1
2
3
47
6
Journal of Integer Sequences, Vol. 15 (2012),
Article 12.2.8
23 11
Alternating Sums of the Reciprocals
of Binomial Coefficients
Hacène Belbachir and Mourad Rahmani
University of Sciences and Technology Houari Boumediene
Faculty of Mathematics
P. O. Box 32
El Alia, Bab-Ezzouar 16111
Algiers
Algeria
hacenebelbachir@gmail.com
mrahmani@usthb.dz
B. Sury
Statistics & Mathematics Unit
Indian Statistical Institute
8th Mile Mysore Road
Bangalore 560059
India
sury@isibang.ac.in
Abstract
P
k m n−1
. We give closed
We investigate summations of the form
0≤k≤n (−1) k
k
formulae in terms of the Akiyama-Tanigawa matrix. Recurrence formulae, ordinary
generating functions and some other results are also given.
1
Introduction and notation
Binomial coefficients play an important role in many areas of mathematics, such as combinatorics, number theory and special functions. The inverse binomial coefficients have an
1
integral representation in terms of beta function
−1
Z1
n
= (n + 1) xk (1 − x)n−k dx.
k
(1)
0
and one can obtain (see [18])
n −1
X
n
k=0
k
n+1
=
n + 1 X 2k
.,
2n+1 k=1 k
There are many papers dealing with sums involving inverses of binomial coefficients, see for
instance [3, 11, 12, 14, 15, 17, 18, 21, 22]. For nonnegative integers n, m and p we consider
the sums
−1
n
X
k m p+n
(m,p)
.
(2)
Tn
:=
(−1) k
p
+
k
k=0
These sums have been studied by many authors. Trif [20], using (1) proved for m = 0
that
−1 !
p+n+1
p
+
n
+
1
Tn(0,p) = (−1)n +
.
(3)
p
p+n+2
Sury, Wang and Zhao [19], studied (2) for m = 1 and m = 2, they obtain
−1
p + n + 1 (−1)n (n + 1)(p + n + 3)
p+n+2
(1,p)
n
−
Tn =
− (−1) ,
p+n+3
p+n+2
p+1
(4)
and
Tn(2,p)
(−1)n (n + 1)2 (−1)n (2n + 3)
−
= (p + n + 1)
p+n+2
p+n+3
−1
−1 2
p+n+3
p+n+2
1
n
+
.
−
+ (−1)
p+n+4
p+n+3
p+2
p+1
(5)
Our aim is to give a closed form and recurrence relation for the sums (2). In order to
(m)
(m,0)
(m,p)
investigate the summation of the form Sn := Tn
and Tn , we shall use the following
tools [1, 6]:
• The Stirling numbers of the first kind s(n, k) (see A008275 in [13]), are defined by the
generating function
X
x (x − 1) · · · (x − n + 1) =
s(n, k)xk ,
k≥0
and satisfy the recurrence relation
s(n + 1, k) = s(n, k − 1) − ns(n, k), (1 ≤ k ≤ n) ,
with s(n, n) = 1, s(n, 0) = 0 for n ≥ 1 and s(n, k) = 0 for k < 0 or k > n.
2
• The Stirling numbers of the second kind nk (see A008277 in [13]), are defined by the
generating function
k
Y
Xn
x
xn ,
=
k
1
−
jx
j=1
n≥k
and satisfy the recurrence relation
n
n
n+1
,
+k
=
k
k−1
k
with n1 = nn = 1.
They also verify the following important identity
n
X
n+k n
n
x (x + 1) · · · (x + k − 1) .
x =
(−1)
k
k=0
• The Eulerian numbers
n
k
(6)
(see A008292 in [13]) are defined by
X
k
n
i
n n+1
, (1 ≤ k ≤ n) ,
=
(−1) (k − i)
i
k
i=0
and satisfy the recursive identity
n
n−1
n−1
=k
+ (n − k + 1)
,
k
k
k−1
with 11 = 1.
• The Worpitzky numbers Wn,k (see A028246 in [13]), are defined by
Wn,k =
k
X
i+k
(−1)
i=0
k
.
(i + 1)
i
n
They can also be expressed through the Stirling numbers of the second kind as follows
n+1
.
(7)
Wn,k = k!
k+1
The Worpitzky numbers satisfy the recursive relation
Wn,k = (k + 1) Wn−1,k + kWn−1,k−1 (n ≥ 1, k ≥ 1) .
(8)
Some simple properties are given
n
n X
n k X
x =
(x − 1)n−k kWn−1,k−1 ,
k
k=0
k=0
3
(9)
n X
n k
k
k=0
and
t
n+1
=
,
t+1
n X
n k+1
k
k=0
t
= Wn,n−t .
(10)
(11)
• The Bernoulli numbers Bn are defined by the exponential generating function
X xn
x
=
Bn .
1 − e−x
n!
n≥0
The recursive relation is
B0 = 1,
Bn = 1 −
n−1 X
n
k=0
Bk
, (n ≥ 1) .
k n−k+1
Thus we have B1 = 21 , B2 = 61 , B3 = 0, and so on, they can also be expressed through
the Worpitzky numbers
n
X
Wn,k
Bn =
(−1)k
.
k+1
k=0
• The Akiyama-Tanigawa matrix (An,k )n,k≥0 associated with initial sequence A0,k =
is defined by (see [2, 5, 8, 10])
1
k+1
An,k = (k + 1) (An−1,k − An−1,k+1 ) ,
or equivalently by [7]
k
1X
(−1)k s(k + 1, i + 1)Bn+i ,
k! i=0
−1
n
X
i−1 k + i + 1
Wn,i .
=
(−1)
k
+
1
i=1
An,k =
4
(12)
The Akiyama-Tanigawa matrix An,k is then

1
1
1
1
1
2
3
4
5


1
1
1
1
 1
 2
3
4
5
6


1
3
2
5
 1
 6
6
20
15
42


1
2
5
1
An,k =  0

30
20
35
84


− 1 − 1 − 3 − 1
0
 30
30
140
105


 0 −1 −1 − 4 −1

42
28
105
28
..
..
..
..
..
.
.
.
.
.
2
···



· · ·



· · ·



.
· · ·



· · ·



· · ·

(m)
Explicit formula for Sn
For any nonnegative integer m, we consider the sums
Sn(m)
n
X
−1
n
:=
(−1) k
.
k
k=0
k
m
(13)
(0)
n+1
(identity (14) of [19] or identity (3) here for p = 0). The
Note that Sn = (1 + (−1)n ) n+2
(m)
following expression for Sn holds and, in terms of computational complexity, it is better
(m)
than the sum defining Sn for n ≥ m as the expression in the theorem involves O(m)
operations.
Theorem 1. For any nonnegative integers n and m, we have
Sn(m)
= (n + 1)
m
X
(−1)m+j
j=0
(m)
Proof. We can write Sn
Sn(m)
n+j+2
n
1 + (−1)
n+j+1
Wm,j .
j
as follows
−1
n
X
k n
=
(−1)
((k + 1) − 1)m
k
k=0
−1 X
m
n
X
m−i m
k n
(k + 1)i ,
=
(−1)
(−1)
i
k
i=0
k=0
5
(14)
and with (6), we obtain
Sn(m)
X
−1 X
i
m
i
n
i+j
m−i m
(k + 1) · · · (k + j)
(−1)
(−1)
=
(−1)
j
i
k
j=0
i=0
k=0
m
i
n
XXX (−1)k
i
m
=
k! (k + 1) · · · (k + j) (n − k)!
(−1)m+j
j
i
n!
k=0 i=0 j=0
−1
m X
n
i
X
X
i
m+j m
k n+j
=
(n + 1) · · · (n + j)
(−1)
(−1)
.
i
j
k
+
j
i=0 j=0
k=0
n
X
k
Now, from (3) and after some rearrangement, we get
X
m m
X
i
m
(−1)m+j
n n+j +1
(m)
.
j!
1 + (−1)
Sn = (n + 1)
j
i
j
n
+
j
+
2
i=0
j=0
From (7) and (10), the result holds.
3
(m)
Recurrence relation for Sn
Theorem 2. For any nonnegative integers m and n, we have
1
Sn(m+1) − (−1)n (n + 1)m .
n+1
−1
Proof. The proof is based on the Sprugnoli [16] observation n+1
=
k
(m)
Sn+1 = Sn(m) −
(m)
The recurrence relation for Sn
is given in the following
(15)
n −1
n −1
k
.
− n+1
k
k
Theorem 3. For any nonnegative integers m and n, we have
(m)
Sn+1
=
Sn(m)
m+1
X
(−1)m+j
− (−1) (n + 1) +
n+j+2
j=0
n
m
(m)
with the initial condition S0
1 + (−1)
n
n+j+1
Wm+1,j ,
j
= δ0m , where δij is the Kronecker symbol.
Proof. This follows immediately from (14) and (15).
Setting m = 1 in (16), we have the following
Corollary 4. If n is nonnegative integer, then
(1)
Sn+1 = Sn(1) +
n − (−1)n (2n4 + 17n3 + 49n2 + 57n + 24)
.
(n + 2) (n + 3) (n + 4)
(m)
Our next goal is to calculate the ordinary generating functions of Sn .
6
(16)
4
(m)
Ordinary generating functions of Sn
In 2002, Mansour [9], generalized the idea of Sury [18] and gave an approach based on
calculus to obtain the generating function for some combinatorial identities.
Theorem 5 (Mansour [9]). Let r, n ≥ k be any nonnegative integer numbers, and let fr (n, k)
be given by
Zu2
(n + r)!
fr (n, k) =
pk (t) q n−k (t) dt,
(17)
n!
u1
where p (t) and q (t) are two functions defined on [u1 , u2 ] . Let {an }n≥0 and {bn }n≥0 be any
two sequences, and let A (x) , B (x) be the corresponding ordinary generating functions. Then

 u
#
" n
2
Z
∞
r
X X
d
fr (n, k) ak bn−k xn = r xr A (xp (t)) B (xq (t)) dt .
(18)
dx
n=0 k=0
u1
n
We apply Theorem 5, for an = (−1) nm (m ≥ 1) and bn = 1, we have
m X
1
m
(−x)k+1
A (x) =
m+1
k
(1 + x)
k=0
=
m
X
(−1)m+k
Wm,k ,
(1 + x)k+1
X
1
.
B (x) =
xn =
1
−
x
n≥0
k=0
For r = 1, formula (18) gives:
X
Sn(m) xn
n≥0

d 
x
=
dx 
Z1
0
m P
m
k=0
k

k+1
(−xt)

.
dt
(1 + xt)m+1 (1 − x + xt) 
(19)
Making the substitution xt = y in the right-hand side of (19), we obtain


m P
k+1
m
x
(−y)
Z
k
X

d 
k=0
(m) n

,
dy
Sn x =
m+1
 (1 + y)

dx
(1
−
x
+
y)
n≥0
0
Since the degree of the denominator is at least one higher than that of the numerator, this
fraction decomposes into partial fractions of the form
m P
m
k=0
k
(−y)k+1
(1 + y)m+1 (1 − x + y)
m
=
(m)
X αs (x)
α(m) (x)
+
,
1 − x + y s=0 (1 + y)m−s+1
7
(20)
We note in passing that (20) is equivalent to
m X
m
k=0
k
(−y)k+1 = (1 + y)m+1 α(m) (x) + (1 − x + y)
m
X
(1 + y)s αs(m) (x)
(21)
s=0
=
m
X
(−1)m+k+1 y (1 + y)m−k Wm−1,k−1 .
k=0
For y = −1 and using the fact that Wp,p = p! for p ≥ 0, we immediately obtain the wellknown identity
m X
m
= m!.
k
k=0
(m)
Next, if we set y = 0 in (21) then we obtain a relation between α(m) (x) and αs
m
X
αs(m) (x) =
s=0
(x)
α(m) (x)
.
x−1
(22)
Proposition 6. For m ≥ 1, we have
αs(m)
m s
X
(−1)i+s+1 X m k + 1
(x) =
k
s−i
xi+1
i=0
k=0
(23)
m
X
(−1)m+j+1
=
Wm,j ,
xs−m+1+j
j=m−s
and
α
(m)
(x) =
=
1
xm+1
m X
m
k=0
k
m
X
(−1)m+j
j=0
xj+1
(1 − x)k+1
(24)
Wm,j ,
(m)
= −αm
(x) .
Proof. We verify that (23) and (24) satisfy (21). Denote the right-hand side of (21) by
R(m) (y), after some rearrangement, we get
"
m X
(1 + y)m+1
m
(m)
(1 − x)k+1
R (y) =
m+1
x
k
k=0
+ (1 − x + y)
m
X
s=0
8
(1 + y)
#
s
j+1 X
k
+
1
(−1)
s
j=0
xs−j+1
j
,
using binomial formula and for k ≤ m, we obtain
R
(m)
(y) =
m X
m
k=0
=
m X
m
k=0
k
"
k
"
m+1
X
s (1 + y)s X k + 1
(−1)j xj
s
j
x
s=m+1
j=0
#
s
m
(1 − x + y) X (1 + y)s X
j j k+1
−
(−1) x
j
x
xs
s=0
j=0
m+1
X
s (1 + y)s X k + 1
(−1)j xj
s
j
x
s=m+1
j=0
+
s
m
X
(1 + y)s X
xs
s=0
j
(−1) x
j
j=0
−
m
s
X
(1 + y)s+1 X
s=0
"
m m+1
s
X (1 + y)s X
X
m
j j k+1
=
(−1) x
s
k
x
j
s=0
j=0
k=0
−
s
X
j
(−1) x
j=0
j
xs+1
(−1)j xj
#
k+1
j
(−1)j xj
#
k+1
j
j=0
s
m
X
(1 + y)s+1 X
s=0
"
m m+1
X (1 + y)s
X
m
=
k
xs
s=0
k=0
k+1
j
xs+1
j=0
k+1
j
−
s−1
X
(−1)j xj
j=0
Finally,
R(m) (y) =
=
" k+1
m X
X
m
k=0
m X
k=0
k
(1 + y)s
s=0
m
(−y)k+1 .
k
9
#
k
+
1
(−1)s
s
!#
k+1
.
j
According to (7) and (11), we have
αs(m)
s
m X
(−1)i+s+1 X m k + 1
(x) =
s−i
k
xi+1
i=0
k=0
s
X
(−1)i+s+1 (m − s + i)!
m+1
=
m−s+i+1
xi+1
i=0
=
=
s
X
(−1)i+s+1
(x)i+1
i=0
m
X
Wm,m−s+i
(−1)m+j+1
Wm,j .
s−m+1+j
x
j=m−s
It follows from (9) that
α
(m)
(x) =
=
1
xm+1
1−x
xm+1
m X
m
k=0
m
X
k
(1 − x)k+1
(−1)m+k xm−k kWm−1,k−1
k=1
m
X
= (1 − x)
(−1)m+k
k=0
=
=
m
X
k=0
m
X
m+k
(−1)
(−1)m+k
k=0
k
xk+1
k
xk+1
k
xk+1
Wm−1,k−1
Wm−1,k−1 −
Wm−1,k−1 +
m
X
k=0
m−1
X
(−1)m+k
k
Wm−1,k−1
xk
(−1)m+k
k+1
Wm−1,k .
xk+1
k=0
Using (8), we get α(m) (x) as desired. This completes the proof.
Now, integrating the right-hand side of (20) over y, we obtain
Zx
0
m P
m
k=0
k
(−y)k+1
(1 + y)m+1 (1 − x + y)
(m)
dy = αm
(x) ln(1 − x2 ) +
m−1
X
s=0
(m)
αs (x) 1 − (1 + x)s−m .
m−s
(25)
(m)
By differentiating (25) we get the ordinary generating function of Sn
m−1
X
X
(m)
(m) n
2 d
Sn x = ln(1 − x ) αm (x) +
dx
s=0
n≥0
+
m−1
X
s=0
d (m)
α
dx s
(1 + x)s−m−1 αs(m) (x) −
10
(x) 1 − (1 + x)s−m
m−s
2x
α(m) (x) ,
1 − x2 m
(26)
with
m
X
d (m)
(s − m + 1 + j) (−1)m+j
α (x) =
Wm,j .
dx s
xs−m+2+j
j=m−s
With Proposition 6, we can now rewrite (26) as follows
Theorem 7. For any real numbers x and for all nonnegative integer m, we have
X
Sn(m) xn =
m
X
(1 + j) (−1)m+j
x2+j
j=0
n≥0
(−1)j
+
Wm,m−j
xs−j+2
0≤j≤s≤m−1
X
Wm,j
!
ln(1 − x2 )
s−j+1
1 − (1 + x)s−m − x (1 + x)s−m−1
m−s
m
2 X (−1)m+j
Wm,j . (27)
+
1 − x2 j=0
xj
In particular for m = 0 and m = 1, we get
X
ln (1 − x2 )
2
+
,
Sn(0) xn =
2
2
1
−
x
x
n≥0
and
X
Sn(1) xn =
n≥0
5
2 + 3x
2−x
2
.
ln
1
−
x
2 +
x3
x (1 + x)
The asymptotic expansion
(m)
In the previous sections, Sn becomes more complex when, m grows, so it is important to
(m)
have asymptotic expansion of Sn .
Theorem 8. For m > 0, we have
(m)
(m)
S2n ∼ (2n)m and S2n+1 ∼ − (2n + 1)m .
Proof. Write
k m = c0 + c1 (k + 1) + c2 (k + 1) (k + 2) · · · + cm (k + 1) · · · (k + m) ,
where ci ’s depend on m (cm = 1). we immediately have
−1
−1
−1
n
n
n
X
X
X
n
n
k n
k
k
(m)
Sn = c 0
(−1)
+c1
(−1) (k + 1)
+· · ·+
(−1) (k + 1) · · · (k + m)
.
k
k
k
k=0
k=0
k=0
−1
−1
, we have
= (n + 1) n+1
After some rearrangement and using the fact that (k + 1) nk
k+1
Sn(m) = c0 Tn(0,0) + c1 (n + 1) Tn(0,1) + c2 (n + 1) (n + 2) Tn(0,2) + · · · + (n + 1) · · · (n + m) Tn(0,m) .
(0,p)
Since T2n
(0,p)
→ 1 and T2n+1 → −1, the result holds.
11
6
A connection to the Akiyama-Tanigawa matrix
(m,p)
In this section we consider Tn
main theorem of this section.
. The following lemma will be useful in the proof of the
Lemma 9. For m ≥ 1, we have
k+1
n
m
X
X
z
m k
k z =
Wm,k
1−z
k=0
k=0
m s
X
X
m
m−s
n+1
−z
(−1)s+k Ws,k (1 − z)−k−1 .
(n + 1)
s
s=0
k=0
Proof. Recall that, for m ≥ 1
∞
X
m k
k z =
k=0
m
X
Wm,k
k=0
z
1−z
k+1
=
m
X
(−1)m+k Wm,k (1 − z)−k−1 ,
k=0
we have
n
X
m k
k z =
k=0
=
∞
X
k=0
m
X
k=0
m
X
m k
k z −
∞
X
kmz k
k=n+1
Wm,k
z
1−z
k+1
−z
k+1
n+1
∞
X
(i + n + 1)m z i
i=0
∞ X
m X
m
=
Wm,k
−z
(n + 1)m−s is z i
s
i=0 s=0
k=0
∞
m
m
k+1
X
X
X
z
m
(n + 1)m−s
is z i ,
=
Wm,k
− z n+1
s
1
−
z
s=0
i=0
k=0
z
1−z
n+1
as desired.
For an alternative proof see Boyadzhiev [4]. The main result of this section is to prove
the following theorem which expresses explicitly the alternating sums of the reciprocals of
(m,p)
binomial coefficients, Tn , in terms of the Akiyama-Tanigawa matrix An,k .
Theorem 10. For nonnegative integers n, m and p, we have
Tn(m,p)
=
n+p
p
−1
m
X
m
(n + 1)m−s As,n+p+1
δ0m + (n + p + 1)
(−1)
s
s=0
−1
m
n + p + 1X
s n+s+p+2
(−1)
Wm,s , (28)
−
n + 1 s=0
p+s+1
n+s
where Ai,j is the Akiyama-Tanigawa matrix.
12
Proof. By the Beta function we can write
Tn(m,p)
=
n
X
k
m
(−1) k (p + n + 1)
k=0
Z1
xp+k (1 − x)n−k dx
0
Z1
= (p + n + 1)
n
p
x (1 − x)
n
X
k
m
k=0
0
−x
1−x
k
dx.
Using the lemma, we get
Tn(m,p)
= (p + n + 1)
Z1
p
x (1 − x)
n
−
s=0
=
s
Wm,k (−x)k+1
k=0
0
m X
m
m
X
!
s
X
(−1)s+k Ws,k (−x)n+1 (1 − x)k−n dx
(n + 1)m−s
k=0
m
(p + n + 1) X
n+1
k+1
(−1)
k=0
− (p + n + 1)
m X
s X
m
s=0 k=0
−1
n+k+p+2
Wm,k
p+k+1
s
m−s
(n + 1)
−1
(−1)n+s+k+1 n + k + 2 + p
Ws,k .
k+1
k+1+p
Finally, from (12) we obtain
Tn(m,p)
−1
m
(p + n + 1) X
k+1 n + k + p + 2
=
Wm,k
(−1)
p+k+1
n + 1 k=0
n+p+1
m X
(−1)n+s X
m
m−s
(−1)k s(n + p + 2, k + 1)Bs+k .
+ (n + p + 1)
(n + 1)
(n
+
p
+
1)!
s
s=0
k=0
As desired, this completes the proof.
Setting p = 0 in (28) we can rewrite (14) as follows
Corollary 11.
Sn(m)
= δ0m − Am+1,n +
m
X
(−1)
s=0
7
n+s
m
(n + 1)m−s+1 As,n+1 .
s
(m,p)
A recurrence relation For Tn
Theorem 12. For any nonnegative integers m, n and p
(m,p)
Tn+1 =
n+1
1
Tn(m,p) −
T (m+1,p) − (−1)n (n + 1)m .
n+p+1
n+p+1 n
13
(29)
Proof. Using the identity
−1
−1
−1
n+1
n+p+1
k
n+p
n+p
=
−
,
k+p
n+p+1 k+p
n+p+1 k+p
we get the relation (29).
(m,p)
Now, from (28) and (29), we have the recurrence relation for Tn
Theorem 13. For nonnegative integers n, m and p, we have
(m,p)
Tn+1
m+1
X (−1)s m + 1
n+1
m+1
(m,p)
n
=
As,n+p+1
T
− (−1) (n + 1)
s
n+p+1 n
(n + 1)s
s=0
−1
m+1
1 X
s n+s+p+2
Wm+1,s − (−1)n (n + 1)m ,
+
(−1)
p+s+1
n + 1 s=0
(m,p)
with the initial condition T0
8
=
n+p −1
δ0m .
p
Acknowledgments
The authors are grateful to the anonymous referee for providing a number of constructive
comments and illuminating suggestions, almost all of which we have tried to carry out. The
first and the second author were partially supported by PNR project 8/u160/664.
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2000 Mathematics Subject Classification: Primary 11B65; Secondary 05A10, 05A16.
Keywords: Binomial coefficient, Akiyama-Tanigawa matrix, recurrence relation, generating
function.
(Concerned with sequences A008275, A008277, A008292, and A028246.)
Received July 27 2011; revised version received January 13 2012. Published in Journal of
Integer Sequences, January 14 2012.
Return to Journal of Integer Sequences home page.
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