 
 u n
 n→∞
 uX

k=1
 t
pi qi 
(2)
cn,k (t)
i=1
24
Ioan T
. incu
Next, we will study the case t = 0; transformations from (12) and (13)
become:
(14)
 
n
  pk · (1 − p)n−k sk ,
Tn(1) (0) = 2 ·
k
k=0
[np]
X
n
P
pi
n
X P (k) (0)
Y
=2·
· sk , P (x) =
(pi x + qi ) ,
k!
i=1
k=0
n
X
pi ∈ (0, 1) , qi = 1 − pi for i = 1, n and lim
pk = ∞.
i=1
(15)
p ∈ (0, 1),
Tn(2) (0)
n→∞
k=1
From (7), for t = 0, we have

 
n

X n



  pk q n−k · θ(np − k) = 1
lim


 n→∞
2
k
k=0


(16)
n

X

n


  pn−k · q k · θ(np − n + k) = 1 ,
lim


 n→∞
2
k
k=0
or
(17)

 
[np]

X

n


  pk q n−k = 1
lim


n→∞

2
k

k=0






n

X
n
  pn−k · q k = 1 , nq 6∈ Z , q ∈ (0, 1) .
lim
n→∞

2
k

k=[nq]+1


 


n

X

n


  pn−k · q k = 1 , nq ∈ Z , q ∈ (0, 1)
lim

 n→∞

2
k
k=nq
We consider the following cases:
Case 1. [np] < [nq] where nq 6∈ Z, q ∈ (0, 1), p = 1 − q. From the
properties i) and ii) we get
(18)
p<
n−1
.
2n
Section of Euler summability method
25
Using Theorem 1, (17) and (18), we get the following regular transformation of the sequence of real numbers (sn )n∈N :
[nαn ]
(19)
Tn(3) (αn )
=
X
(3)
cn,k (α)
n
X
· sk +
k=0
(3)
cn,k (α) · sk where
k=[n(1−αn )]+1
  


n


  αnk (1 − αn )n−k , k




k



(3)
0
, k
cn,k (α) =








n

  αnn−k (1 − αn )k , k



 k
µ
¶
n−1
αn ∈ 0,
and n(1 − αn ) 6∈ Z.
2n
1
Case 2. [np] = [nq], p 6= , nq 6∈ Z,
2
properties i) and ii), we get
[np] =
(20)
= 0, [nαn ]
= [nαn ] + 1, [n(1 − αn )]
= [n(1 − αn )] + 1, n ,
q ∈ (0, 1), p = 1 − q. From
n−1
.
2
Equality from (20) is valid if n is odd number, i.e. n = 2m + 1, m ∈ N.
m
m+1
[(2m + 1)p] = m, m ≤ (2m + 1)p < m + 1,
≤p<
.
2m + 1
2m + 1
1
Remark 2.2. For m → ∞ results p = .
2
Case 3. [np] = nq − 1, nq ∈ Z, q ∈ (0, 1), p = 1 − q.
From the fact that nq ∈ Z, it follows that np ∈ Z.
We have: np = nq − 1, np = n − np − 1, 2np = n − 1. Equality is valid
if n is an odd number, that is n = 2m + 1, m ∈ N.
Consequently
(21)
p=
m
,
2m + 1
q=
m+1
.
2m + 1
26
Ioan T
. incu
Using Theorem 1, (17) and (21), we get the following regular transfor-
mation of the sequence of real numbers (sn )n∈N :
(4)
T2m+1
(22)



m

X
2m + 1
1

 mk · (m + 1)2m+1−k · sk +
=
·
2m+1

(2m + 1)
k
k=0



2m+1

X
2m + 1
2m+1−k
k


+
m
· (m + 1) · sk .

k
k=m+1
Case 4. [np] < nq − 1, nq ∈ Z, q ∈ (0, 1), p = 1 − q. From the fact that
nq ∈ Z it follows that np ∈ Z.
We obtain: np < n − np − 1,
p<
(23)
n−1
.
2n
By using Theorem 1, (16) and (22), we get the following regular transformation of the sequence of real numbers (sn )n∈N :
Tn(5) (αn ) =
(24)
nαn
X
(5)
cn,k (α) · sk +
k=0
n
X
(5)
cn,k (α) · sk , where
k=n(1−αn )
  

n


  αnk (1 − αn )n−k , k = 0, nαn




k




(5)
0
, k = nαn + 1, n(1 − αn ) − 1
cn,k (α) =


 




n


  αnn−k (1 − αn )k , k = n(1 − αn ), n,


 k
µ
αn ∈
n−1
0,
2n
¶
and n · αn ∈ Z.
Section of Euler summability method
27
From (9), for t = 0, we have

à n
!
n
(k)
X
X

1
P
n (0)


pi − k =
θ
lim


 n→∞
k!
2
i=1
k=0
(25)
à n
!
n
(n−k)

X
X

P
(0)
1
n


lim
θ
pi − n + k =

 n→∞
(n − k)!
2
i=1
k=0
where Pn (x) =
n
Y
(pi x + qi ), pi ∈ (0, 1), qi = 1 − pi for i = 1, n or
i=1
(26)

n
P

p

i

i=1

X Pn(k) (0)

1


lim
=


n→∞

k!
2

k=0



"
#

n
n
(n−k)

X
X

(0)
Pn
1
lim
= , n−
pi 6∈ Z
n→∞
(n − k)!
2

n
i=1
P


k= n−
pi +1


i=1


n
n

X Pn(n−k) (0)
X

1


lim
= , n−
pi ∈ Z


n→∞

(n − k)!
2

n
P
i=1


k=n−
pi
i=1
We consider
the following
cases:
" n #
"
#
n
n
X
X
X
Case 5.
pi < n −
pi where n −
pi 6∈ Z, pi ∈ (0, 1), i = 1, n.
i=1
i=1
i=1
From properties i) and ii), we get
n
X
(27)
i=1
pi <
n−1
.
2
By using Theorem 1, and (25) and (26), we get the following regular
transformation of the sequence of real numbers (sn )n∈N :
n
P
X
i=1
(28)
Tn(6) (p) =
pi
k=0
cn,k (p)sk +
n
X
n
P
k= n−
pi +1
i=1
cn,k (p)sk
28
Ioan T
. incu














(k)
Pn (0)
k!
" n #
X
, k = 0,
pi
i=1
" n #
"
#
n
X
X
, k=
pi + 1 , n −
pi ,
where cn,k (p) =
0



i=1



"
#

n
(n−k)

X

P
(0)
n


pi + 1, n,

 (n − k)! , k = n −
i=1
n
n
X
X
n−1
0<
pi <
, n−
pi 6∈ Z, pi ∈ (0, 1), i = 1, n.
2
i=1
"
#
" n # i=1
n
n
X
X
X
pi = n −
pi , n −
pi 6∈ Z.
Case 6.
i=1
i=1
i=1
" n #
X
n−1
pi =
. It follows that
2
i=1
(29)



n = 2m + 1 , m ∈ N


n
n−1 X
n+1


pi <
,

 2 ≤
2
i=1
m≤
2m+1
X
i=1
We have
pi < m + 1.
i=1
Consequently, we consider the following regular transformation of the
sequence (sn )n∈N :
2m+1
P
X
i=1
(30)
(7)
T2m+1 (p) =
pi
k=0
c2m+1,k (p)sk +
2m+1
X
2m+1 P
k=
pi +1
c2m+1,k (p)sk
i=1
where
" 2m+1 #

X

1
(k)


P
(0),
k
=
pi , m ∈ N
0,

2m+1
 k!
i=1
"2m+1 #
c2m+1,k (p) =

X

1
(2m+1−k)


pi + 1, 2m + 1
P
(0),
k
=
 (2m + 1 − k)! 2m+1
i=1
.
Section of Euler summability method
29
" n #
à n
!
n
X
X
X
Case 7.
pi = n −
pi + 1 , n −
pi ∈ Z.
i=1
We obtain
n
X
i=1
pi =
i=1
(31)
i=1
n−1
, n = 2m + 1, m ∈ N i.e.
2



n = 2m + 1




,m ∈ N
.
2m+1

X



pi = m


i=1
We consider the following regular transformation of the sequence (sn )n∈N :
(8)
T2m+1 (p)
=
m
X
c2m+1,k (p)sk +
2m+1
X
c2m+1,k (p)sk
k=m+1
k=0
where
 (k)
P2m+1 (0)



, k = 0, m
2m+1

X
k!
c2m+1,k (p) =
,
pi ∈ Z, m ∈ N.
(2m+1−k)


P
(0)
i=1

 2m+1
, k = m + 1, 2m + 1
(2m + 1 − k)!
" n #
à n
! n
n
X
X
X
X
n−1
Case 8.
pi < n −
pi + 1 ,
pi ∈ Z. We obtain
pi <
.
2
i=1
i=1
i=1
i=1
We consider the following regular transformation of the sequence (sn )n∈N :
n
P
Tn(9) (p)
(32)
=
pi
X
i=1
k=0
n
X
cn,k (p)sk +
k=n−
n
P
i=1
cn,k (p)sk ,
pi
 X
n


pi ∈ Z , pi ∈ (0, 1) , i = 1, n


where
i=1
n
X

n−1


pi <

2
i=1
and
30
Ioan T
. incu

n
(k)
X

Pn (0)


, k = 0,
pi


k!
i=1
cn,k (p) =
.
n
n
(n−k)
X
X

P
(0)
 n


pi + 1, n −
pi − 1
 (n − k)! , k =
i=1
i=1
Example 1. In (13) we consider p = 1/2; it follows that
 
n
  sk .
Tn(1) (0) = 2−n+1
k
k=0
[ n2 ]
X
References
[1] B. V. Gnedenko, The Theory of Probability, MIR Publishers, Moskow
1976.
[2] G. H. Hardy, Divergent Series, Oxford 1949.
[3] G. H. Mihoc, G. Ciucu, V. Craiu, Teoria probabilitǎt.ilor .si statisticǎ
matematicǎ, E.D.P. Bucures.ti, 1970.
“Lucian Blaga” University of Sibiu
Faculty of Sciences
Department of Mathematics
Str. I. Raţiu, no. 5-7
550012 - Sibiu, Romania
E-mail:tincuioan@yahoo.com