Alternating Sums of the Reciprocals of Binomial Coefficients

Hacène Belbachir and Mourad Rahmani University of Sciences and Technology Houari Boumediene

Faculty of Mathematics P. O. Box 32 El Alia, Bab-Ezzouar 16111 Algiers Algeria

    hacenebelbachir@gmail.com
        mrahmani@usthb.dz

B. Sury Statistics & Mathematics Unit Indian Statistical Institute 8th Mile Mysore Road Bangalore 560059 India

        sury@isibang.ac.in

Abstract

We investigate summations of the form ∑_{0≤k≤n} (−1)^{k} k^{m}\binom{n}{k}^{−1}. We give closed formulae in terms of the Akiyama-Tanigawa matrix. Recurrence formulae, ordinary generating functions and some other results are also given.

1 Introduction and notation

Binomial coefficients play an important role in many areas of mathematics, such as combinatorics, number theory and special functions. The inverse binomial coefficients have an integral representation in terms of beta function

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

and one can obtain (see [18])

∑_{k=0}^{n}\binom{n}{k}^{−1} = \frac{n + 1}{2^{n+1}} ∑_{k=1}^{n+1} \frac{2^{k}}{k} .,

There are many papers dealing with sums involving inverses of binomial coefficients, see for instance [3, 11, 12, 14, 15, 17, 18, 21, 22]. For nonnegative integers n, m and p we consider the sums

T_{n}^{(m,p)} := ∑_{k=0}^{n} (−1)^{k} k^{m}\binom{p + n}{p + k}^{−1}.  (2)

These sums have been studied by many authors. Trif [20], using (1) proved for m = 0 that

T_{n}^{(0,p)} = ((−1)^{n} + \binom{p + n + 1}{p}^{−1}) \frac{p + n + 1}{p + n + 2}.  (3)

Sury, Wang and Zhao [19], studied (2) for m = 1 and m = 2, they obtain

T_{n}^{(1,p)} = \frac{p + n + 1}{p + n + 3}(\frac{(−1)^{n}(n + 1)(p + n + 3)}{p + n + 2} − \binom{p + n + 2}{p + 1}^{−1} − (−1)^{n}),  (4)

and

T_{n}^{(2,p)} = (p + n + 1)(\frac{(−1)^{n}(n + 1)^{2}}{p + n + 2} − \frac{(−1)^{n}(2n + 3)}{p + n + 3}
+ \frac{2}{p + n + 4}(\binom{p + n + 3}{p + 2}^{−1} + (−1)^{n}) − \frac{1}{p + n + 3}\binom{p + n + 2}{p + 1}^{−1}).  (5)

Our aim is to give a closed form and recurrence relation for the sums (2). In order to investigate the summation of the form S_{n}^{(m)} := T_{n}^{(m,0)} and T_{n}^{(m,p)}, we shall use the following tools [1, 6]:

• The Stirling numbers of the first kind s(n, k) (see A008275 in [13]), are defined by the generating function

x (x − 1) ⋯ (x − n + 1) = ∑_{k≥0}s(n, k)x^{k},

and satisfy the recurrence relation

s(n + 1, k) = s(n, k − 1) − ns(n, k), (1 ≤ k ≤ n) ,

with s(n, n) = 1, s(n, 0) = 0 for n ≥ 1 and s(n, k) = 0 for k < 0 or k > n.

• The Stirling numbers of the second kind \left\{{n \atop k}\right\} (see A008277 in [13]), are defined by the generating function

∏_{j=1}^{k} \frac{x}{1 − jx} = ∑_{n≥k}\left\{{n \atop k}\right\}x^{n},

and satisfy the recurrence relation

\left\{{n + 1 \atop k}\right\} = \left\{{n \atop k − 1}\right\} + k\left\{{n \atop k}\right\},

with \left\{{n \atop 1}\right\} = \left\{{n \atop n}\right\} = 1.

They also verify the following important identity

x^{n} = ∑_{k=0}^{n} (−1)^{n+k} \left\{{n \atop k}\right\}x (x + 1) ⋯ (x + k − 1) .  (6)

• The Eulerian numbers \left⟨{n \atop k}\right⟩ (see A008292 in [13]) are defined by

\left⟨{n \atop k}\right⟩ = ∑_{i=0}^{k} (−1)^{i} (k − i)^{n} \binom{n + 1}{i}, (1 ≤ k ≤ n) ,

and satisfy the recursive identity

\left⟨{n \atop k}\right⟩ = k\left⟨{n − 1 \atop k}\right⟩ + (n − k + 1) \left⟨{n − 1 \atop k − 1}\right⟩,

with \left⟨{1 \atop 1}\right⟩ = 1.

• The Worpitzky numbers W_{n,k} (see A028246 in [13]), are defined by

W_{n,k} = ∑_{i=0}^{k} (−1)^{i+k} (i + 1)^{n} \binom{k}{i}.

They can also be expressed through the Stirling numbers of the second kind as follows

W_{n,k} = k!\left\{{n + 1 \atop k + 1}\right\}.  (7)

The Worpitzky numbers satisfy the recursive relation

W_{n,k} = (k + 1) W_{n−1,k} + kW_{n−1,k−1} (n ≥ 1, k ≥ 1) .  (8)

Some simple properties are given

∑_{k=0}^{n}\left⟨{n \atop k}\right⟩x^{k} = ∑_{k=0}^{n} (x − 1)^{n−k} kW_{n−1,k−1},  (9)

∑_{k=0}^{n}\binom{n}{k}\left\{{k \atop t}\right\} = \left\{{n + 1 \atop t + 1}\right\},  (10)

and

∑_{k=0}^{n}\left⟨{n \atop k}\right⟩\binom{k + 1}{t} = W_{n,n−t}.  (11)

• The Bernoulli numbers B_{n} are defined by the exponential generating function

\frac{x}{1 − e^{−x}} = ∑_{n≥0}B_{n} \frac{x^{n}}{n!} .

The recursive relation is

B_{0} = 1,
B_{n} = 1 − ∑_{k=0}^{n−1}\binom{n}{k}\frac{B_{k}}{n − k + 1}, (n ≥ 1) .

Thus we have B_{1} = \frac{1}{2}, B_{2} = \frac{1}{6}, B_{3} = 0, and so on, they can also be expressed through the Worpitzky numbers

B_{n} = ∑_{k=0}^{n} (−1)^{k} \frac{W_{n,k}}{k + 1}.

• The Akiyama-Tanigawa matrix (A_{n,k})_{n,k≥0} associated with initial sequence A_{0,k} = \frac{1}{k+1} is defined by (see [2, 5, 8, 10])

A_{n,k} = (k + 1) (A_{n−1,k} − A_{n−1,k+1}) ,

or equivalently by [7]

A_{n,k} = \frac{1}{k!}∑_{i=0}^{k} (−1)^{k} s(k + 1, i + 1)B_{n+i},

= ∑_{i=1}^{n} (−1)^{i−1} \binom{k + i + 1}{k + 1}^{−1}W_{n,i}.  (12)

The Akiyama-Tanigawa matrix A_{n,k} is then

A_{n,k} = (\begin{matrix}1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & ⋯ \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & ⋯ \\ \frac{1}{6} & \frac{1}{6} & \frac{3}{20} & \frac{2}{15} & \frac{5}{42} & ⋯ \\ 0 & \frac{1}{30} & \frac{1}{20} & \frac{2}{35} & \frac{5}{84} & ⋯ \\ −\frac{1}{30} & −\frac{1}{30} & −\frac{3}{140} & −\frac{1}{105} & 0 & ⋯ \\ 0 & −\frac{1}{42} & −\frac{1}{28} & −\frac{4}{105} & −\frac{1}{28} & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \end{matrix}) .

2 Explicit formula for S_{n}^{(m)}

For any nonnegative integer m, we consider the sums

S_{n}^{(m)} := ∑_{k=0}^{n} (−1)^{k} k^{m}\binom{n}{k}^{−1}.  (13)

Note that S_{n}^{(0)} = (1 + (−1)^{n})\frac{n+1}{n+2} (identity (14) of [19] or identity (3) here for p = 0). The following expression for S_{n}^{(m)} holds and, in terms of computational complexity, it is better than the sum defining S_{n}^{(m)} for n ≥ m as the expression in the theorem involves O(m) operations.

Theorem 1. For any nonnegative integers n and m, we have

S_{n}^{(m)} = (n + 1) ∑_{j=0}^{m} \frac{(−1)^{m+j}}{n + j + 2} (1 + (−1)^{n} \binom{n + j + 1}{j}) W_{m,j}.  (14)

Proof. We can write S_{n}^{(m)} as follows

S_{n}^{(m)} = ∑_{k=0}^{n}(−1)^{k}\binom{n}{k}^{−1} ((k + 1) − 1)^{m}

= ∑_{k=0}^{n}(−1)^{k}\binom{n}{k}^{−1}∑_{i=0}^{m} (−1)^{m−i} \binom{m}{i} (k + 1)^{i} ,

and with (6), we obtain

S_{n}^{(m)} = ∑_{k=0}^{n}(−1)^{k}\binom{n}{k}^{−1}∑_{i=0}^{m} (−1)^{m−i} \binom{m}{i}∑_{j=0}^{i} (−1)^{i+j} \left\{{i \atop j}\right\} (k + 1) ⋯ (k + j)

= ∑_{k=0}^{n}∑_{i=0}^{m}∑_{j=0}^{i} \frac{(−1)^{k}}{n!} (−1)^{m+j} \binom{m}{i}\left\{{i \atop j}\right\}k! (k + 1) ⋯ (k + j) (n − k)!

= ∑_{i=0}^{m}∑_{j=0}^{i} (−1)^{m+j} \binom{m}{i}\left\{{i \atop j}\right\} (n + 1) ⋯ (n + j) ∑_{k=0}^{n}(−1)^{k}\binom{n + j}{k + j}^{−1}.

Now, from (3) and after some rearrangement, we get

S_{n}^{(m)} = (n + 1) ∑_{j=0}^{m} \frac{(−1)^{m+j}}{n + j + 2} (1 + (−1)^{n} \binom{n + j + 1}{j}) j!∑_{i=0}^{m}\binom{m}{i}\left\{{i \atop j}\right\}.

From (7) and (10), the result holds.

3 Recurrence relation for S_{n}^{(m)}

Theorem 2. For any nonnegative integers m and n, we have

S_{n+1}^{(m)} = S_{n}^{(m)} − \frac{1}{n + 1}S_{n}^{(m+1)} − (−1)^{n} (n + 1)^{m} .  (15)

Proof. The proof is based on the Sprugnoli [16] observation \binom{n+1}{k}^{−1} = \binom{n}{k}^{−1} − \frac{k}{n+1} \binom{n}{k}^{−1}.

The recurrence relation for S_{n}^{(m)} is given in the following Theorem 3. For any nonnegative integers m and n, we have

S_{n+1}^{(m)} = S_{n}^{(m)} − (−1)^{n} (n + 1)^{m} + ∑_{j=0}^{m+1} \frac{(−1)^{m+j}}{n + j + 2} (1 + (−1)^{n} \binom{n + j + 1}{j}) W_{m+1,j}, (16)

with the initial condition S_{0}^{(m)} = δ_{0m}, where δ_{ij} is the Kronecker symbol. Proof. This follows immediately from (14) and (15).

Setting m = 1 in (16), we have the following Corollary 4. If n is nonnegative integer, then

S_{n+1}^{(1)} = S_{n}^{(1)} + \frac{n − (−1)^{n} (2n^{4} + 17n^{3} + 49n^{2} + 57n + 24)}{(n + 2) (n + 3) (n + 4)}.

Our next goal is to calculate the ordinary generating functions of S_{n}^{(m)}.

4 Ordinary generating functions of S_{n}^{(m)}

In 2002, Mansour [9], generalized the idea of Sury [18] and gave an approach based on calculus to obtain the generating function for some combinatorial identities.

Theorem 5 (Mansour [9]). Let r, n ≥ k be any nonnegative integer numbers, and let f_{r} (n, k) be given by

f_{r} (n, k) = \frac{(n + r)!}{n!}∫_{u_{1}}^{u_{2}}p^{k} (t) q^{n−k} (t) dt,  (17)

where p (t) and q (t) are two functions defined on [u_{1}, u_{2}] . Let {a_{n}}_{n≥0} and {b_{n}}_{n≥0} be any two sequences, and let A (x) , B (x) be the corresponding ordinary generating functions. Then

∑_{n=0}^{∞} [∑_{k=0}^{n}f_{r} (n, k) a_{k}b_{n−k}] x^{n} = \frac{d^{r}}{dx^{r}} [x^{r}∫_{u_{1}}^{u_{2}}A (xp (t)) B (xq (t)) dt] .  (18)

We apply Theorem 5, for a_{n} = (−1)^{n} n^{m} (m ≥ 1) and b_{n} = 1, we have

A (x) = \frac{1}{(1 + x)^{m+1}} ∑_{k=0}^{m} \left⟨{m \atop k}\right⟩ (−x)^{k+1}

= ∑_{k=0}^{m} \frac{(−1)^{m+k}}{(1 + x)^{k+1}} W_{m,k},

B (x) = ∑_{n≥0}x^{n} = \frac{1}{1 − x}.

For r = 1, formula (18) gives:

∑_{n≥0}S_{n}^{(m)}x^{n} = \frac{d}{dx} [x∫_{0}^{1} \begin{matrix}∑_{k=0}^{m} & \left⟨{m \atop k}\right⟩ & (−xt) & k+1 \\ (1 + xt) & m+1 & (1 − x + xt) \end{matrix}dt] .  (19)

Making the substitution xt = y in the right-hand side of (19), we obtain

∑_{n≥0}S_{n}^{(m)}x^{n} = \frac{d}{dx} [∫_{0}^{x} \begin{matrix}∑_{k=0}^{m} & \left⟨{m \atop k}\right⟩ & (−y) & k+1 \\ (1 + y) & m+1 & (1 − x + y) \end{matrix}dy] ,

Since the degree of the denominator is at least one higher than that of the numerator, this fraction decomposes into partial fractions of the form

\begin{matrix}∑_{k=0}^{m} & \left⟨{m \atop k}\right⟩ & (−y) & k+1 \\ (1 + y) & m+1 & (1 − x + y) \end{matrix} = \frac{α^{(m)} (x)}{1 − x + y} + ∑_{s=0}^{m} \frac{α_{s}^{(m)} (x)}{(1 + y)^{m−s+1}} ,  (20)

We note in passing that (20) is equivalent to

∑_{k=0}^{m} \left⟨{m \atop k}\right⟩ (−y)^{k+1} = (1 + y)^{m+1} α^{(m)} (x) + (1 − x + y) ∑_{s=0}^{m} (1 + y)^{s} α_{s}^{(m)} (x)  (21)

= ∑_{k=0}^{m} (−1)^{m+k+1} y (1 + y)^{m−k} W_{m−1,k−1}.

For y = −1 and using the fact that W_{p,p} = p! for p ≥ 0, we immediately obtain the wellknown identity

∑_{k=0}^{m} \left⟨{m \atop k}\right⟩ = m!.

Next, if we set y = 0 in (21) then we obtain a relation between α^{(m)} (x) and α_{s}^{(m)} (x)

∑_{s=0}^{m} α_{s}^{(m)} (x) = \frac{α^{(m)} (x)}{x − 1}.  (22)

Proposition 6. For m ≥ 1, we have

α_{s}^{(m)} (x) = ∑_{i=0}^{s} \frac{(−1)^{i+s+1}}{x^{i+1}} ∑_{k=0}^{m} \left⟨{m \atop k}\right⟩\binom{k + 1}{s − i}  (23)

= ∑_{j=m−s}^{m} \frac{(−1)^{m+j+1}}{x^{s−m+1+j}} W_{m,j},

and

α^{(m)} (x) = \frac{1}{x^{m+1}} ∑_{k=0}^{m} \left⟨{m \atop k}\right⟩ (1 − x)^{k+1}  (24)

= ∑_{j=0}^{m} \frac{(−1)^{m+j}}{x^{j+1}} W_{m,j},

= −α_{m}^{(m)} (x) .

Proof. We verify that (23) and (24) satisfy (21). Denote the right-hand side of (21) by R^{(m)}(y), after some rearrangement, we get

R^{(m)}(y) = ∑_{k=0}^{m} \left⟨{m \atop k}\right⟩ [\frac{(1 + y)^{m+1}}{x^{m+1}} (1 − x)^{k+1}

+ (1 − x + y) ∑_{s=0}^{m} (1 + y)^{s} ∑_{j=0}^{s} \frac{(−1)^{j+1}}{x^{s−j+1}} \binom{k + 1}{j}] ,

using binomial formula and for k ≤ m, we obtain

R^{(m)}(y) = ∑_{k=0}^{m} \left⟨{m \atop k}\right⟩ [ ∑_{s=m+1}^{m+1} \frac{(1 + y)^{s}}{x^{s}} ∑_{j=0}^{s} \binom{k + 1}{j} (−1)^{j} x^{j}

− \frac{(1 − x + y)}{x} ∑_{s=0}^{m} \frac{(1 + y)^{s}}{x^{s}} ∑_{j=0}^{s} (−1)^{j} x^{j}\binom{k + 1}{j}]

= ∑_{k=0}^{m} \left⟨{m \atop k}\right⟩ [ ∑_{s=m+1}^{m+1} \frac{(1 + y)^{s}}{x^{s}} ∑_{j=0}^{s} \binom{k + 1}{j} (−1)^{j} x^{j}

+ ∑_{s=0}^{m} \frac{(1 + y)^{s}}{x^{s}} ∑_{j=0}^{s} (−1)^{j} x^{j}\binom{k + 1}{j}

− ∑_{s=0}^{m} \frac{(1 + y)^{s+1}}{x^{s+1}} ∑_{j=0}^{s} (−1)^{j} x^{j}\binom{k + 1}{j}]

= ∑_{k=0}^{m} \left⟨{m \atop k}\right⟩ [∑_{s=0}^{m+1} \frac{(1 + y)^{s}}{x^{s}} ∑_{j=0}^{s} (−1)^{j} x^{j}\binom{k + 1}{j}

− ∑_{s=0}^{m} \frac{(1 + y)^{s+1}}{x^{s+1}} ∑_{j=0}^{s} (−1)^{j} x^{j}\binom{k + 1}{j}]

= ∑_{k=0}^{m} \left⟨{m \atop k}\right⟩ [∑_{s=0}^{m+1} \frac{(1 + y)^{s}}{x^{s}} (∑_{j=0}^{s} (−1)^{j} x^{j}\binom{k + 1}{j}

− ∑_{j=0}^{s−1} (−1)^{j} x^{j}\binom{k + 1}{j})] .

Finally,

R^{(m)}(y) = ∑_{k=0}^{m} \left⟨{m \atop k}\right⟩ [∑_{s=0}^{k+1} (1 + y)^{s} ((−1)^{s} \binom{k + 1}{s})]

= ∑_{k=0}^{m} \left⟨{m \atop k}\right⟩ (−y)^{k+1} .

According to (7) and (11), we have

α_{s}^{(m)} (x) = ∑_{i=0}^{s} \frac{(−1)^{i+s+1}}{x^{i+1}} ∑_{k=0}^{m} \left⟨{m \atop k}\right⟩\binom{k + 1}{s − i}

= ∑_{i=0}^{s} \frac{(−1)^{i+s+1} (m − s + i)!}{x^{i+1}} \left\{{m + 1 \atop m − s + i + 1}\right\}

= ∑_{i=0}^{s} \begin{matrix}(−1) & i+s+1 \\ (x) & i+1 \end{matrix} W_{m,m−s+i}

= ∑_{j=m−s}^{m} \frac{(−1)^{m+j+1}}{x^{s−m+1+j}} W_{m,j}.

It follows from (9) that

α^{(m)} (x) = \frac{1}{x^{m+1}} ∑_{k=0}^{m} \left⟨{m \atop k}\right⟩ (1 − x)^{k+1}

= \frac{1 − x}{x^{m+1}} ∑_{k=1}^{m} (−1)^{m+k} x^{m−k}kW_{m−1,k−1}

= (1 − x) ∑_{k=0}^{m} (−1)^{m+k} \frac{k}{x^{k+1}} W_{m−1,k−1}

= ∑_{k=0}^{m} (−1)^{m+k} \frac{k}{x^{k+1}} W_{m−1,k−1} − ∑_{k=0}^{m} (−1)^{m+k} \frac{k}{x^{k}} W_{m−1,k−1}

= ∑_{k=0}^{m} (−1)^{m+k} \frac{k}{x^{k+1}} W_{m−1,k−1} + ∑_{k=0}^{m−1} (−1)^{m+k} \frac{k + 1}{x^{k+1}}W_{m−1,k}.

Using (8), we get α^{(m)} (x) as desired. This completes the proof. Now, integrating the right-hand side of (20) over y, we obtain

∫_{0}^{x} \begin{matrix}∑_{k=0}^{m} & \left⟨{m \atop k}\right⟩ & (−y) & k+1 \\ (1 + y) & m+1 & (1 − x + y) \end{matrix}dy = α_{m}^{(m)} (x) ln(1 − x^{2}) + ∑_{s=0}^{m−1} \frac{α_{s}^{(m)} (x)}{m − s} [1 − (1 + x)^{s−m}] . (25)

By differentiating (25) we get the ordinary generating function of S_{n}^{(m)}

∑_{n≥0}S_{n}^{(m)}x^{n} = ln(1 − x^{2})\frac{d}{dx}α_{m}^{(m)} (x) + ∑_{s=0}^{m−1} \frac{\frac{d}{dx}α_{s}^{(m)} (x)}{m − s} ([1 − (1 + x)^{s−m}])
+ ∑_{s=0}^{m−1} ((1 + x)^{s−m−1} α_{s}^{(m)} (x)) − \frac{2x}{1 − x^{2}} α_{m}^{(m)} (x) ,  (26)

with

\frac{d}{dx}α_{s}^{(m)} (x) = ∑_{j=m−s}^{m} \frac{(s − m + 1 + j) (−1)^{m+j}}{x^{s−m+2+j}} W_{m,j}.

With Proposition 6, we can now rewrite (26) as follows Theorem 7. For any real numbers x and for all nonnegative integer m, we have

∑_{n≥0}S_{n}^{(m)}x^{n} = (∑_{j=0}^{m} \frac{(1 + j) (−1)^{m+j}}{x^{2+j}} W_{m,j}) ln(1 − x^{2})

+ ∑_{0≤j≤s≤m−1} \frac{(−1)^{j}}{x^{s−j+2}} W_{m,m−j} (\frac{s − j + 1}{m − s} (1 − (1 + x)^{s−m}) − x (1 + x)^{s−m−1})

+ \frac{2}{1 − x^{2}} ∑_{j=0}^{m} \frac{(−1)^{m+j}}{x^{j}} W_{m,j}. (27)

In particular for m = 0 and m = 1, we get

∑_{n≥0}S_{n}^{(0)}x^{n} = \frac{2}{1 − x^{2}} + \frac{ln (1 − x^{2})}{x^{2}},

and

∑_{n≥0}S_{n}^{(1)}x^{n} = \frac{2 + 3x}{x (1 + x)^{2}} + \frac{2 − x}{x^{3}} ln (1 − x^{2}) .

5 The asymptotic expansion

In the previous sections, S_{n}^{(m)} becomes more complex when, m grows, so it is important to have asymptotic expansion of S_{n}^{(m)}. Theorem 8. For m > 0, we have

S_{2n}^{(m)} ∼ (2n)^{m} and S_{2n+1}^{(m)} ∼ − (2n + 1)^{m} .

Proof. Write

k^{m} = c_{0} + c_{1} (k + 1) + c_{2} (k + 1) (k + 2) ⋯ + c_{m} (k + 1) ⋯ (k + m) ,

where c_{i}’s depend on m (c_{m} = 1). we immediately have

S_{n}^{(m)} = c_{0}∑_{k=0}^{n} (−1)^{k} \binom{n}{k}^{−1}+c_{1}∑_{k=0}^{n} (−1)^{k} (k + 1) \binom{n}{k}^{−1}+⋯+∑_{k=0}^{n} (−1)^{k} (k + 1) ⋯ (k + m) \binom{n}{k}^{−1}.

After some rearrangement and using the fact that (k + 1) \binom{n}{k}^{−1} = (n + 1) \binom{n+1}{k+1}^{−1}, we have S_{n}^{(m)} = c_{0}T_{n}^{(0,0)} + c_{1} (n + 1) T_{n}^{(0,1)} + c_{2} (n + 1) (n + 2) T_{n}^{(0,2)} + ⋯ + (n + 1) ⋯ (n + m) T_{n}^{(0,m)}. Since T_{2n}^{(0,p)} → 1 and T_{2n+1}^{(0,p)} → −1, the result holds.

6 A connection to the Akiyama-Tanigawa matrix

In this section we consider T_{n}^{(m,p)}. The following lemma will be useful in the proof of the main theorem of this section.

Lemma 9. For m ≥ 1, we have

∑_{k=0}^{n}k^{m}z^{k} = ∑_{k=0}^{m}W_{m,k} (\frac{z}{1 − z} )^{k+1}

− z^{n+1} ∑_{s=0}^{m}\binom{m}{s} (n + 1)^{m−s} ∑_{k=0}^{s} (−1)^{s+k} W_{s,k}(1 − z)^{−k−1}.

Proof. Recall that, for m ≥ 1

∑_{k=0}^{∞}k^{m}z^{k} = ∑_{k=0}^{m}W_{m,k} (\frac{z}{1 − z} )^{k+1} = ∑_{k=0}^{m} (−1)^{m+k} W_{m,k}(1 − z)^{−k−1},

we have

∑_{k=0}^{n}k^{m}z^{k} = ∑_{k=0}^{∞}k^{m}z^{k} − ∑_{k=n+1}^{∞}k^{m}z^{k}

= ∑_{k=0}^{m}W_{m,k} (\frac{z}{1 − z} )^{k+1} − z^{n+1}∑_{i=0}^{∞} (i + n + 1)^{m} z^{i}

= ∑_{k=0}^{m}W_{m,k} (\frac{z}{1 − z} )^{k+1} − z^{n+1}∑_{i=0}^{∞}∑_{s=0}^{m}\binom{m}{s} (n + 1)^{m−s} i^{s}z^{i}

= ∑_{k=0}^{m}W_{m,k} (\frac{z}{1 − z} )^{k+1} − z^{n+1} ∑_{s=0}^{m}\binom{m}{s} (n + 1)^{m−s} ∑_{i=0}^{∞}i^{s}z^{i},

as desired.

For an alternative proof see Boyadzhiev [4]. The main result of this section is to prove the following theorem which expresses explicitly the alternating sums of the reciprocals of binomial coefficients, T_{n}^{(m,p)}, in terms of the Akiyama-Tanigawa matrix A_{n,k}.

Theorem 10. For nonnegative integers n, m and p, we have

T_{n}^{(m,p)} = \binom{n + p}{p}^{−1}δ_{0m} + (n + p + 1) ∑_{s=0}^{m} (−1)^{n+s} \binom{m}{s} (n + 1)^{m−s} A_{s,n+p+1}

− \frac{n + p + 1}{n + 1}∑_{s=0}^{m} (−1)^{s} \binom{n + s + p + 2}{p + s + 1}^{−1}W_{m,s}, (28)

where A_{i,j} is the Akiyama-Tanigawa matrix.

Proof. By the Beta function we can write

T_{n}^{(m,p)} = ∑_{k=0}^{n} (−1)^{k} k^{m} (p + n + 1) ∫_{0}^{1}x^{p+k} (1 − x)^{n−k} dx

= (p + n + 1) ∫_{0}^{1}x^{p} (1 − x)^{n} ∑_{k=0}^{n}k^{m} (\frac{−x}{1 − x})^{k} dx.

Using the lemma, we get

T_{n}^{(m,p)} = (p + n + 1) ∫_{0}^{1}x^{p} (1 − x)^{n} (∑_{k=0}^{m}W_{m,k} (−x)^{k+1}

− ∑_{s=0}^{m}\binom{m}{s} (n + 1)^{m−s} ∑_{k=0}^{s} (−1)^{s+k} W_{s,k} (−x)^{n+1} (1 − x)^{k−n}) dx

= \frac{(p + n + 1)}{n + 1}∑_{k=0}^{m} (−1)^{k+1} \binom{n + k + p + 2}{p + k + 1}^{−1}W_{m,k}

− (p + n + 1) ∑_{s=0}^{m}∑_{k=0}^{s}\binom{m}{s} (n + 1)^{m−s} \frac{(−1)^{n+s+k+1}}{k + 1} \binom{n + k + 2 + p}{k + 1 + p}^{−1}W_{s,k}.

Finally, from (12) we obtain

T_{n}^{(m,p)} = \frac{(p + n + 1)}{n + 1}∑_{k=0}^{m} (−1)^{k+1} \binom{n + k + p + 2}{p + k + 1}^{−1}W_{m,k}

+ (n + p + 1) ∑_{s=0}^{m}\binom{m}{s} (n + 1)^{m−s} \frac{(−1)^{n+s}}{(n + p + 1)!} ∑_{k=0}^{n+p+1} (−1)^{k} s(n + p + 2, k + 1)B_{s+k}.

As desired, this completes the proof.

Setting p = 0 in (28) we can rewrite (14) as follows Corollary 11.

S_{n}^{(m)} = δ_{0m} − A_{m+1,n} + ∑_{s=0}^{m} (−1)^{n+s} \binom{m}{s} (n + 1)^{m−s+1} A_{s,n+1}.

7 A recurrence relation For T_{n}^{(m,p)}

Theorem 12. For any nonnegative integers m, n and p

T_{n+1}^{(m,p)} = \frac{n + 1}{n + p + 1}T_{n}^{(m,p)} − \frac{1}{n + p + 1}T_{n}^{(m+1,p)} − (−1)^{n} (n + 1)^{m} .  (29)

Proof. Using the identity

\binom{n + p + 1}{k + p}^{−1} = \frac{n + 1}{n + p + 1}\binom{n + p}{k + p}^{−1} − \frac{k}{n + p + 1}\binom{n + p}{k + p}^{−1},

we get the relation (29).

Now, from (28) and (29), we have the recurrence relation for T_{n}^{(m,p)}

Theorem 13. For nonnegative integers n, m and p, we have

T_{n+1}^{(m,p)} = \frac{n + 1}{n + p + 1}T_{n}^{(m,p)} − (−1)^{n} (n + 1)^{m+1} ∑_{s=0}^{m+1} \begin{matrix}(−1) & s \\ (n + 1) & s \end{matrix} \binom{m + 1}{s}A_{s,n+p+1}

+ \frac{1}{n + 1} ∑_{s=0}^{m+1} (−1)^{s} \binom{n + s + p + 2}{p + s + 1}^{−1}W_{m+1,s} − (−1)^{n} (n + 1)^{m} ,

with the initial condition T_{0}^{(m,p)} = \binom{n+p}{p}^{−1}δ_{0m}.

8 Acknowledgments

The authors are grateful to the anonymous referee for providing a number of constructive comments and illuminating suggestions, almost all of which we have tried to carry out. The first and the second author were partially supported by PNR project 8/u160/664.

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2000 Mathematics Subject Classification: Primary 11B65; Secondary 05A10, 05A16. Keywords: Binomial coefficient, Akiyama-Tanigawa matrix, recurrence relation, generating function.

(Concerned with sequences \underline{A008275}, \underline{A008277}, \underline{A008292}, and \underline{A028246}.)

Received July 27 2011; revised version received January 13 2012. Published in Journal of Integer Sequences, January 14 2012.

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