American Journal of Computational Mathematics, 2013, 3, 31-37
doi:10.4236/ajcm.2013.33B006 Published Online September 2013 (http://www.scirp.org/journal/ajcm)
The Series of Reciprocals of Non-central Binomial
Coefficients
Laiping Zhang, Wanhui Ji
Yinchan Energy College, Yinchuan, China
Email: [email protected], [email protected]
Received 2013
ABSTRACT
Utilizing Gamma-Beta function, we can build one series involving reciprocal of non-central binomial coefficients, then
We can structure several new series of reciprocals of non-central binomial coefficients by item splitting, these new created denominator of series contain 1 to 4 odd factors of binomial coefficients. As the result of splitting items, some
identities of series of numbers values of reciprocals of binomial coefficients are given. The method of splitting terms
offered in this paper is a new combinatorial analysis way and elementary method to create new series.
Keywords: Binomial Coefficients; Split Terms; Reciprocals; Series; Non-central; Closed Form
1. Introduction
Binomial coefficient series plays an important role among
Number theory, Graph Theory, mathematical statistics
and Probability Theory, etc mathematics branches. Binomial coefficient series conversion also plays key role
in research areas of combinatorial mathematics and
mathematical analysis, it aroused great attention and can
be referred to a large number of literature [1-7]. Paper
[2-5] the author used is called Lehmer series identity


n 1
(2 x)
n
2n
 
2n
n

2 x arcsin x
1  x2
, | x | 1,
Some authors differential, integral, generating function,
the white to Beta-gamma function, hyper geometric
function series, recurrence and other mathematical tools
to get important results on series of reciprocals of binomial coefficients. Paper [8] obtains alternating series
involving reciprocals of central binomial coefficients.
Paper [9] obtains series involving reciprocals of central
binomial coefficients. We did the research the series of
reciprocal of non-central binomial coefficient. Utilizing
Gamma-beta function, we obtain one series involving
reciprocal of non-central binomial coefficients. We can
structure several new series of reciprocal of non-central
binomial coefficients by splitting item; the new created
denominators of series contain 1 to 4 odd factors of
non-central binomial coefficients.
By continuously using the method of splitting items,
we can get the series involving reciprocal of non-central.
Binomial coefficients of which denominator contain
5odd factor, 6 odd factor, … and p odd factor.
Copyright © 2013 SciRes.
Therefore splitting term method in this paper is an
elementary to build new series and also provide a new
combinatorial analysis method. As the result of analysis
some identities of series of number values of reciprocal
of non-central binomial coefficients also provided in this
paper.
2. Main Results and Proof of Theorem
Theorem 1 one series of reciprocals of non-central binomial coefficients


k 0
xk
 
2 k 1
k
=D
8
x
 2
arctan
,0  x  4
4  x 
2
2
(4
)
4
4
x
x
x
x
x





4
4 x  x 2  x
 2 
ln
, 4  x  0
4  x
4 x  x 2  x
(4  x) 4 x  x 2

(0)
Proof From power series theories easy to understand,
the series


k 0
xk
 
2 k 1
k


k 0
xk
 
2 k 1
k 1
is convergence.
Over interval (-4, 4). Utilizing Gamma-Beta function
relation and binomial coefficients,
 
n 1
k
1
 (n  1)  t k (1  t ) n  k dt
0
AJCM
L. P. ZHANG, W. H. JI
32
we calculation sum function of series


k 0
x

k
2 k 1
k
1 


1
k 0
0
2) The denominator contains 2 odd factors of binomial
coefficients in the series
  x k (2k  2)  t k 1 (1  t ) k dt
=2  kx t
k k 1
0 k 0
1 
k k 1
(1  t ) dt
k
1 
=2  k  xt (1  t )  tdt +2   xt (1  t )  tdt '
k
0 k 0
k


0 k 0
m0
xt (1  t )
tdt
dt  2 
 2
2
xt
1

(1  t )
xt
t
[1

(1

)]
0
0
1
1
1
1  xt (1  t )  1
tdt
tdt

2

2
0 [1  xt (1  t )]
0 1  xt (1  t )
1

m0
m0

2 m 1
m
m0


m 0

m
4
4
=(  1) D 
x
(2m  3) x
xm
 (2m  5)
x
2 m 1
m
m
 (2m  7)
=(
32 12
32 4
  1) D  2 
3x
x2 x
x
(
512 64 4 1
   )
3 x3 x 2 x 3
D


m0

x
2 m 1
m
m
 (2m  9)
(
Copyright © 2013 SciRes.
(1)
(2)
4096 512
32
4



4
3
2
35 x
5x
15 x
25 x

 (2m  3)(2m  9)


2 m 1
m
 (2m  5)(2m  9)
2 m 1
m
 (2m  7)(2m  9)
2048 3584 208 8 1


  )D
5 x 4 15 x 3 5 x 2 5 x 15
2048 512
656
8

 3

4
2
5x
5x
225 x 105 x
 (
(10)
3) The denominator contains 3 odd factors of binomial
coefficients in the series
m 0

xm
2 m 1
m
 (2m  3)(2m  5)(2m  7)
(11)
64 16 4 1
64 80
4
 ( 3  2   )D  3  2 
x 3
x
3x
3x 9 x 5 x


m0
(4)
(9)
xm

(3)
(8)
xm


512 64
4
 2
3
3x
9 x 15 x
4096 1536 96
4 1
 3  2  )
4
5x 5
5x
5x
5x
D
2 m 1
m
1024 384 16 16 1


  )D
5 x 4 5 x3 5 x 2 5 x 5
1024 128 208
32




5 x 4 15 x 3 25 x 2 105 x
m 0
(7)
xm
 (
for 4  x  0 . This completes proof of Theorem 1.
Theorem 2 The series of reciprocals of non-central
binomial coefficients
1) The denominator contains 1 odd factor of binomial
coefficients in the series



4 x  x  x
2
4


ln
,
2
4  x (4  x) 4 x  x
4 x  x 2  x



2

2 m 1
m
2048 256 16
8 1
 3  2
 )D
4
15 x
5x
5 x 15 x 5
2048 256
16
24




4
3
2
15 x
45 x 75 x 35 x
2
8
x
arctan
,

4  x (4  x) 4 x  x 2
4 x  x2

(6)
xm
 (
for 0  x  4
m0
 (2m  3)(2m  7)

 1

1 1
1
dt 
 2


2
 4  x 4  x 0 1  xt  xt 
2
2 1
1
dt



4  x 4  x 0 1  xt  xt 2
x
2 m 1
m
(2m  5)(2m  7)
256 48 8 1
256 176
8
 ( 3  2   ) D  3  2 
x 3
3x
3x
9 x 15 x
x
Using recursion formulas of integral [7]
2 m 1
m


m0
tdt
 2
2 2
0 [1  xt  xt ]

xm


1

(5)
128 16 1
128 16 16
 ( 3  2  ) D  3  2 
3x
3x 9 x 15x
x 3
 2 

 (2m  3)(2m  5)
2 m 1
m
16 8
16 8
 ( 2   1)D  2 
x
x
x 3x
0 k 0
1 
xm

m 0
(1  t ) dt +2  x t
k



xm
2 m 1
m
 ( 2 m  3 )( 2 m  5)( 2 m  9 )
512
64
16
28
1



 )D
15 x 4 5 x3 5 x 2 15 x 5
512
64
304
12




15 x 4
45 x3
75 x 2 105 x
 (
(12)
AJCM
L. P. ZHANG, W. H. JI


m0

sides of (0) splitting terms
xm
2 m 1
m
 (2m  3)(2m  7)(2m  9)
1024 704 48
4
1
(



 )D
15 x 4 15 x3 5 x 2 15 x 15
1
(13)
1024 1088 176
4



4
3
2
21x
15 x
45 x 225 x



m0
(

 (2m  5)(2m  7)(2m  9)
1
(14)
512 704 1228
4



4
3
2
35
x
5 x 15 x 225 x
4) The denominator contains 4 odd factors of binomial
coefficients in the series


m 0

xm
2 m 1
m
(n  2)!(n  1)!(n  1)nn(n  1) x n
n  0 (2n  4  1)!(2n  3  1)(2n  2  1)(2n  1  1)((2n  1)


put n  2  m
512 1216 112 12 1


  )D
5 x 4 15 x3 5 x 2 5 x 15

x
3
 D,
xm
2 m 1
m
33
 (2m  3)(2m  5)(2m  7)(2m  9)
256 256 32 16 1
 (



 )D
15 x 4 15 x3 5 x 2 15 x 15
256 512
544
16




4
3
2
15 x
45 x
225 x 105 x
x
3
m !(m  1)!(m  1)(m  2)(m  2)(m  3) x m  2
m  0 (2m  1)!(2m  2)(2m  3)(2m  4)(2m  5)
 D,


Multiply both sides by
16 16  m !(m  1)! x m
 
x 2 3x m  0 (2m  1)!
(1 
(15)
1
1
16
)(1 
) 2 D
2m  3
2m  5
x
16 16  m !(m  1)! x m
 
x 2 3 x m  0 (2m  1)!
(1 
Proof of Theorem
1) Left sides of (0) splitting terms
16
, arrive to
x2
1
1
1
16


)= 2 D
2m  3 2m  5 (2m  3)(2m  5) x
16 16  m !(m  1)! x m
 
x 2 3 x m  0 (2m  1)!
n !(n  1)! x n
 (2n  1)!  D
n 0

(1 
n !(n  1) ! x n
 (2n  1)!
n 0

(n  1)!n !(n)(n  1) x n
 1 
 D,
n  0 (2n  2  1)!(2n  1  1)( 2n  1)

3/ 2
1/ 2
16

) 2 D
2m  3 2m  5
x
Multiply both sides by 2 , arrive to
32 32
  2D  3D3
x 2 3x
put n  1  m ,
32
m !(m  1)! x m
 2D
x
m  0 (2m  1)!(2m  5)


m 1
m !(m  1)!(m  1)(m  2) x
 D,
m  0 (2m  1)!(2m  2)(2m  3)

1 
Multiply both sides by
4
, arrive to
x
4  m !(m  1)!(2m  3  1) x m 4

 D,
x m0
(2m  1)!(2m  3)
x
4  m !(m  1)! x m
1
4

(1 
)  D,
x m  0 (2m  1) !
2m  3 x
We obtain (1). Let right sides (1) be D3 .(2). Left
1
We obtain (2). Let right sides (2) be D5 .
3) Left sides of (0) splitting terms
1
x x2

3 10
(n  3)!(n  2)!(n  2)( n  1) n(n  1) n( n  1) x n
n  0 (2 n  6  1)!(2n  5  1)(2n  4  1)  (2n  1)


 D,
put n  3  m
x x 2  m !(m  1)!(m  1)(m  2)(m  3)(m  2)(m  3)(m  4) x m  3
 
 D,
3 10 m  0 (2m  1)!(2m  2)(2m  3)(2m  4)(2m  5)(2m  6)(2m  7)
Copyright © 2013 SciRes.
AJCM
L. P. ZHANG, W. H. JI
34
1
Multiply both sides by
x x 2  m !(m  1)![(m  2)(m  3)](m  4) x m
 
 D,
3 10 m  0 8(2m  1)!(2m  3)(2m  5)(2m  7)
64
,
x3
64 64 16  m !(m  1)! x m
3/ 2
1/ 2
1
64
 2  
[1 

](1 
) 3 D
3
5 x m  0 (2m  1)!
2m  3 2m  5
2m  7
x 3x
x
3/ 2
1/ 2
64 64 16  m !(m  1)! x m
3/ 2
1/ 2
1
64

 
[1 




] 3 D
2m  3 2m  5 2m  7 (2m  3)(2m  7) (2m  5)(2m  7)
x3 3 x 2 5 x m  0 (2m  1)!
x
15 / 8
3/ 4
3/8
64
64 64 16  m !(m  1)! x m


(1 
)  3 D,
 2  
3
2 m  3 2 m  5 2m  7
5 x m  0 (2m  1)!
x
x 3x
Multiply both sides by
8
, arrive to
3

512 512 256 8
m !(m  1)! x m
512
 2
 D  5D3  2 D5  
 3 D,
3
(2
m

1)!(2
m

7)
3x
9 x 15 x 3
3x
m0
we obtain (3).
4) Let right sides of (3) be D7 , Left sides of (0) splitting terms
1
x x 2 x3  (n  4)!(n  3)!(n  3)(n  2)(n  1)n(n  2)( n  1)n( n  1) x n
  
D
3 10 35 n  0
(2n  8  1)!(2n  7  1)(2n  6  1) (2n)((2n  1)
Put n  4  m ，
1
Multiply both sides by
x x 2 x3  m !(m  1)![(m  2)(m  3)(m  4)](m  5) x m  4
  
 D.
3 10 35 m  0 16(2m  1)!(2m  3)(2m  5)(2m  7)(2m  9)
256
, arrive to
x4
256 256 128 256  m !(m  1)! x m
1
1
1
1
256




(1 
)(1 
)(1 
)(1 
)  4 D,
2m  3
2m  5
2m  7
2m  9
x 4 3 x3 5 x 2 35 x m  0 (2m  1)!
x
expand to
256 256 128 256  m !(m  1)! x m
1
1
1
1
1
 3  2

[1 




4
35 x m  0 (2m  1)!
2m  3 2m  5 2m  7 2m  9 (2m  3)(2m  5)
x
3x
5x
1
1
1
1
1





(2m  3)(2m  7) (2m  3)(2m  9) (2m  5)(2m  7) (2m  5)(2m  9) (2m  7)(2m  9)
1
1
1



(2m  3)(2m  5)(2m  7) (2m  3)(2m  5)(2m  9) (2m  5)(2m  7)(2m  9)
1
256
] 4 D

(2m  3)(2m  5)(2m  7)(2m  9)
x
In (0.1), there are 10 fractions containing 2 factors, 10
fractions containing 3 factors, 5 fractions containing 4
and one fraction containing 5 factors.
After the below calculation to (0.1), arrive one series
of reciprocals of non-central binomial coefficients with
the denominators 1, 2, 3, 4, 5 factors
1) All fractions of (0.1) divided into partial fractions,
arrive to
256 256 128 256  m !(m  1)! x m
15 / 8
3/ 4
3/8
1
15 / 8
 3  2

(1 




4
35 x m 0 (2m  1)!
2m  3 2m  5 2m  7 2m  9 (2m  3)(2m  9)
x
3x
5x
3/8
3/ 4
256


 4 D,
(2m  5)(2m  9) (2m  7)(2m  9)
x
Copyright © 2013 SciRes.
AJCM
L. P. ZHANG, W. H. JI
35
arrive to
256 256 128 256  m !(m  1)! x m
35 /16 15 / 16 9 / 16
5 /16
256




[1 



] 4 D
2m  3 2m  5 2m  7 2m  9
x 4 3 x3 5 x 2 35 x m  0 (2m  1)!
x
16
，arrive to
5

4096 4096 2048 4096 16
9
m !(m  1)! x m
4096
D
7
D
3
D
D
D,










3
5
7
4
3
2
5
5x
15 x
25 x 175 x 5
5x4
m  0 (2m  1)!(2m  9)
Multiply both sides by
We obtain (4), Let right sides (4) be D9
2) In (0.1), the factions of 2 factors retained, other
fractions divided partial fractions. Then spit term the
fraction containing 2 factions, reserve one during each
split term, what left has been split into partial fraction,
arrive to
256 256 128 256
27



 D35  D 
D3
16
x 4 3 x3 5 x 2 35 x
A)
23
9
5
256
 D5  D7  D9  4 D
16
16
16
x
256 256 128 256
31



 D3 7  D  D3
16
x 4 3x3 5 x 2 35 x
B)
15
13
5
256
 D5  D7  D9  4 D
16
16
16
x
256 256 128 256
97
 3  2
 D39  D  D3
4
35 x
48
3x
5x
x
C)
15
9
23
256
 D5  D7  D9  4 D
16
16
48
x
256 256 128 256
35



 D5 7  D  D3
16
x 4 3 x3 5 x 2 35 x
D)
5
7
17
256
 D5  D7  D9  4 D.
16
16
16
x
256 256 128 256
35
 3  2
 D59  D  D3
4
35 x
16
x
3x
5x
E)
11
9
9
256
 D5  D7  D9  4 D.
16
16
16
x
256 256 128 256
35
 3  2
 D7 9  D  D3
4
35 x
16
x
3x
5x
F)
1
13
1024
15
 D5  D7  D9  5 D.
16
16
16
x
Because D, D3 , D5 , D7 , D9 is known, the calculation of
(A) – (F), we get (5) – (10),
3) In (0.1), the factions of 3 factors retained, other
fractions divided partial fractions.
Then spit term the fraction containing 3 factions,
reserve one during each split term, what left has been
split into partial fraction, arrive to
A)
256 256 128 256
33



 D35 7  D  D3
16
x 4 3x3 5 x 2 35 x
Copyright © 2013 SciRes.

19
7
5
256
D5  D7  D9  4 D.
16
16
16
x
256 256 128 256
101
D3



 D359  D 
48
x 4 3x3 5 x 2 35 x
B)
17
9
13
256
 D5  D7  D9  4 D.
16
16
48
x
256 256 128 256
103
D3
 3  2
 D3 7 9  D 
4
35 x
48
x
3x
5x
C)
15
11
11
256
 D5  D7  D9  4 D.
16
16
48
x
256 256 128 256
35
 3  2
 D5 7 9  D  D3
4
35 x
16
x
3x
5x
D)
13
13
3
256
 D5  D7  D9  4 D.
16
16
16
x
Because D, D3 , D5 , D7 , D9 is known, the calculation
of (A) – (D), we get (11) – (14),
4) In (0.1), the factions of 4 factors retained, other fractions divided partial fractions.
Then spit term the fraction containing 4 factions,
reserve one during each split term, what left has been
split into partial fraction, arrive to
256 256 128 256
13
 3  2
 D35 7 9  D  D3
4
35 x
6
x
3x
5x
1
1
256
 D5  D7  D9  4 D.
2
3
x
Because D, D3 , D5 , D7 , D9 , is known, the calculation
of this expression , we get (15).
This completes proof of Theorem 2
3. Some Series of Number Values
In (1) - (15) of theorem 2, put x  1,
D


m0

1
2 m 1
m


2 4 3

3
27
Put x  1 ,
D


m0
(1) m

2 m 1
m


2 8 5
5 1
ln  ,  

;
5 25
2
we have
AJCM
L. P. ZHANG, W. H. JI
36
Corollary 1 The series of number values of reciprocal
of non-central binomial coefficients

1
4 3
1)  2 m 1

2 ;
9
(2m  3)
m0 m


m 0

2 m 1
m


2 m 1
m
m 0

m0

2 m 1
m

m 0

2 m 1
m

m 0

2 m 1
m

m 0

2 m 1
m

m0




m0
2 m 1
m

m 0
2 m 1
m




11)
m0
1
1078
 4 3 
;
45
(2m  3)(2m  7)
20 3 1634


;
3
45
(2m  5)(2m  7)
1
 (2m  3)(2m  9)


10)
1
1


1
196 3 112054

;
15
1575
284 3 18084


;
15
175
(2m  5)(2m  9)
1
2 m 1
m
 (2m  7)(2m  9)
156 3 267422

;
1575
5
1

2 m 1
m

 (2m  3)(2m  5)(2m  7)

2 m 1
m
 (2m  3)(2m  5)(2m  9)
44 3 45316
;

15
1575

1

m0

2 m 1
m
 (2m  3)(2m  7)(2m  9)
68 3 38842

;
15
1575

1

14)

2 m 1
m
 (2m  5)(2m  9)(2m  9)
92 3 52306


;
15
1575
Copyright © 2013 SciRes.

m0



4)
m0

(1) m
2 m 1
m
 (2m  5)
(1) m
2 m 1
m
 (2m  7)
(1) m
2 m 1
m


5)
m0

2 m 1
m
m0


m0


8)
m0

2 m 1
m
m 0

2 m 1
m


m 0
1832 5
206938
ln  
;
5
525
(1) m
(2m  5)(2m  7)
(1)m
2 m 1
m
 (2m  3)(2m  9)
(1) m
2 m 1
m
(1)m
  (2m  7)(2m  9)
2m1
m
 8 5 ln  
26
;
3

56 5
902
ln  
;
3
45

136 5
2194
ln  
;
3
45

 (2m  5)(2m  9)

10)

1144 5
3698
ln  
;
15
45
 (2m  3)(2m  7)


9)

(1) m

7)
72 5
46
ln   ;
5
3
 (2m  3)(2m  5)


6)
(2m  9)

(1) m

184 5
152294
ln  
;
3
1575
 88 5 ln  

16574
;
175
664 5
375122
ln  
;
3
1575
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
m0


1

13)
m 0
4 3 326


;
3
45
m0



1

12)

 (2m  3)(2m  5)(2m  7)(2m  9)
4 3 762

;
5
175


3)

9)
148 3 41058

;
9
45
4 3 22
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

3
3
(2m  3)(2m  5)

8)



7)
1

1
2 m 1
m
Corollary 2 Alternating series of number values of reciprocals of non-central binominal coefficients

(1) m
8 5
1)  2 m 1

ln   2;
5
(2m  3)
m0 m
2)

6)
28 3 50
 ;
9
3
m0

2348 3 225158
;


9
525
(2m  9)

5)

 (2m  7)

4)
1
 (2m  5)

3)
15)


2)


AJCM
L. P. ZHANG, W. H. JI
2005, p. A22.
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37
Positive and Negative Involving Reciprocals of Binominal Coefficients,” Pure Mathematics, Vol. 2, No. 4, 2012,
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AJCM