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2
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Journal of Integer Sequences, Vol. 9 (2006),
Article 06.4.2
23 11
Sums Involving the Inverses of Binomial
Coefficients
Jin-Hua Yang
Department of Applied Mathematics
Dalian University of Technology
Dalian, 116024
China
and
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 compute certain sums involving the inverses of binomial coefficients. We derive the recurrence formulas for certain infinite sums related to the
inverses of binomial coefficients.
1
Introduction
As usual, the binomial coefficient
¡n¢
m
µ ¶
n
=
m
is defined by
(
n!
, if n ≥ m;
m!(n−m)!
0,
if n < m;
1
where n and m are nonnegative integers.
There are many identities involving binomial coefficients. However, computations related
to the inverses of binomial coefficients are difficult. For some results involving the inverses
of binomial coefficients, see [5, 4, 1, 6, 8, 9, 7, 10]. In order to compute sums involving the
inverses of binomial coefficients, using integrals is an effective approach. This idea is based
on Euler’s well-known Beta function defined by (see [8])
Z 1
B(n, m) =
tn−1 (1 − t)m−1 dt
0
(n − 1)!(m − 1)!
Γ(n)Γ(m)
=
, the inverse
for all positive integers n and m. Since B(n, m) =
Γ(n + m)
(n + m − 1)!
¡ n ¢−1
satisfies the identity
binomial coefficient m
µ ¶−1
Z 1
n
= (n + 1)
tm (1 − t)n−m dt.
m
0
(1)
with this method, a series of identities related to the inverses of binomial coefficients is
obtained (see [8, 9, 7, 10]). In this paper, we also use Eq. (1) to evaluate the sums
∞
X
n=1
∞
X
n=1
∞
X
εn
¡ ¢,
n(n + k) 2n
n
n=1
εn
¡ ¢,
n2 (n + k) 2n
n
∞
X
εn
εn
¡2n+k¢, and
¡2n+2k¢,
n(n + k) n
n(n
+
k)
n+k
n=1
where |ε| = 1, and k is an arbitrary positive integer with k > 1. For convenience, we put
S1 (k) =
T1 (k) =
Q1 (k) =
∞
X
n=1
∞
X
n=1
∞
X
n=1
R1 (k) =
∞
X
n=1
1
¡ ¢,
n(n + k) 2n
n
S2 (k) =
∞
X
(−1)n
¡ ¢,
n(n + k) 2n
n
n=1
∞
X
1
¡ ¢,
2
n (n + k) 2n
n
T2 (k) =
1
¡
¢,
n(n + k) 2n+k
n
n=1
∞
X
(−1)n
¡2n+k¢ ,
n(n
+
k)
n
n=1
∞
n
X
(−1)
¢.
¡
R2 (k) =
n(n + k) 2n+2k
n+k
n=1
Q2 (k) =
1
¢,
¡
n(n + k) 2n+2k
n+k
(−1)n
¡ ¢,
n2 (n + k) 2n
n
In the next section, we evaluate the sums above. In the third section, we define Wk =
∞
r
X
1
1X
¡2n¢ and Xr =
Wk ; our aim is to compute lim Xr .
r→+∞
r k=1
nk n
n=1
2
2
Some Results For Si(k) And Ti(k) (1 ≤ i ≤ 2)
Theorem 2.1. Let k be a positive integer with k ≥ 2. Then
1 − 2k
S1 (k) =
k
1
Z
ln[1 − t(1 − t)] +
tk (1
0
1
i=1
− t)k
ti (1 − t)i /i
ln[1 + t(1 − t)] +
(1 − 2k)
S2 (k) = (−1)k
k
√ 0
µ
¶
2 √
5+1
5 ln
−
−1 ,
k
2
S1 (k)
π2
−
,
T1 (k) =
18k
k
µ √
¶2
5−1
2
S2 (k)
.
T2 (k) = −
ln
−
k
2
k
Z
k
X
k
X
tk (1
i=1
√ ¶
µ
3π
1
dt +
2−
,
k
3
(−1)i ti (1 − t)i /i
− t)k
(2)
dt
(3)
(4)
(5)
Proof. It follows from Eq. (1) that
¶X
∞ R1 n
t (1 − t)n dt
1
0
tn (1 − t)n dt + 2 −
k n=1
n+k
0
R
¶ ∞
µ
Z
1
∞
1 X 0 tn−k (1 − t)n−k dt
1X1 1 n
n
,
t (1 − t) dt + 2 −
=
k n=1 n 0
k n=k+1
n
∞
1X1
S1 (k) =
k n=1 n
∞
1 X (−1)n
S2 (k) =
k n=1
Z
R1
0
1
µ
R1
¶ ∞
µ
tn (1 − t)n dt
1 X (−1)n−k 0 tn−k (1 − t)n−k dt
+ 2−
.
n
k n=k+1
n
It is well known that
∞
X
un
n
n=1
= − ln(1 − u), for u ∈ [−1, 1).
(6)
On the other hand,
√
1
3π
ln(1 − t + t )dt = −2 +
,
3
0
√
¶
µ
Z 1
√
5+1
2
−1 .
ln(1 + t − t )dt = 2
5 ln
2
0
Z
2
From Eqs. (6-8) we get Eqs. (2) and (3).
3
(7)
(8)
Now we give the proofs of Eqs. (4-5).
One can verify that
∞
∞
∞
1
1X
1
1X 1
1X 1
¡2n¢ −
¡2n¢ − S1 (k),
¡2n¢ =
T1 (k) =
2
2
k n=1 n n
k n=1 n(n + k) n
k n=1 n n
k
∞
1
1 X (−1)n
¡2n¢ − S2 (k).
T2 (k) =
2
k n=1 n n
k
It follows from [2, 10] that
∞
X
n=1
π2
1
¡2n¢ = ,
18
n2 n
Hence Eqs. (4-5) are valid.
µ √
¶2
∞
X
5−1
(−1)n
¡ ¢ = −2 ln
.
2
n2 2n
n
n=1
Next, we extend Si (k) and Ti (k)(i = 1, 2) to the following forms:
S1 (k, m) =
∞
X
n=1
T1 (k, m) =
∞
X
n=1
1
¡ ¢,
n(n + k) 2mn
mn
S2 (k, m) =
∞
X
n=1
(−1)n
¡ ¢,
n(n + k) 2mn
mn
∞
X
1
(−1)n
¡2mn¢ , and T2 (k, m) =
¡ ¢.
2 (n + k) 2mn
n2 (n + k) mn
n
mn
n=1
we give the corresponding results as a corollary:
Corollary 2.1. Let k and m be positive integers. Then
m
S1 (k, m) =
µ
¶Z
1
m
ln[1 − t (1 − t) ] +
k
X
tmi (1 − t)mi /i
1
i=1
− 2m
mk
k
t (1 − t)mk
0
Z 1
1
−
ln[1 − tm (1 − t)m ]dt,
k 0
m
µ
¶Z
1
m
ln[1 + t (1 − t) ] +
(9)
k
X
(−1)i tmi (1 − t)mi /i
1
i=1
− 2m
S2 (k, m) = (−1)k
mk (1 − t)mk
k
t
0
Z
1 1
−
ln[1 + tm (1 − t)m ]dt,
k 0
Z
m 1 ln[1 − tm (1 − t)m ]dt 1
− S1 (k, m),
T1 (k, m) = −
2k 0
t(1 − t)
k
Z 1
m
m
ln[1 + t (1 − t) ]dt 1
m
− S2 (k, m).
T2 (k, m) = −
2k 0
t(1 − t)
k
4
dt
dt
(10)
(11)
(12)
Proof. We only give the proofs of Eqs. (11-12), and leave the proofs of Eqs. (9-10) to the
reader. We can immediately obtain that
∞
∞
∞
1
1
1
1X
1
1X
1X
¡2mn¢ −
¡2mn¢ − S1 (k, m),
¡2mn¢ =
T1 (k, m) =
2
2
k n=1 n mn
k n=1 n(n + k) mn
k n=1 n mn
k
∞
1
1 X (−1)n
¡2mn¢ − S2 (k, m).
T2 (k, m) =
2
k n=1 n mn
k
Owing to the conclusions (see [10]):
∞
X
n=1
m
1
¡2mn¢ = −
2
2
n mn
Z
1
0
ln[1 − tm (1 − t)m ]dt
t(1 − t)
Z
∞
X
m 1 ln[1 + tm (1 − t)m ]dt
(−1)n
¡2mn¢ = −
,
2 0
t(1 − t)
n2 mn
n=1
we can show that Eqs. (11-12) hold.
It is evident that Eqs. (9-12) are the generalizations of Eqs. (2-5), respectively. Now we
give the recurrence relation for Si (k) and Ti (k).
Theorem 2.2. Let k be a positive integer with k ≥ 2. Then
√
3π
2(2k + 1)
1
S1 (k + 1) =
S1 (k) +
−
,
2
k+1
(k + 1)
3(k + 1)
√
√
5+1
1
2 5
2(2k + 1)
S2 (k) +
−
ln
,
S2 (k + 1) = −
2
k+1
(k + 1)
k+1
2
√
3π
1
(3k + 1)π 2 2k(2k + 1)
+
T1 (k) −
+
,
T1 (k + 1) = −
2
2
3
18(k + 1)
(k + 1)
(k + 1)
3(k + 1)2
¶2
µ √
5−1
2(5k + 3)
2k(2k + 1)
1
T2 (k + 1) = −
ln
−
T2 (k) −
2
2
(k + 1)
2
(k + 1)
(k + 1)3
√
√
5+1
2 5
+
ln
.
2
(k + 1)
2
Proof. For 0 < a ≤ 1, we consider the integrals:
Ik (a) =
µ
Jk (a) = (−1)k
k
X
ai ti (1 − t)i /i
¶ Z 1 ln[1 − at(1 − t)] +
1
i=1
−2
dt
k (1 − t)k
k
t
0
µ
k
X
(−1)i ai ti (1 − t)i /i
¶ Z 1 ln[1 + at(1 − t)] +
1
i=1
−2
dt,
k
k
t (1 − t)k
0
5
(13)
(14)
(15)
(16)
where Ik (0) = 0 and Jk (0) = 0. It is clear that
√
√ ¶
µ
µ
¶
3π
5+1
2 √
1
5 ln
2−
and S2 (k) = Jk (1) −
−1 .
S1 (k) = Ik (1) +
k
3
k
2
When 0 < a ≤ 1, we have
¶
µ
¶
Z 1
dt
1
k−1
.
1−
=
−2 a
2
k
0 1 − at + at
r
r
¶
µ
¶µ
a
a
1
k−1
k−2
,
=
−2 a
− 4a
arctan
k
4−a
4−a
¶Z 1 k
µ
a t(1 − t)dt
1
′
−2
Jk (a) =
k
0 1 + at(1 − t)
r
a+4
µ
¶µ
√
+ 1¶
k−2
a
1
2a
a
k−1
ln r
.
=
−2 a
− √
k
a+4
a+4
−1
a
Ik′ (a)
µ
Hence
(2k − 1)ak 4(2k − 1)
Ik (a) = −
+
k2
k
Jk (a) =
Let u =
r
a
and
4−a
Ik (a)
Jk (a)
r
a
a
arctan
da.
a
4−a
4−a
r
a+4
r
Z
+1
k
a
(2k − 1)a
2(2k − 1)
a
k−2
da.
−
+
ln r
a
k2
k
a+4
a+4
−1
a
r
a+4
v=
. Then we get
a
Z 2k−2
u
arctan u
(2k − 1)ak (2k − 1)22k+1
+
du,
= −
2
k
k
(1 + u2 )k
Z
(2k − 1)ak (2k − 1)4k
1
v+1
= −
−
ln
dv.
2
2
k
k
k
(v − 1)
v−1
Z
k−2
r
It is well known that
Z 2k
Z 2k−2
u arctan u
2k − 1
u2k−1 arctan u
u
arctan u
du
=
du
−
(1 + u2 )k+1
2k
(1 + u2 )k
2k(1 + u2 )k
Z
u2k−1
1
du,
+
2k
(1 + u2 )k+1
Z
Z
1
1
v+1
2k − 1
v+1
1
ln
dv = −
ln
dv + 2 2
2
k+1
2
k
(v − 1)
v−1
2k
(v − 1)
v−1
2k (v − 1)k
v+1
v
ln
.
−
2
k
2k(v − 1)
v−1
6
(17)
(18)
In the meantime, we note that
Z
Z
2
u2k−1
du = k+1 ak−1 da and Ik (0) = Jk (0) = 0.
(1 + u2 )k+1
4
Therefore Ik (a) and Jk (a) satisfy that
√
r
ak+1
4ak 4 4 − aak−1/2
k+1
a
Ik+1 (a) = 2Ik (a) −
+
−
arctan
,
2k + 1
k+1
k
k
4−a
√
√
√
a+4+ a
k+1
ak+1
4ak 2ak−1/2 a + 4
Jk+1 (a) = −2Jk (a) −
−
+
ln √
√ .
2k + 1
k+1
k
k
a+4− a
(19)
(20)
From Eqs. (19-20) we can derive Eqs. (13-14). According to Eqs. (4-5) we can obtain Eqs.
(15-16).
We note that the recurrences given in Eqs. (13-14) are similar to [3, Eq. (28)].
Theorem 2.3. Let k be a positive integer with k ≥ 2. Then
k
X
ti (1 − t)i /i
¶ Z 1 ln[1 − t(1 − t)] +
1
i=1
−1
dt
Q1 (k) =
k
k
t
0
µ
¶Z 1
1
− 1+
(1 − t)k ln[1 − t(1 − t)]dt,
k
0
k
X
(−1)i ti (1 − t)i /i
µ
¶ Z 1 ln[1 + t(1 − t)] +
1
i=1
Q2 (k) = (−1)k
−1
dt
k
k
t
0
µ
¶Z 1
1
− 1+
(1 − t)k ln[1 + t(1 − t)]dt,
k
0
¾
Z 1½
k
X
ti (1 − t)i
1
dt
ln[1 − t(1 − t)] +
R1 (k) =
k 0
i
i=1
µ
¶Z 1
1
− 2+
tk (1 − t)k ln[1 − t(1 − t)]dt,
k
0
µ
¾
Z ½
k
X
(−1)k 1
(−1)i ti (1 − t)i
R2 (k) =
dt
ln[1 + t(1 − t)] +
k
i
0
i=1
¶Z 1
µ
1
tk (1 − t)k ln[1 + t(1 − t)]dt.
− 2+
k
0
7
(21)
(22)
(23)
(24)
Proof. We only give the proof of Eq. (21). The proofs of Eqs. (22-24) follow the same pattern
and are omitted here. It follows from Eq. (1) and Eq. (6) that
Z
∞
X
2n + k + 1
1
tn (1 − t)n+k dt
n(n
+
k)
0
n=1
µ
¶ X
¶ ∞ Z
µ
∞ Z 1 n−k
1
t (1 − t)n
1 X 1 tn (1 − t)n+k
=
1−
dt + 1 +
dt.
k n=k+1 0
n
k n=1 0
n
Q1 (k) =
k
X
ti (1 − t)i /i
¶Z 1
1
i=1
=
−1
dt
k
k
t
0
¶Z 1
µ
1
(1 − t)k ln[1 − t(1 − t)]dt.
− 1+
k
0
µ
ln[1 − t(1 − t)] +
Hence Eq. (21) holds.
By computing integrals of Eqs. (17-18) and Eqs. (21-24), we can establish a series of
identities involving inverses of binomial coefficients. In the final part of this section, we
evaluate special cases of Si (k), Ti (k), and Ri (k)(1 ≤ i ≤ 2) according to the particular
choice of k. For example, when k = 2 in Eqs. (17-18), we have
Z 2
3a2
u arctan udu
I2 (a) = −
+ 48
4
(1 + u2 )2
12
3a2 24u arctan u
−
+ 12(arctan u)2 −
+ c2
= −
2
4
1+u
1 + u2
r
r
µ
¶2
p
3a2
a
a
− 6 a(4 − a) arctan
+ 12 arctan
= −
− 12 + 3a + c2 .
4
4−a
4−a
Z
3a2
1
v+1
J2 (a) = −
− 24
ln
dv
4
(v 2 − 1)2 v − 1
¶
µ
3a2
v+1
12v
v+1
12
2
= −
+ c′2
+ 2
ln
− 2
− 3 ln
4
v −1 v−1 v −1
v−1
√
µ√
√ ¶
√
2
p
a+4+ a
3a
a+4+ a
2
+ 3 a(a + 4) ln √
= −
+ c′2 .
√ − 3a − 3 ln √
√
4
a+4− a
a+4− a
Since I2 (0) = 0 and J2 (0) = 0, then c2 = 12, c′2 = 0,
√
15
9 √
π2
I2 (1) = − 3π + , J2 (1) = − + 6 5 ln
4
3
4
8
√
5+1
− 12 ln2
2
µ√
¶
5+1
.
2
Hence, we get
√
√
7 3π 13 5π 2
13 7 3π π 2
−
+ , T1 (2) =
−
−
,
S1 (2) =
4
6
3
12
8
36
√
¶
µ√
√
5+1
5+1
11
2
S2 (2) = − + 5 5 ln
,
− 12 ln
4
2
2
√
√
µ√
¶
5+1
5+1
11 5 5
2
−
ln
+ 5 ln
.
T2 (2) =
8
2
2
2
By means of Si (2), Ti (2), and Eqs. (13-16), we can compute other values of Si (k) and
Ti (k)(1 ≤ i ≤ 2, k > 2).
If k = 2 in Eqs. (23-24), we can obtain
¸
Z ·
t2 (1 − t)2
1 1
2
ln(1 − t + t ) + t(1 − t) +
dt
R1 (2) =
2 0
2
Z
5 1 2
−
t (1 − t)2 ln(1 − t + t2 )dt
2 0
√
3π
17
−
,
=
36Z ·12
¸
1 1
t2 (1 − t)2
2
R2 (2) =
dt
ln(1 + t − t ) − t(1 − t) +
2 0
2
Z
5 1 2
t (1 − t)2 ln(1 + t − t2 )dt
−
2 0
√
√
5−1
1
5 5
=
+
ln
.
36
6
2
3
The Value of lim Xr
r→+∞
√
3π
π2
17π 4
, W2 =
, and W4 =
(see [2]). However, we do not
We know that W1 =
9
18
3240
know how to evaluate Wk in closed form when k ≥ 5. In this section, we are interested in
the average Xr of Wk . We compute lim Xr by Eq. (1).
r→+∞
1
Theorem 3.1. Let r be a positive integer with r > 4. Then lim Xr = .
r→+∞
2
Proof. One can verify that
1 1 X 1 − n1r
¡ ¢.
Xr = +
2 r n=2 (n − 1) 2n
n
∞
9
Since 1 −
1
nr+1
< 1, the series
∞
X
n=2
∞
X
n=2
It follows from Eq. (1) that
∞
X
n=2
Then
∞
X
n=2
1
n¡r ¢ satisfies that
(n − 1) 2n
n
1−
X
1 − n1r
1
¡2n¢ ≤
¡ ¢.
(n − 1) n
(n − 1) 2n
n
n=2
∞
R1
∞
X
(2n + 1) 0 tn (1 − t)n dt
1
¡ ¢=
.
n−1
(n − 1) 2n
n
n=2
R1
∞
X
(2n + 3) 0 tn+1 (1 − t)n+1 dt
1
¡ ¢ =
n
(n − 1) 2n
n
n=1
Z 1 2
Z 1
t (1 − t)2 dt
= 2
t(1 − t) ln[1 − t(1 − t)]dt.
−3
2
0 [1 − t(1 − t)]
0
1
Hence lim Xr = .
r→+∞
2
4
Acknowledgments
The authors are grateful to the referee for a careful reading and for numerous suggestions,
all of which have helped improve the presentation of this article.
References
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433–437.
[2] L. Comtet, Advanced Combinatorics, Reidel, 1974.
[3] C. Elsner, On recurrence formulae for sums involving binomial coefficients, Fibonacci
Quart. 43 (2005), 31-45.
[4] P. Juan, The sum of inverses of binomial coefficients revisited, Fibonacci Quart. 35
(1997), 342–345.
[5] P. Nicolae, Problem C:1280, Gaz. Mat. 97 (1992), 230.
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[7] B. Sury, Tianming Wang, and Feng-Zhen Zhao, Some identities involving reciprocals of
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2000 Mathematics Subject Classification: Primary 11B65.
Keywords: binomial coefficients, integral, recurrence relation.
Received June 26 2006; revised version received September 3 2006. Published in Journal of
Integer Sequences, September 3 2006.
Return to Journal of Integer Sequences home page.
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