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.

For nonnegative integers m and n, the binomial coefficient \binom{n}{m} is defined by

\binom{n}{m} = \begin{cases}\frac{n!}{m!(n − m)!}, & if n ≥ m, \\ 0, & if n < m. \end{cases}

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])

B(n, m) = ∫_{0}^{1} t^{n−1}(1 − t)^{m−1}dt

for all positive integers n and m. Since B(n, m) = \frac{Γ(n)Γ(m)}{Γ(n + m)} = \frac{(n − 1)!(m − 1)!}{(n + m − 1)!}, the

binomial coefficient \binom{n}{m} satisfies the equation

\binom{n}{m}^{−1} = (n + 1) ∫_{0}^{1} t^{m}(1 − t)^{n−m}dt.  (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=0}^{∞} \frac{\binom{n+k}{n}f_{n}}{\binom{2n}{n}} and ∑_{n=0}^{∞} \frac{f_{n}}{\binom{2n}{n}\binom{n+k}{n}},

where f_{n} is a rational function of n. In Section 2, we derive some relations for ∑_{n=0}^{∞} \frac{\binom{n+k}{n}f_{n}}{\binom{2n}{n}} by

means of (1) and other integrals. In the meantime, we express the series ∑_{n=0}^{∞} \frac{1}{\binom{2n}{n}2} by some

double integrations. In addition, we discuss the approximate value of the series ∑_{n=1}^{∞} \frac{1}{n^{s}\binom{2n}{n}},

where s is a positive integer.

2 Main Results

Lemma 1. For the integrals ∫_{0}^{1} \frac{dx}{[1 − x(1 − x)]^{k+1}} and ∫_{0}^{1} \frac{dx}{[1 + x(1 − x)]^{k+1}} , we have

∫_{0}^{1} \frac{dx}{[1 − x(1 − x)]^{k+2}} = \frac{2}{3(k + 1)} + \frac{2(2k + 1)}{3(k + 1)} ∫_{0}^{1} \frac{dx}{[1 − x(1 − x)]^{k+1}} ,  (2)

∫_{0}^{1} \frac{dx}{[1 + x(1 − x)]^{k+2}} = \frac{2}{5(k + 1)} + \frac{2(2k + 1)}{5(k + 1)} ∫_{0}^{1} \frac{dx}{[1 + x(1 − x)]^{k+1}} .  (3)

Proof. The proofs of (2)-(3) are simple and are omitted here. Theorem 2. Assume that

A_{k} = ∑_{n=0}^{∞} \begin{matrix}( & {n+k \atop n} & ) \\ ( & {2n \atop n} & ) \end{matrix} and B_{k} = ∑_{n=0}^{∞} \frac{(−1)^{n}\binom{n+k}{n}}{\binom{2n}{n}},

where k is a nonnegative integer. Then we have the following recurrence relations:

A_{k+1} = −\frac{2}{3(k + 1)} + \frac{2(2k + 3)}{3(k + 1)}A_{k},  (4)
B_{k+1} = −\frac{2}{5(k + 1)} + \frac{2(2k + 3)}{5(k + 1)}B_{k}.  (5)

Proof. It follows from (1) that

A_{k} = ∑_{n=0}^{∞} \binom{n + k}{n}(2n + 1) ∫_{0}^{1} x^{n}(1 − x)^{n}dx.

Since ∑_{n=0}^{∞} \binom{n + k}{n}(2n + 1) ∫_{0}^{1} x^{n}(1 − x)^{n}dx converges uniformly for x ∈ [0, 1], we have

A_{k} = ∫_{0}^{1} [ ∑_{n=0}^{∞} \binom{n + k}{n}(2n + 1)x^{n}(1 − x)^{n}]dx.

It is well known that

∑_{n=0}^{∞} \binom{n + k}{n}u^{n} = \frac{1}{(1 − u)^{k+1}} , for |u| < 1,  (6)

and

∑_{n=0}^{∞} n\binom{n + k}{n}u^{n} = \frac{(k + 1)u}{(1 − u)^{k+2}} , for |u| < 1.  (7)

From (6) and (7) we have

A_{k} = ∫_{0}^{1} \frac{dx}{[1 − x(1 − x)]^{k+1}} + 2(k + 1) ∫_{0}^{1} \frac{x(1 − x)dx}{[1 − x(1 − x)]^{k+2}}
= −(2k + 1) ∫_{0}^{1} \frac{dx}{[1 − x(1 − x)]^{k+1}} + 2(k + 1) ∫_{0}^{1} \frac{dx}{[1 − x(1 − x)]^{k+2}} .

Using the same method, we have

B_{k} = −(2k + 1) ∫_{0}^{1} \frac{dx}{[1 + x(1 − x)]^{k+1}} + 2(k + 1) ∫_{0}^{1} \frac{dx}{[1 + x(1 − x)]^{k+2}} .

It follows from (2) and (3) that

A_{k} = \frac{4}{3} + \frac{2k + 1}{3} ∫_{0}^{1} \frac{dx}{[1 − x(1 − x)]^{k+1}}  (8)
B_{k} = \frac{4}{5} − \frac{2k + 1}{5} ∫_{0}^{1} \frac{dx}{[1 + x(1 − x)]^{k+1}} .  (9)

From (8) and (9) we have recurrence relations (4) and (5). Corollary 3. Let m be a positive integer. Define

A_{k,m} = ∑_{n=0}^{∞} \begin{matrix}( & {n+k \atop n} & ) \\ ( & {2mn \atop mn} & ) \end{matrix} and B_{k,m} = ∑_{n=0}^{∞} \frac{(−1)^{n}\binom{n+k}{n}}{\binom{2mn}{mn}}.

Then

A_{k,m} = −(2mk + 2m − 1) ∫_{0}^{1} \frac{dx}{[1 − x^{m}(1 − x)^{m}]^{k+1}}

+2m(k + 1) ∫_{0}^{1} \frac{dx}{[1 − x^{m}(1 − x)^{m}]^{k+2}}

B_{k,m} = −(2mk + 2m − 1) ∫_{0}^{1} \frac{dx}{[1 + x^{m}(1 − x)^{m}]^{k+1}}

+2m(k + 1) ∫_{0}^{1} \frac{dx}{[1 + x^{m}(1 − x)^{m}]^{k+2}} .

Theorem 4. Suppose that

C_{k} = ∑_{n=1}^{∞} \frac{\binom{n+k}{n}}{n\binom{2n}{n}} and D_{k} = ∑_{n=1}^{∞} \frac{\binom{n+k−1}{k}}{n^{2}\binom{2n}{n}}.

Then we have

C_{k} = \frac{1}{2} ∑_{i=1}^{k+1} ∫_{0}^{1} \frac{dx}{[1 − x(1 − x)]^{i}} ,  (10)
D_{k} = \frac{1}{2k} ∑_{i=1}^{k} ∫_{0}^{1} \frac{dx}{[1 − x(1 − x)]^{i}} ,  (11)
C_{k+1} = \frac{1}{3(k + 1)} + \frac{7k + 5}{3(k + 1)}C_{k} − \frac{4k + 2}{3(k + 1)}C_{k−1},  (12)
D_{k+1} = \frac{1}{3k(k + 1)} + \frac{7k − 2}{3(k + 1)}D_{k} − \frac{2(2k − 1)(k − 1)}{3k(k + 1)}D_{k−1}.  (13)

Proof. It is evident that

C_{k} = \frac{1}{2} ∑_{n=1}^{∞} \frac{\binom{n+k}{n}}{(2n − 1)\binom{2n−2}{n−1}}
D_{k} = \frac{1}{2} ∑_{n=1}^{∞} \frac{\binom{n+k−1}{k}}{n(2n − 1)\binom{2n−2}{n−1}}.

It follows from (1) that

C_{k} = \frac{1}{2} ∑_{n=1}^{∞} \binom{n + k}{n} ∫_{0}^{1} x^{n−1}(1 − x)^{n−1}dx = \frac{1}{2} ∫_{0}^{1} ∑_{n=1}^{∞} \binom{n + k}{n}x^{n−1}(1 − x)^{n−1}dx.

D_{k} = \frac{1}{2} ∫_{0}^{1} ∑_{n=1}^{∞} \frac{\binom{n+k−1}{k}}{n}x^{n−1}(1 − x)^{n−1}dx.

Owing to (6), (10) holds.

We note that

∑_{n=1}^{∞} \frac{\binom{n+k−1}{k}}{n}u^{n} = \frac{1}{k}[\frac{1}{(1 − u)^{k}} − 1], for |u| < 1.  (14)

It follows from (14) that

D_{k} = \frac{1}{2k} ∫_{0}^{1} \frac{1}{x(1 − x)}\{\frac{1}{[1 − x(1 − x)]^{k}} − 1\}dx.

Hence, (11) holds.

It is clear that

C_{k+1} − C_{k} = \frac{1}{2} ∫_{0}^{1} \frac{dx}{[1 − x(1 − x)]^{k+2}}
2(k + 1)D_{k+1} − 2kD_{k} = ∫_{0}^{1} \frac{1}{[1 − x(1 − x)]^{k+1}} dx

Using (2), we can obtain (12)-(13).

Remark: By using the same method, the reader can consider the sums

∑_{n=1}^{∞} \frac{(−1)^{n}\binom{n+k}{n}}{n\binom{2n}{n}} and ∑_{n=1}^{∞} \frac{(−1)^{n}\binom{n+k−1}{k}}{n^{2}\binom{2n}{n}}.

Theorem 5. Let

E_{k} = ∑_{n=0}^{∞} \frac{1}{\binom{2n}{n}\binom{n+k}{n}}, F_{k} = ∑_{n=1}^{∞} \frac{1}{n\binom{2n}{n}\binom{n+k}{n}}, G_{k} = ∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2n}{n}\binom{n+k}{n}}.

Then

E_{k} = 12 ∑_{i=0}^{k} \binom{k}{i}(−3)^{i}d_{k−1,i} + 4(2k − 3) ∑_{i=0}^{k} \binom{k}{i}(−3)^{i}d_{k,i}

−8k ∑_{i=0}^{k} \binom{k}{i}(−3)^{i}d_{k+1,i} + 24 ∫_{0}^{1} \frac{(1 − y)^{k}dy}{(4 − y)^{2}}

+(4k − 2) ∫_{0}^{1} \frac{(1 − y)^{k}dy}{4 − y} ,  (15)

F_{k} = −8 ∑_{i=0}^{k} \binom{k}{i}(−3)^{i}d_{k+1,i} + 8 ∑_{i=0}^{k} \binom{k}{i}(−3)^{i}d_{k,i} + 4 ∫_{0}^{1} \frac{(1 − y)^{k}}{4 − y} dy

−\frac{1}{k + 1} + 16(k + 1) ∑_{i=0}^{k} \binom{k}{i}(−3)^{i}d_{k+2,i+1},  (16)

G_{k} = 16(k + 1) ∑_{i=0}^{k} \binom{k}{i}(−3)^{i}g_{k+2,i} + 16 ∑_{i=0}^{k} \binom{k}{i}(−3)^{i}d_{k+2,i+1},  (17)

where

d_{m,i} = ∫0^{\frac{1}{\sqrt{3}}} \frac{u^{2i} arctan u}{(1 + u^{2})^{m}}du, g_{m,i} = ∫0^{\frac{1}{\sqrt{3}}} \frac{u^{2i+1}(arctan u)^{2}}{(1 + u^{2})^{m}} du,

and they satisfy the equations

d_{m,i} + d_{m,i+1} = d_{m−1,i},  (18)
g_{m,i} + g_{m,i+1} = g_{m−1,i}.  (19)

Proof. It follows from (1) that

E_{k} = ∑_{n=0}^{∞}(2n + 1)(n + k + 1) ∫_{0}^{1} x^{n}(1 − x)^{n}dx ∫_{0}^{1} y^{n}(1 − y)^{k}dy

= ∑_{n=0}^{∞}(2n + 1)(n + 1 + k) ∫_{0}^{1} ∫_{0}^{1} x^{n}(1 − x)^{n}y^{n}(1 − y)^{k}dxdy

= 2 ∑_{n=0}^{∞} n^{2} ∫_{0}^{1} ∫_{0}^{1} x^{n}(1 − x)^{n}y^{n}(1 − y)^{k}dxdy

+(2k + 3) ∑_{n=0}^{∞} n ∫_{0}^{1} ∫_{0}^{1} x^{n}(1 − x)^{n}y^{n}(1 − y)^{k}dxdy

+(k + 1) ∑_{n=0}^{∞} ∫_{0}^{1} ∫_{0}^{1} x^{n}(1 − x)^{n}y^{n}(1 − y)^{k}dxdy.

By using (6)-(7) and

∑_{n=0}^{∞} n^{2}u^{n} = \frac{1}{1 − u} − \frac{3}{(1 − u)^{2}} + \frac{2}{(1 − u)^{3}} ,  (20)

we obtain

E_{k} = 4 ∫_{0}^{1} ∫_{0}^{1} \frac{(1 − y)^{k}}{[1 − x(1 − x)y]^{3}} dxdy + (2k − 3) ∫_{0}^{1} ∫_{0}^{1} \frac{(1 − y)^{k}}{[1 − x(1 − x)y]^{2}} dxdy

−k ∫_{0}^{1} ∫_{0}^{1} \frac{(1 − y)^{k}}{1 − x(1 − x)y} dxdy.

When |y| ≤ 1, put I_{n}(y) = ∫_{0}^{1} \frac{dx}{[1 − x(1 − x)y]^{n}}. One can verify that

I_{n+1}(y) = \frac{2(2n − 1)}{n(4 − y)}I_{n}(y) + \frac{2}{n(4 − y)}.  (21)

From (21), we have

E_{k} = 4 ∫_{0}^{1}(1 − y)^{k}[\frac{6}{(4 − y)^{2}} I_{1}(y) + \frac{6}{(4 − y)^{2}} + \frac{1}{4 − y} ]dy

+(2k − 3) ∫_{0}^{1}(1 − y)^{k}[\frac{2}{4 − y} I_{1}(y) + \frac{2}{4 − y} ]dy − k ∫_{0}^{1}(1 − y)^{k}I_{1}(y)dy.

Noting that

I_{1}(y) = \frac{4}{\sqrt{y(4 − y)}} arctan \sqrt{\frac{y}{4 − y}} ,

E_{k} = 96 ∫_{0}^{1} \frac{(1 − y)^{k}}{(4 − y)^{2}\sqrt{y(4 − y)}} arctan \sqrt{\frac{y}{4 − y}} dy + 24 ∫_{0}^{1} \frac{(1 − y)^{k}}{(4 − y)^{2}} dy

+(4k − 2) ∫_{0}^{1} \frac{(1 − y)^{k}}{4 − y} dy + 8(2k − 3) ∫_{0}^{1} \frac{(1 − y)^{k}}{(4 − y)\sqrt{y(4 − y)}} arctan \sqrt{\frac{y}{4 − y}} dy

−4k ∫_{0}^{1} \frac{(1 − y)^{k}}{\sqrt{y(4 − y)}} arctan \sqrt{\frac{y}{4 − y}} dy.

Put \sqrt{\frac{y}{4 − y}} = u. Then we get y = \frac{4u^{2}}{1 + u^{2}} , dy = \frac{8udu}{(1 + u^{2})^{2}} , and

E_{k} = 12 ∫0^{\frac{1}{\sqrt{3}}} \frac{(1 − 3u^{2})^{k} arctan u}{(1 + u^{2})^{k−1}}du + 4(2k − 3) ∫0^{\frac{1}{\sqrt{3}}} \frac{(1 − 3u^{2})^{k} arctan u}{(1 + u^{2})^{k}}du

−8k ∫0^{\frac{1}{\sqrt{3}}} \frac{(1 − 3u^{2})^{k} arctan u}{(1 + u^{2})^{k+1}}du + 24 ∫_{0}^{1} \frac{(1 − y)^{k}dy}{(4 − y)^{2}} + (4k − 2) ∫_{0}^{1} \frac{(1 − y)^{k}dy}{4 − y} .

Hence (15) holds.

Using a similar approach, we have

F_{k} = ∑_{n=1}^{∞} 2n ∫_{0}^{1} ∫_{0}^{1}(xy)^{n}(1 − x)^{n}(1 − y)^{k}dxdy

+(2k + 3) ∑_{n=1}^{∞} ∫_{0}^{1} ∫_{0}^{1}(xy)^{n}(1 − x)^{n}(1 − y)^{k}dxdy

+(k + 1) ∑_{n=1}^{∞} \frac{1}{n} ∫_{0}^{1} ∫_{0}^{1}(xy)^{n}(1 − x)^{n}(1 − y)^{k}dxdy

= 2 ∫_{0}^{1} ∫_{0}^{1} \frac{(1 − y)^{k}xy(1 − x)}{[1 − x(1 − x)y]^{2}}dxdy + (2k + 3) ∫_{0}^{1} ∫_{0}^{1} \frac{(1 − y)^{k}xy(1 − x)}{1 − xy(1 − x)}dxdy

−(k + 1) ∫_{0}^{1} ∫_{0}^{1}(1 − y)^{k} ln[1 − xy(1 − x)]dxdy

= − ∫_{0}^{1}(1 − y)^{k}I_{1}(y)dy + 4 ∫_{0}^{1} \frac{(1 − y)^{k}}{4 − y} I_{1}(y)dy + 4 ∫_{0}^{1} \frac{(1 − y)^{k}}{4 − y} dy − \frac{1}{k + 1}

+\frac{k + 1}{2} ∫_{0}^{1}(1 − y)^{k}yI_{1}(y)dy

= −8 ∫0^{\frac{1}{\sqrt{3}}} \frac{(1 − 3u^{2})^{k} arctan u}{(1 + u^{2})^{k+1}}du + 8 ∫0^{\frac{1}{\sqrt{3}}} \frac{(1 − 3u^{2})^{k} arctan u}{(1 + u^{2})^{k}}du

+4 ∫_{0}^{1} \frac{(1 − y)^{k}}{4 − y} dy − \frac{1}{k + 1} + 16(k + 1) ∫0^{\frac{1}{\sqrt{3}}} \frac{u^{2}(1 − 3u^{2})^{k} arctan u}{(1 + u^{2})^{k+2}}du.

Then (16) holds. Now we prove (17). After calculus, we have

G_{k} = \frac{k + 1}{2} ∫_{0}^{1}(1 − y)^{k}dy ∫_{0}^{1} \frac{ln[1 − x(1 − x)y]}{−x(1 − x)}dx + \frac{1}{2} ∫_{0}^{1} y(1 − y)^{k}I_{1}(y)dy.

Let

h(y) = ∫_{0}^{1} \frac{ln[1 − x(1 − x)y]}{−x(1 − x)}dx, 0 ≤ y ≤ 1.

We can verify that

h′(y) = ∫_{0}^{1} \frac{1}{1 − x(1 − x)y} dx
= \frac{4}{y}\sqrt{\frac{y}{4 − y}} arctan \sqrt{\frac{y}{4 − y}} ,
h(y) = 4( arctan \sqrt{\frac{y}{4 − y}} )^{2}.

Then

G_{k} = 2(k + 1) ∫_{0}^{1}(1 − y)^{k}( arctan \sqrt{\frac{y}{4 − y}} )^{2}dy + \frac{1}{2} ∫_{0}^{1} y(1 − y)^{k}I_{1}(y)dy.

By the proof of (15), we get

G_{k} = 16(k + 1) ∫0^{\frac{1}{\sqrt{3}}} \frac{(1 − 3u^{2})^{k}u(arctan u)^{2}}{(1 + u^{2})^{k+2}} du + 16 ∫0^{\frac{1}{\sqrt{3}}} \frac{(1 − 3u^{2})^{k}u^{2} arctan u}{(1 + u^{2})^{k+2}}du.

Then (17) holds. The proofs of (18)-(19) are omitted here.

Now, we express the series ∑_{n=0}^{∞} \frac{1}{\binom{2n}{n}2} by means of some double integrations.

Theorem 6. For the series ∑_{n=0}^{∞} \frac{1}{\binom{2n}{n}2} , we have

∑_{n=0}^{∞} \frac{1}{\binom{2n}{n}2} = ∫_{0}^{1} ∫_{0}^{1} \frac{dxdy}{1 − x(1 − x)y(1 − y)} − 8 ∫_{0}^{1} ∫_{0}^{1} \frac{1}{[1 − x(1 − x)y(1 − y)]^{2}} dxdy
+8 ∫_{0}^{1} ∫_{0}^{1} \frac{1}{[1 − x(1 − x)y(1 − y)]^{3}} dxdy.  (22)

Proof. It follows from (1) that

∑_{n=0}^{∞} \frac{1}{\binom{2n}{n}2} = ∑_{n=0}^{∞}(4n^{2} + 4n + 1) ∫_{0}^{1} x^{n}(1 − x)^{n}dx ∫_{0}^{1} y^{n}(1 − y)^{n}dy
= ∑_{n=0}^{∞}(4n^{2} + 4n + 1) ∫_{0}^{1} ∫_{0}^{1} x^{n}(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 ∑_{n=1}^{∞} \frac{1}{n^{s}\binom{2n}{n}}.

Let

ψ(s) = ∑_{n=1}^{∞} \frac{1}{n^{s}\binom{2n}{n}}.

One knows that (see [3])

ψ(1) = \frac{\sqrt{3}π}{9} , ψ(2) = \frac{π^{2}}{18}, and ψ(4) = \frac{17π^{4}}{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

ψ(s) − \frac{1}{2} − \frac{1}{6 × 2^{s}} = ∑_{n=3}^{∞} \frac{1}{n^{s}\binom{2n}{n}} < ∑_{n=3}^{∞} \frac{1}{n^{s}} = 3^{−s}(1 + \frac{3^{s}}{4^{s}} + \frac{3^{s}}{5^{s}} + ⋯)
< 3^{−s}(1 + ∫_{3}^{∞} \frac{3^{s}}{x^{s}} dx) = 3^{−s}(1 + \frac{3}{s − 1}).

Then we have

lim_{s→+∞} s^{3}(ψ(s) − \frac{1}{2} − \frac{1}{6 × 2^{s}} ) = 0.  (23)

Using a similar approach, we have

lim_{s→+∞} s^{3}(ψ(s − ψ(s + 1) − \frac{1}{6 × 2^{s+1}} ) = 0.  (24)

(23)-(24) provide the following simple approximate

ψ(s) ≈ \frac{1}{2} + \frac{1}{6 × 2^{s}} ,  (25)
ψ(s + 1) ≈ ψ(s) − \frac{1}{6 × 2^{s+1}} .  (26)

By (25)-(26), we obtain

ψ(6) ≈ \frac{193}{384}, ψ(3) ≈ \frac{π^{2}}{18} − \frac{1}{48}, ψ(5) ≈ \frac{17π^{4}}{3240} − \frac{1}{192}.

Similarly, we get

∑_{n=1}^{∞} \frac{1}{n^{s}\binom{2n}{n}^{2}} ≈ \frac{1}{4} + \frac{1}{36 × 2^{s}} , ∑_{n=1}^{∞} \frac{1}{n^{s}\binom{2n}{n}^{r}} ≈ \frac{1}{2^{r}} + \frac{1}{2^{s} × 6^{r}} , r ≥ 3.

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.

[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.

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