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 \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}

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])

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 inverse

binomial coefficient \binom{n}{m}^{−1} satisfies the identity

\binom{n}{m}^{−1} = (n + 1) ∫_{0}^{1} t^{m}(1 − t)^{n−m}dt.  (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

∑_{n=1}^{∞} \frac{ε^{n}}{n(n + k)\binom{2n}{n}}, ∑_{n=1}^{∞} \frac{ε^{n}}{n^{2}(n + k)\binom{2n}{n}},

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

where |ε| = 1, and k is an arbitrary positive integer with k > 1. For convenience, we put

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

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

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

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

In the next section, we evaluate the sums above. In the third section, we define W_{k} =

∑_{n=1}^{∞} \frac{1}{n^{k}\binom{2n}{n}} and X_{r} = \frac{1}{r} ∑_{k=1}^{r} W_{k}; our aim is to compute lim_{r→+∞} X_{r}.

2 Some Results For S_{i}(k) And T_{i}(k) (1 ≤ i ≤ 2)

Theorem 2.1. Let k be a positive integer with k ≥ 2. Then

S_{1}(k) = \frac{1 − 2k}{k} ∫_{0}^{1} \frac{ln[1 − t(1 − t)] + ∑_{i=1}^{k} t^{i}(1 − t)^{i}/i}{t^{k}(1 − t)^{k}}dt + \frac{1}{k}(2 − \frac{\sqrt{3}π}{3} ),  (2)
S_{2}(k) = (−1)^{k} \frac{(1 − 2k)}{k} ∫_{0}^{1} \frac{ln[1 + t(1 − t)] + ∑_{i=1}^{k}(−1)^{i}t^{i}(1 − t)^{i}/i}{t^{k}(1 − t)^{k}}dt
−\frac{2}{k}(\sqrt{5} ln \frac{\sqrt{5} + 1}{2} − 1),  (3)
T_{1}(k) = \frac{π^{2}}{18k} − \frac{S_{1}(k)}{k},  (4)
T_{2}(k) = −\frac{2}{k}( ln \frac{\sqrt{5} − 1}{2})^{2} − \frac{S_{2}(k)}{k}.  (5)

Proof. It follows from Eq. (1) that

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

It is well known that

∑_{n=1}^{∞} \frac{u^{n}}{n} = − ln(1 − u), for u ∈ [−1, 1).  (6)

On the other hand,

∫_{0}^{1} ln(1 − t + t^{2})dt = −2 + \frac{\sqrt{3}π}{3} ,  (7)
∫_{0}^{1} ln(1 + t − t^{2})dt = 2(\sqrt{5} ln \frac{\sqrt{5} + 1}{2} − 1).  (8)

From Eqs. (6-8) we get Eqs. (2) and (3).

Now we give the proofs of Eqs. (4-5). One can verify that

T_{1}(k) = \frac{1}{k} ∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2n}{n}} − \frac{1}{k} ∑_{n=1}^{∞} \frac{1}{n(n + k)\binom{2n}{n}} = \frac{1}{k} ∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2n}{n}} − \frac{1}{k}S_{1}(k),

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

It follows from [2, 10] that

∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2n}{n}} = \frac{π^{2}}{18}, ∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}\binom{2n}{n}} = −2( ln \frac{\sqrt{5} − 1}{2})^{2} .

Hence Eqs. (4-5) are valid.

Next, we extend S_{i}(k) and T_{i}(k)(i = 1, 2) to the following forms:

S_{1}(k, m) = ∑_{n=1}^{∞} \frac{1}{n(n + k)\binom{2mn}{mn}}, S_{2}(k, m) = ∑_{n=1}^{∞} \frac{(−1)^{n}}{n(n + k)\binom{2mn}{mn}},

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

we give the corresponding results as a corollary:

Corollary 2.1. Let k and m be positive integers. Then

S_{1}(k, m) = (\frac{1}{k} − 2m) ∫_{0}^{1} \frac{ln[1 − t^{m}(1 − t)^{m}] + ∑_{i=1}^{k} t^{mi}(1 − t)^{mi}/i}{t^{mk}(1 − t)^{mk}}dt
−\frac{1}{k} ∫_{0}^{1} ln[1 − t^{m}(1 − t)^{m}]dt,  (9)

S_{2}(k, m) = (−1)^{k}(\frac{1}{k} − 2m) ∫_{0}^{1} \frac{ln[1 + t^{m}(1 − t)^{m}] + ∑_{i=1}^{k}(−1)^{i}t^{mi}(1 − t)^{mi}/i}{t^{mk}(1 − t)^{mk}}dt
−\frac{1}{k} ∫_{0}^{1} ln[1 + t^{m}(1 − t)^{m}]dt,  (10)
T_{1}(k, m) = −\frac{m}{2k} ∫_{0}^{1} \frac{ln[1 − t^{m}(1 − t)^{m}]dt}{t(1 − t)} − \frac{1}{k}S_{1}(k, m),  (11)
T_{2}(k, m) = −\frac{m}{2k} ∫_{0}^{1} \frac{ln[1 + t^{m}(1 − t)^{m}]dt}{t(1 − t)} − \frac{1}{k}S_{2}(k, m).  (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

T_{1}(k, m) = \frac{1}{k} ∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2mn}{mn}} − \frac{1}{k} ∑_{n=1}^{∞} \frac{1}{n(n + k)\binom{2mn}{mn}} = \frac{1}{k} ∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2mn}{mn}} − \frac{1}{k}S_{1}(k, m),

T_{2}(k, m) = \frac{1}{k} ∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}\binom{2mn}{mn}} − \frac{1}{k}S_{2}(k, m).

Owing to the conclusions (see [10]):

∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2mn}{mn}} = −\frac{m}{2} ∫_{0}^{1} \frac{ln[1 − t^{m}(1 − t)^{m}]dt}{t(1 − t)}

∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}\binom{2mn}{mn}} = −\frac{m}{2} ∫_{0}^{1} \frac{ln[1 + t^{m}(1 − t)^{m}]dt}{t(1 − t)},

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 S_{i}(k) and T_{i}(k).

Theorem 2.2. Let k be a positive integer with k ≥ 2. Then

S_{1}(k + 1) = \frac{2(2k + 1)}{k + 1}S_{1}(k) + \frac{1}{(k + 1)^{2}} − \frac{\sqrt{3}π}{3(k + 1)},  (13)
S_{2}(k + 1) = −\frac{2(2k + 1)}{k + 1}S_{2}(k) + \frac{1}{(k + 1)^{2}} − \frac{2\sqrt{5}}{k + 1} ln \frac{\sqrt{5} + 1}{2},  (14)
T_{1}(k + 1) = −\frac{(3k + 1)π^{2}}{18(k + 1)^{2}} + \frac{2k(2k + 1)}{(k + 1)^{2}}T_{1}(k) − \frac{1}{(k + 1)^{3}} + \frac{\sqrt{3}π}{3(k + 1)^{2}} ,  (15)
T_{2}(k + 1) = −\frac{2(5k + 3)}{(k + 1)^{2}}( ln \frac{\sqrt{5} − 1}{2})^{2} − \frac{2k(2k + 1)}{(k + 1)^{2}}T_{2}(k) − \frac{1}{(k + 1)^{3}}
+\frac{2\sqrt{5}}{(k + 1)^{2}} ln \frac{\sqrt{5} + 1}{2}.  (16)

Proof. For 0 < a ≤ 1, we consider the integrals:

I_{k}(a) = (\frac{1}{k} − 2) ∫_{0}^{1} \frac{ln[1 − at(1 − t)] + ∑_{i=1}^{k} a^{i}t^{i}(1 − t)^{i}/i}{t^{k}(1 − t)^{k}}dt

J_{k}(a) = (−1)^{k}(\frac{1}{k} − 2) ∫_{0}^{1} \frac{ln[1 + at(1 − t)] + ∑_{i=1}^{k}(−1)^{i}a^{i}t^{i}(1 − t)^{i}/i}{t^{k}(1 − t)^{k}}dt,

where I_{k}(0) = 0 and J_{k}(0) = 0. It is clear that

S_{1}(k) = I_{k}(1) + \frac{1}{k}(2 − \frac{\sqrt{3}π}{3} ) and S_{2}(k) = J_{k}(1) − \frac{2}{k}(\sqrt{5} ln \frac{\sqrt{5} + 1}{2} − 1).

When 0 < a ≤ 1, we have

I′_{k}(a) = (\frac{1}{k} − 2)a^{k−1}(1 − ∫_{0}^{1} \frac{dt}{1 − at + at^{2}} ).
= (\frac{1}{k} − 2)(a^{k−1} − 4a^{k−2}\sqrt{\frac{a}{4 − a}} arctan \sqrt{\frac{a}{4 − a}}),
J′_{k}(a) = (\frac{1}{k} − 2) ∫_{0}^{1} \frac{a^{k}t(1 − t)dt}{1 + at(1 − t)}

= (\frac{1}{k} − 2)(a^{k−1} − \frac{2a^{k−2}\sqrt{a}}{\sqrt{a + 4}} ln \frac{\sqrt{\frac{a + 4}{a}} + 1}{\sqrt{\frac{a + 4}{a}} − 1}).

Hence

I_{k}(a) = −\frac{(2k − 1)a^{k}}{k^{2}} + \frac{4(2k − 1)}{k} ∫ a^{k−2}\sqrt{\frac{a}{4 − a}} arctan \sqrt{\frac{a}{4 − a}}da.

J_{k}(a) = −\frac{(2k − 1)a^{k}}{k^{2}} + \frac{2(2k − 1)}{k} ∫ a^{k−2}\sqrt{\frac{a}{a + 4}} ln \frac{\sqrt{\frac{a + 4}{a}} + 1}{\sqrt{\frac{a + 4}{a}} − 1}da.

Let u = \sqrt{\frac{a}{4 − a}} and v = \sqrt{\frac{a + 4}{a}}. Then we get

I_{k}(a) = −\frac{(2k − 1)a^{k}}{k^{2}} + \frac{(2k − 1)2^{2k+1}}{k} ∫ \frac{u^{2k−2} arctan u}{(1 + u^{2})^{k}}du,  (17)
J_{k}(a) = −\frac{(2k − 1)a^{k}}{k^{2}} − \frac{(2k − 1)4^{k}}{k} ∫ \frac{1}{(v^{2} − 1)^{k}} ln \frac{v + 1}{v − 1}dv.  (18)

It is well known that

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

+\frac{1}{2k} ∫ \frac{u^{2k−1}}{(1 + u^{2})^{k+1}} du,

∫ \frac{1}{(v^{2} − 1)^{k+1}} ln \frac{v + 1}{v − 1}dv = −\frac{2k − 1}{2k} ∫ \frac{1}{(v^{2} − 1)^{k}} ln \frac{v + 1}{v − 1}dv + \frac{1}{2k^{2}(v^{2} − 1)^{k}}

−\frac{v}{2k(v^{2} − 1)^{k}} ln \frac{v + 1}{v − 1}.

In the meantime, we note that

∫ \frac{u^{2k−1}}{(1 + u^{2})^{k+1}} du = \frac{2}{4^{k+1}} ∫ a^{k−1}da and I_{k}(0) = J_{k}(0) = 0.

Therefore I_{k}(a) and J_{k}(a) satisfy that

\frac{k + 1}{2k + 1}I_{k+1}(a) = 2I_{k}(a) − \frac{a^{k+1}}{k + 1} + \frac{4a^{k}}{k} − \frac{4\sqrt{4 − a}a^{k−1/2}}{k} arctan \sqrt{\frac{a}{4 − a}},  (19)
\frac{k + 1}{2k + 1}J_{k+1}(a) = −2J_{k}(a) − \frac{a^{k+1}}{k + 1} − \frac{4a^{k}}{k} + \frac{2a^{k−1/2}\sqrt{a + 4}}{k} ln \frac{\sqrt{a + 4} + \sqrt{a}}{\sqrt{a + 4} − \sqrt{a}}.

(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

Q_{1}(k) = (\frac{1}{k} − 1) ∫_{0}^{1} \frac{ln[1 − t(1 − t)] + ∑_{i=1}^{k} t^{i}(1 − t)^{i}/i}{t^{k}}dt
−(1 + \frac{1}{k}) ∫_{0}^{1}(1 − t)^{k} ln[1 − t(1 − t)]dt,  (21)
Q_{2}(k) = (−1)^{k}(\frac{1}{k} − 1) ∫_{0}^{1} \frac{ln[1 + t(1 − t)] + ∑_{i=1}^{k}(−1)^{i}t^{i}(1 − t)^{i}/i}{t^{k}}dt
−(1 + \frac{1}{k}) ∫_{0}^{1}(1 − t)^{k} ln[1 + t(1 − t)]dt,  (22)
R_{1}(k) = \frac{1}{k} ∫_{0}^{1} \{ ln[1 − t(1 − t)] + ∑_{i=1}^{k} \frac{t^{i}(1 − t)^{i}}{i} \}dt
−(2 + \frac{1}{k}) ∫_{0}^{1} t^{k}(1 − t)^{k} ln[1 − t(1 − t)]dt,  (23)
R_{2}(k) = \frac{(−1)^{k}}{k} ∫_{0}^{1} \{ ln[1 + t(1 − t)] + ∑_{i=1}^{k} \frac{(−1)^{i}t^{i}(1 − t)^{i}}{i} \}dt
−(2 + \frac{1}{k}) ∫_{0}^{1} t^{k}(1 − t)^{k} ln[1 + t(1 − t)]dt.  (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

Q_{1}(k) = ∑_{n=1}^{∞} \frac{2n + k + 1}{n(n + k)} ∫_{0}^{1} t^{n}(1 − t)^{n+k}dt

= (1 − \frac{1}{k}) ∑_{n=k+1}^{∞} ∫_{0}^{1} \frac{t^{n−k}(1 − t)^{n}}{n} dt + (1 + \frac{1}{k}) ∑_{n=1}^{∞} ∫_{0}^{1} \frac{t^{n}(1 − t)^{n+k}}{n} dt.

= (\frac{1}{k} − 1) ∫_{0}^{1} \frac{ln[1 − t(1 − t)] + ∑_{i=1}^{k} t^{i}(1 − t)^{i}/i}{t^{k}}dt

−(1 + \frac{1}{k}) ∫_{0}^{1}(1 − t)^{k} ln[1 − t(1 − t)]dt.

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 S_{i}(k), T_{i}(k), and R_{i}(k)(1 ≤ i ≤ 2) according to the particular choice of k. For example, when k = 2 in Eqs. (17-18), we have

I_{2}(a) = −\frac{3a^{2}}{4} + 48 ∫ \frac{u^{2} arctan udu}{(1 + u^{2})^{2}}
= −\frac{3a^{2}}{4} − \frac{24u arctan u}{1 + u^{2}} + 12(arctan u)^{2} − \frac{12}{1 + u^{2}} + c_{2}
= −\frac{3a^{2}}{4} − 6\sqrt{a(4 − a)} arctan \sqrt{\frac{a}{4 − a}} + 12( arctan \sqrt{\frac{a}{4 − a}})^{2} − 12 + 3a + c_{2}.
J_{2}(a) = −\frac{3a^{2}}{4} − 24 ∫ \frac{1}{(v^{2} − 1)^{2}} ln \frac{v + 1}{v − 1}dv
= −\frac{3a^{2}}{4} + \frac{12v}{v^{2} − 1} ln \frac{v + 1}{v − 1} − \frac{12}{v^{2} − 1} − 3 ln^{2} (\frac{v + 1}{v − 1}) + c′_{2}
= −\frac{3a^{2}}{4} + 3\sqrt{a(a + 4)} ln \frac{\sqrt{a + 4} + \sqrt{a}}{\sqrt{a + 4} − \sqrt{a}} − 3a − 3 ln^{2} (\frac{\sqrt{a + 4} + \sqrt{a}}{\sqrt{a + 4} − \sqrt{a}}) + c′_{2}.

Since I_{2}(0) = 0 and J_{2}(0) = 0, then c_{2} = 12, c′_{2} = 0,

I_{2}(1) = \frac{9}{4} − \sqrt{3}π + \frac{π^{2}}{3} , J_{2}(1) = −\frac{15}{4} + 6\sqrt{5} ln \frac{\sqrt{5} + 1}{2} − 12 ln^{2} (\frac{\sqrt{5} + 1}{2}).

Hence, we get

S_{1}(2) = \frac{13}{4} − \frac{7\sqrt{3}π}{6} + \frac{π^{2}}{3} , T_{1}(2) = \frac{7\sqrt{3}π}{12} − \frac{13}{8} − \frac{5π^{2}}{36} ,
S_{2}(2) = −\frac{11}{4} + 5\sqrt{5} ln \frac{\sqrt{5} + 1}{2} − 12 ln^{2} (\frac{\sqrt{5} + 1}{2}),
T_{2}(2) = \frac{11}{8} − \frac{5\sqrt{5}}{2} ln \frac{\sqrt{5} + 1}{2} + 5 ln^{2} (\frac{\sqrt{5} + 1}{2}).

By means of S_{i}(2), T_{i}(2), and Eqs. (13-16), we can compute other values of S_{i}(k) and T_{i}(k)(1 ≤ i ≤ 2, k > 2).

If k = 2 in Eqs. (23-24), we can obtain

R_{1}(2) = \frac{1}{2} ∫_{0}^{1} [ ln(1 − t + t^{2}) + t(1 − t) + \frac{t^{2}(1 − t)^{2}}{2} ]dt

−\frac{5}{2} ∫_{0}^{1} t^{2}(1 − t)^{2} ln(1 − t + t^{2})dt

= \frac{17}{36} − \frac{\sqrt{3}π}{12} ,
R_{2}(2) = \frac{1}{2} ∫_{0}^{1} [ ln(1 + t − t^{2}) − t(1 − t) + \frac{t^{2}(1 − t)^{2}}{2} ]dt

−\frac{5}{2} ∫_{0}^{1} t^{2}(1 − t)^{2} ln(1 + t − t^{2})dt

= \frac{1}{36} + \frac{5\sqrt{5}}{6} ln \frac{\sqrt{5} − 1}{2}.

3 The Value of lim_{r→+∞} X_{r}

We know that W_{1} = \frac{\sqrt{3}π}{9} , W_{2} = \frac{π^{2}}{18}, and W_{4} = \frac{17π^{4}}{3240} (see [2]). However, we do not know how to evaluate W_{k} in closed form when k ≥ 5. In this section, we are interested in the average X_{r} of W_{k}. We compute lim_{r→+∞} X_{r} by Eq. (1).

Theorem 3.1. Let r be a positive integer with r > 4. Then lim_{r→+∞} X_{r} = \frac{1}{2}.

Proof. One can verify that

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

Since 1 − \frac{1}{n^{r+1}} < 1, the series ∑_{n=2}^{∞} \frac{1 − \frac{1}{n^{r}}}{(n − 1)\binom{2n}{n}} satisfies that

∑_{n=2}^{∞} \frac{1 − \frac{1}{n^{r}}}{(n − 1)\binom{2n}{n}} ≤ ∑_{n=2}^{∞} \frac{1}{(n − 1)\binom{2n}{n}}.

It follows from Eq. (1) that

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

Then

∑_{n=2}^{∞} \frac{1}{(n − 1)\binom{2n}{n}} = ∑_{n=1}^{∞} \frac{(2n + 3) ∫_{0}^{1} t^{n+1}(1 − t)^{n+1}dt}{n}

= 2 ∫_{0}^{1} \frac{t^{2}(1 − t)^{2}dt}{[1 − t(1 − t)]^{2}} − 3 ∫_{0}^{1} t(1 − t) ln[1 − t(1 − t)]dt.

Hence lim_{r→+∞} X_{r} = \frac{1}{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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[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.

[6] B. Sury, Sum of the reciprocals of the binomial coefficients, European J. Combin. 14 (1993), 351–353.

[7] B. Sury, Tianming Wang, and Feng-Zhen Zhao, Some identities involving reciprocals of binomial coefficients, J. Integer Sequences 7 (2004), Article 04.2.8.

[8] T. Tiberiu, Combinatorial sums and series involving inverses of binomial coefficients, Fibonacci Quart. 38 (2000), 79–84.

[9] WMC Problems Group, Problem 10494, Amer. Math. Monthly 103 (1996), 74.

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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.

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