Notes on Sums Involving the Inverses of Binomial Coefficients

Jin-Hua Yang

Department of Applied Mathematics, Dalian University of Technology, Dalian,

116024, China Zhoukou Normal University, Zhoukou, 466001, China

Feng-Zhen Zhao Department of Applied Mathematics, Dalian University of Technology, Dalian 116024, China.

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. keywords: binomial coefficients, integral, recurrence relation. AMS Classification Number: 11B65.

1. Introduction

As usual, the binomial coefficient \binom{n}{m} is defined by

\binom{n}{m} = \{ \begin{matrix}\frac{n!}{m!(n − m)!}, n ≥ m, \\ 0, n < m, \end{matrix}

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 references [2]–[9]. Computing sums involving the inverses of binomial coefficients, integral is an effective approach. It is based on Euler’s well-known Beta function defined by (see reference [6])

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 that

\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 are obtained (see references [6]-[9]). In this paper, we also use (1) to discuss the computation of 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 calculate above sums. In the third section, we denote

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 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 (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 (6-8) we have (2) and (3). Now we give the proofs of (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 references [1] and [9] 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 (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 : 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 (11-12), and leave the proofs of (9-10) to the readers. 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 reference [9]):

∑_{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 (11-12) hold.

It is evident that (9-12) are the generalizations of (2-5), respectively. Now we give the recurrence relation for S_{i}(k) and T_{i}(k). Theorem 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 (19-20) we can derive (13-14). According to (4-5) we can obtain (15-16).

We note that the recurrences (13-14) are similar to (28) of reference [10]. Theorem 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 (21). The proofs of (22-24) follow the same pattern and are omitted here. It follows from (1) and (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 (21) holds.

By computing integrals of (17-18) and (21-24), we can establish a series of identities involving inverses of binomial coefficients. In the final of this section, we give 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 (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 (13-16), we can compute other values of S_{i}(k) and T_{i}(k)(1 ≤ i ≤ 2, k > 2).

If k = 2 in (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 reference [1]). However we cannot give the accurate value of W_{k} when k ≥ 5. In this section, we are interested in the average X_{r} of W_{k}. We compute lim_{r→+∞} X_{r} by (1).

Theorem 4: 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 (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}.

Acknowledgment The authors wish to thank the anonymous referee for his/her valuable suggestions for this paper.

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