1
2
3
47
6
Journal of Integer Sequences, Vol. 10 (2007),
Article 07.8.7
23 11
Certain Sums Involving Inverses of Binomial
Coefficients and Some Integrals
Jin-Hua Yang
Zhoukou Normal University
Zhoukou, 466001
China
Feng-Zhen Zhao
Department of Applied Mathematics
Dalian University of Technology
Dalian, 116024
China
fengzhenzhao@yahoo.com.cn
Abstract
In this paper, we are concerned with sums involving inverses of binomial coefficients.
We study certain sums involving reciprocals of binomial coefficients by using some
integrals. Some recurrence relations related to inverses of binomial coefficients are
obtained. In addition, we give the approximate values of certain sums involving the
inverses of binomial coefficients.
1
Introduction
It is well known that binomial coefficients play an important role in various subjects such
as combinatorics, number theory, and probability. There are many results for sums related
to binomial coefficients. Sums involving inverses of binomial coefficients have been receiving
much attention. For example, see [1]-[2] or [4]-[14]. In this paper, we are still interested in
sums involving inverses of binomial coefficients, and we investigate these kinds of sums by
using some integrals. For convenience, we first give the definition of binomial coefficients.
1
n
is defined by
For nonnegative integers m and n, the binomial coefficient m

n!

, if n ≥ m,
n
= m!(n − m)!

m
0,
if n < m.
We know that integral is an effective method for computing sums involving inverses of binomial coefficients (see [8, 11]). It is based on Euler’s well-known Beta function defined by
(see [11])
Z
1
B(n, m) =
0
tn−1 (1 − t)m−1 dt
Γ(n)Γ(m)
(n − 1)!(m − 1)!
for all positive integers n and m. Since B(n, m) =
=
, the
Γ(n + m)
(n + m − 1)!
n
satisfies the equation
binomial coefficient m
−1
Z 1
n
= (n + 1)
tm (1 − t)n−m dt.
m
0
(1)
the equation It is clear that integrals have connections with inverses of binomial coefficients.
The purpose of this paper is to study sums of the following forms:
∞
∞
n+k
X
X
fn
fn
n
,
and
2n
2n n+k
n
n=0
n=0
n
n
where fn is a rational function of n. In Section 2, we derive some relations for
∞
X
n=0
means of (1) and other integrals. In the meantime, we express the series
∞
X
n=0
n
∞
X
n=1
where s is a positive integer.
by
1
by some
2n 2
double integrations. In addition, we discuss the approximate value of the series
2
n+k
fn
n
2n
n
1
ns
,
2n
n
Main Results
Z
1
dx
, we have
[1 + x(1 − x)]k+1
dx
2
2(2k + 1)
=
+
[1 − x(1 − x)]k+2
3(k + 1)
3(k + 1)
Z
1
dx
,
[1 − x(1 − x)]k+1
(2)
2
2(2k + 1)
dx
=
+
k+2
[1 + x(1 − x)]
5(k + 1)
5(k + 1)
Z
1
dx
.
[1 + x(1 − x)]k+1
(3)
Lemma 1. For the integrals
0
Z
Z
1
0
0
1
1
dx
and
[1 − x(1 − x)]k+1
Z
2
0
0
0
Proof. The proofs of (2)-(3) are simple and are omitted here.
Theorem 2. Assume that
Ak =
∞
X
n=0
n+k
n
2n
n
and
Bk =
∞
X
(−1)n
n=0
2n
n
n+k
n
,
where k is a nonnegative integer. Then we have the following recurrence relations:
Ak+1 = −
2(2k + 3)
2
+
Ak ,
3(k + 1)
3(k + 1)
(4)
Bk+1 = −
2
2(2k + 3)
+
Bk .
5(k + 1)
5(k + 1)
(5)
Proof. It follows from (1) that
Ak =
∞ X
n+k
n=0
n
(2n + 1)
Z
0
1
xn (1 − x)n dx.
Z 1
∞ X
n+k
Since
xn (1 − x)n dx converges uniformly for x ∈ [0, 1], we have
(2n + 1)
n
0
n=0
Ak =
Z
0
1
X
∞ n=0
n+k
n
n
(2n + 1)x (1 − x) dx.
n
It is well known that
∞ X
n+k
1
, for |u| < 1,
(1 − u)k+1
(6)
∞
X
(k + 1)u
n+k n
, for |u| < 1.
n
u =
k+2
(1
−
u)
n
n=0
(7)
n=0
n
un =
and
From (6) and (7) we have
Z 1
Z 1
x(1 − x)dx
dx
+ 2(k + 1)
Ak =
k+1
k+2
0 [1 − x(1 − x)]
0 [1 − x(1 − x)]
Z 1
Z 1
dx
dx
+ 2(k + 1)
.
= −(2k + 1)
k+1
k+2
0 [1 − x(1 − x)]
0 [1 − x(1 − x)]
Using the same method, we have
Z 1
Z 1
dx
dx
Bk = −(2k + 1)
+
2(k
+
1)
.
k+1
k+2
0 [1 + x(1 − x)]
0 [1 + x(1 − x)]
3
It follows from (2) and (3) that
Ak
Bk
Z
4 2k + 1 1
dx
=
+
k+1
3
3
0 [1 − x(1 − x)]
Z 1
dx
4 2k + 1
−
.
=
k+1
5
5
0 [1 + x(1 − x)]
(8)
(9)
From (8) and (9) we have recurrence relations (4) and (5).
Corollary 3. Let m be a positive integer. Define
∞
∞
n+k
X
X
(−1)n n+k
n
n
and Bk,m =
.
Ak,m =
2mn
2mn
mn
n=0
mn
n=0
Then
1
dx
Ak,m = −(2mk + 2m − 1)
m
m k+1
0 [1 − x (1 − x) ]
Z 1
dx
+2m(k + 1)
m
m k+2
0 [1 − x (1 − x) ]
Z 1
dx
Bk,m = −(2mk + 2m − 1)
m
m k+1
0 [1 + x (1 − x) ]
Z 1
dx
+2m(k + 1)
.
m
m k+2
0 [1 + x (1 − x) ]
Z
Theorem 4. Suppose that
n+k
n
2n
n
n
n=1
∞
X
Ck =
Then we have
k+1
Ck
Dk
1X
=
2 i=1
Z
0
1
and
Dk =
n+k−1
k
2n
2
n
n
n=1
∞
X
.
dx
,
[1 − x(1 − x)]i
(10)
k Z
dx
1 X 1
,
=
2k i=1 0 [1 − x(1 − x)]i
(11)
1
7k + 5
4k + 2
+
Ck −
Ck−1 ,
3(k + 1) 3(k + 1)
3(k + 1)
1
7k − 2
2(2k − 1)(k − 1)
=
+
Dk −
Dk−1 .
3k(k + 1) 3(k + 1)
3k(k + 1)
Ck+1 =
(12)
Dk+1
(13)
Proof. It is evident that
n+k
∞
Ck
1X
n
=
2 n=1 (2n − 1)
Dk =
1
2
2n−2
n−1
∞
n+k−1
X
k
2n−2 .
n(2n
−
1)
n−1
n=1
4
It follows from (1) that
Ck
Dk
Z 1
Z ∞ ∞ 1X n+k
1 1 X n + k n−1
n−1
n−1
=
x (1 − x)n−1 dx.
x (1 − x) dx =
n
n
2 n=1
2
0
0 n=1
Z 1X
∞
n+k−1
1
k
=
xn−1 (1 − x)n−1 dx.
2 0 n=1
n
Owing to (6), (10) holds.
We note that
n+k−1
k
∞
X
n
n=1
1
1
u =
− 1 , for |u| < 1.
k (1 − u)k
n
(14)
It follows from (14) that
1
Dk =
2k
Z
1
0
1
1
− 1 dx.
x(1 − x) [1 − x(1 − x)]k
Hence, (11) holds.
It is clear that
1
=
2
Z
=
Z
1
dx
k+2
0 [1 − x(1 − x)]
1
1
dx
k+1
0 [1 − x(1 − x)]
Ck+1 − Ck
2(k + 1)Dk+1 − 2kDk
Using (2), we can obtain (12)-(13).
Remark: By using the same method, the reader can consider the sums
∞
∞
n n+k−1
X
X
(−1)
(−1)n n+k
k
n
.
and
2n
2 2n
n
n
n
n
n=1
n=1
Theorem 5. Let
Ek =
∞
X
n=0
Then
1
2n
n
n+k
n
,
Fk =
∞
X
n=1
∞
n
X
1
,
G
=
k
2n n+k
n
5
n
1
n2 2n
n
n=1
n+k
n
.
Gk
k X
k
(−3)i dk,i
(−3) dk−1,i + 4(2k − 3)
i
i
i=0
i=0
Z
k
1
X k
(1 − y)k dy
−8k
(−3)i dk+1,i + 24
(4 − y)2
i
0
i=0
Z 1
(1 − y)k dy
+(4k − 2)
,
4−y
0
Z 1
k k X
X
k
(1 − y)k
k
i
i
= −8
dy
(−3) dk+1,i + 8
(−3) dk,i + 4
4−y
i
i
0
i=0
i=0
k X
k
1
+ 16(k + 1)
(−3)i dk+2,i+1 ,
−
k+1
i
i=0
k k X
X
k
k
i
(−3)i dk+2,i+1 ,
(−3) gk+2,i + 16
= 16(k + 1)
i
i
i=0
i=0
Ek = 12
Fk
k X
k
i
(15)
(16)
(17)
where
dm,i =
Z
1
√
3
0
u2i arctan u
du, gm,i =
(1 + u2 )m
1
√
3
Z
0
u2i+1 (arctan u)2
du,
(1 + u2 )m
and they satisfy the equations
dm,i + dm,i+1 = dm−1,i ,
gm,i + gm,i+1 = gm−1,i .
(18)
(19)
Proof. It follows from (1) that
Ek
Z
∞
X
=
(2n + 1)(n + k + 1)
=
n=0
∞
X
(2n + 1)(n + 1 + k)
n=0
∞
X
= 2
1
n
0
Z
0
n2
n=0
1
Z
0
+(2k + 3)
+(k + 1)
Z
1Z
0
1
0
y n (1 − y)k dy
1
xn (1 − x)n y n (1 − y)k dxdy
xn (1 − x)n y n (1 − y)k dxdy
n
Z
1
Z
1
0
0
n=0
Z
Z
∞
X 1 1
n
n=0
x (1 − x) dx
Z
1
0
∞
X
n
0
0
xn (1 − x)n y n (1 − y)k dxdy
x (1 − x)n y n (1 − y)k dxdy.
By using (6)-(7) and
∞
X
n=0
n 2 un =
1
3
2
−
+
,
1 − u (1 − u)2 (1 − u)3
6
(20)
we obtain
1
1
Z 1Z 1
(1 − y)k
(1 − y)k
Ek = 4
dxdy
+
(2k
−
3)
dxdy
3
2
0
0 [1 − x(1 − x)y]
0
0 [1 − x(1 − x)y]
Z 1Z 1
(1 − y)k
−k
dxdy.
0
0 1 − x(1 − x)y
Z 1
dx
. One can verify that
When |y| ≤ 1, put In (y) =
n
0 [1 − x(1 − x)y]
Z
Z
In+1 (y) =
2
2(2n − 1)
In (y) +
.
n(4 − y)
n(4 − y)
(21)
From (21), we have
Z 1
6
6
1
k
Ek = 4
(1 − y)
dy
I1 (y) +
+
(4 − y)2
(4 − y)2 4 − y
0
Z 1
Z 1
2
2
k
+(2k − 3)
(1 − y)
dy − k
I1 (y) +
(1 − y)k I1 (y)dy.
4−y
4−y
0
0
Noting that
I1 (y) = p
Ek
Put
r
Ek
4
y(4 − y)
arctan
r
y
,
4−y
1
r
Z 1
(1 − y)k
y
(1 − y)k
p
arctan
= 96
dy + 24
dy
2
2
4−y
y(4 − y)
0 (4 − y)
0 (4 − y)
r
Z 1
Z 1
y
(1 − y)k
(1 − y)k
p
dy + 8(2k − 3)
dy
+(4k − 2)
arctan
4−y
4−y
0 (4 − y) y(4 − y)
0
r
Z 1
y
(1 − y)k
p
−4k
dy.
arctan
4−y
y(4 − y)
0
Z
4u2
8udu
y
= u. Then we get y =
, dy =
, and
2
4−y
1+u
(1 + u2 )2
1
√
3
Z √1
2 k
3 (1 − 3u ) arctan u
(1 − 3u2 )k arctan u
= 12
du
+
4(2k
−
3)
du
(1 + u2 )k−1
(1 + u2 )k
0
0
Z √1
Z 1
Z 1
2 k
3 (1 − 3u ) arctan u
(1 − y)k dy
(1 − y)k dy
−8k
du
+
24
+
(4k
−
2)
.
(1 + u2 )k+1
(4 − y)2
4−y
0
0
0
Z
Hence (15) holds.
7
Using a similar approach, we have
Z 1Z 1
∞
X
Fk =
2n
(xy)n (1 − x)n (1 − y)k dxdy
0
n=1
0
+(2k + 3)
+(k + 1)
∞ Z 1
X
n=1
1Z
0
n=1
∞
X
1
n
Z
0
1
Z
(xy)n (1 − x)n (1 − y)k dxdy
0
1
Z
1
0
(xy)n (1 − x)n (1 − y)k dxdy
1
Z 1Z 1
(1 − y)k xy(1 − x)
(1 − y)k xy(1 − x)
= 2
dxdy
+
(2k
+
3)
dxdy
[1 − x(1 − x)y]2
1 − xy(1 − x)
0
0
0
0
Z 1Z 1
(1 − y)k ln[1 − xy(1 − x)]dxdy
−(k + 1)
0
0
Z 1
Z 1
Z 1
(1 − y)k
1
(1 − y)k
k
(1 − y) I1 (y)dy + 4
= −
I1 (y)dy + 4
dy −
4−y
4−y
k+1
0
0
0
Z 1
k+1
(1 − y)k yI1 (y)dy
+
2
0
Z √1
Z √1
2 k
2 k
3 (1 − 3u ) arctan u
3 (1 − 3u ) arctan u
du + 8
du
= −8
(1 + u2 )k+1
(1 + u2 )k
0
0
Z 1
Z √1 2
2 k
3 u (1 − 3u ) arctan u
(1 − y)k
1
+4
dy −
+ 16(k + 1)
du.
4−y
k+1
(1 + u2 )k+2
0
0
Z
Then (16) holds.
Now we prove (17). After calculus, we have
Z 1
Z
Z
1 1
ln[1 − x(1 − x)y]
k+1 1
k
(1 − y) dy
dx +
y(1 − y)k I1 (y)dy.
Gk =
2
−x(1
−
x)
2
0
0
0
Let
h(y) =
Z
1
0
We can verify that
ln[1 − x(1 − x)y]
dx, 0 ≤ y ≤ 1.
−x(1 − x)
1
1
dx
0 1 − x(1 − x)y
r
r
y
y
4
=
arctan
,
y 4−y
4−y
2
r
y
h(y) = 4 arctan
.
4−y
′
h (y) =
Z
Then
Gk = 2(k + 1)
Z
1
0
(1 − y)
k
arctan
r
8
y
4−y
2
1
dy +
2
Z
0
1
y(1 − y)k I1 (y)dy.
By the proof of (15), we get
Gk = 16(k + 1)
Z
1
√
3
0
(1 − 3u2 )k u(arctan u)2
du + 16
(1 + u2 )k+2
Z
√1
3
0
(1 − 3u2 )k u2 arctan u
du.
(1 + u2 )k+2
Then (17) holds. The proofs of (18)-(19) are omitted here.
∞
X
Now, we express the series
1
by means of some double integrations.
2n 2
n=0
Theorem 6. For the series
∞
X
n=0
∞
X
n=0
n
1
, we have
2n 2
n
Z 1Z 1
1
dxdy
−8
dxdy
2
0
0 [1 − x(1 − x)y(1 − y)]
0
0 1 − x(1 − x)y(1 − y)
Z 1Z 1
1
+8
dxdy.
(22)
3
0
0 [1 − x(1 − x)y(1 − y)]
1
=
2n 2
Z
n
1
Z
1
Proof. It follows from (1) that
∞
X
n=0
Z
∞
X
1
2
(4n + 4n + 1)
=
2n 2
n
=
n=0
∞
X
1
n
x (1 − x) dx
0
(4n2 + 4n + 1)
Z
0
n=0
n
1
Z
Z
0
1
y n (1 − y)n dy
1
0
xn (1 − x)n y n (1 − y)n dxdy.
From (6)-(7) and (20) we can prove that (23) holds.
Finally, we discuss the computation of the series
Let
ψ(s) =
∞
X
1
.
ns 2n
n
n=1
.
ns 2n
n
n=1
One knows that (see [3])
ψ(1) =
1
∞
X
√
3π
π2
17π 4
, ψ(2) = , and ψ(4) =
.
9
18
3240
But we do not know the accurate value of ψ(s) (s = 3 or s ≥ 5). Now we give the approximate
value of ψ(s). It is clear that
∞
∞
X
X
3s 3s
1
1
1
1
−s
<
1 + s + s + ···
=
=3
ψ(s) − −
s
s 2n
2 6 × 2s
n
4
5
n
n
n=3
n=3
Z ∞ s 3
3
−s
−s
.
1+
1+
< 3
dx = 3
xs
s−1
3
9
Then we have
lim s
s→+∞
3
1
1
ψ(s) − −
2 6 × 2s
Using a similar approach, we have
3
lim s ψ(s − ψ(s + 1) −
s→+∞
= 0.
(23)
(24)
1
6 × 2s+1
= 0.
(23)-(24) provide the following simple approximate
ψ(s) ≈
1
1
+
,
2 6 × 2s
ψ(s + 1) ≈ ψ(s) −
(25)
1
.
6 × 2s+1
(26)
By (25)-(26), we obtain
ψ(6) ≈
π2
1
17π 4
1
193
, ψ(3) ≈
− , ψ(5) ≈
−
.
384
18 48
3240 192
Similarly, we get
∞
X
n=1
1
1
≈
+
,
2
4 36 × 2s
ns 2n
n
1
∞
X
1
s 2n r
n
n
n=1
≈
1
1
+ s
, r ≥ 3.
r
2
2 × 6r
References
[1] S. Amghibech, On sums involving binomial coefficients. J. Integer Sequences 10 (2007),
Article 07.2.1.
[2] M. R. Andrew, Sums of the inverses of binomial coefficients. Fib. Quart. 19 (1981),
433–437.
[3] L. Comtet, Advanced Combinatorics. Reidel, 1974.
[4] C. Elsner, On recurrence formula for sums involving binomial coefficients. Fib. Quart.
43 (2005), 31–45.
[5] P. Juan, The sum of inverses of binomial coefficients revisited. Fib. Quart. 35 (1997),
342–345.
[6] P. Nicolae, Problem C:1280. Gaz. Mat. 97 (1992), 230.
[7] A. Sofo, General properties involving reciprocals of binomial coefficients. J. Integer Sequences 9 (2004), Article 06.4.5.
[8] R. Sprugnoli, Sums of the reciprocals of central binomial coefficients. Integers 6 (2006),
A27.
10
[9] B. Sury, Sum of the reciprocals of the binomial coefficients. European J. Combin. 14
(1993), 351–353.
[10] B.
Sury,
Tianming
Wang,
and
Feng-Zhen
Zhao.
Some identities involving reciprocals of binomial coefficients. J. Integer Sequences 7
(2004), Article 04.2.8.
[11] T. Tiberiu, Combinatorial sums and series involving inverses of binomial coefficients.
Fib. Quart. 38 (2000), 79–84.
[12] WMC Problems Group. Problem 10494. Amer. Math. Monthly 103(1996), 74.
[13] Jin-Hua Yang and Feng-Zhen Zhao.
Sums involving the inverses of binomial coefficients. J. Integer Sequences Vol.9 (2006),
Article 06.4.2.
[14] Feng-Zhen Zhao and Tianming Wang. Some results for sums of the inverses of binomial
coefficients. Integers 5 (2005), A22.
2000 Mathematics Subject Classification: Primary 11B65.
Keywords: binomial coefficient, integral, recurrence relation, generating function.
Received March 15 2007; revised version received August 16 2007. Published in Journal of
Integer Sequences, August 16 2007.
Return to Journal of Integer Sequences home page.
11