Formulas for Bernoulli Numbers and Polynomials

Abstract

We present several formulas involving the classical Bernoulli numbers and polynomials. Among others, we extend an identity for Bernoulli polynomials published by Wu et al. (Fibonacci Quart 42:295-299, 2004).

1 Introduction and Statement of the Results

In 2001, Momiyama [7] used methods from p-adic analysis to prove the following remarkable identity for the classical Bernoulli numbers $B_{\nu }$, defined by

$$\begin{aligned} \frac{z}{e^z-1}=\sum _{\nu =0}^\infty B_{\nu } \frac{z^{\nu }}{\nu !}. \end{aligned}$$

Proposition 1

For all nonnegative integers m and n with $m+n>0$, we have

$$\begin{aligned} & (-1)^m \sum _{\nu =0}^m (\nu +n+1) {m+1\atopwithdelims ()\nu }B_{\nu +n}\nonumber \\ & \quad =(-1)^{n+1} \sum _{\nu =0}^n (\nu +m+1) {n+1\atopwithdelims ()\nu }B_{\nu +m}. \end{aligned}$$

(1.1)

From (1.1) with $m=n$ we obtain

$$\begin{aligned} \sum _{\nu =0}^m (\nu +m+1) {m+1\atopwithdelims ()\nu } B_{\nu +m}=0, \end{aligned}$$

which is attributed to von Ettingshausen [8, pp. 284-285]; see also Kaneko [5]. In 2004, Wu et al. [9] used properties of power series to prove an interesting generalization of (1.1) involving the Bernoulli polynomials

$$\begin{aligned} B_n(z)= \sum _{\nu =0}^n {n\atopwithdelims ()\nu } B_{n-\nu } z^{\nu }. \end{aligned}$$

Proposition 2

For all nonnegative integers m and n and complex numbers z, we have

$$\begin{aligned} & (-1)^m \sum _{\nu =0}^m (\nu +n+1) {m+1\atopwithdelims ()\nu }B_{\nu +n}(z)+ (-1)^{n} \sum _{\nu =0}^n (\nu +m+1)\nonumber \\ & \quad \ \times {n+1\atopwithdelims ()\nu }B_{\nu +m}(-z) \nonumber \\ & \quad = (-1)^m (m+n+1) (m+n+2) z^{m+n}. \end{aligned}$$

(1.2)

If $m+n>0$ and $z=0$, then (1.2) reduces to (1.1). Chen and Sun [2] applied Zeilberger’s algorithm to prove (1.2). We show that (1.2) can be written in a slightly shorter and more elegant form as follows.

Theorem 1

Let m and n be nonnegative integers. Then, for $z\in {\mathbb {C}}$,

$$\begin{aligned} & (-1)^m \sum _{\nu =0}^{m+1} (\nu +n+1) {m+1\atopwithdelims ()\nu }B_{\nu +n}(z)\nonumber \\ & \quad =(-1)^{n+1} \sum _{\nu =0}^ {n+1} (\nu +m+1) {n+1\atopwithdelims ()\nu }B_{\nu +m}(-z). \end{aligned}$$

(1.3)

We apply some basic properties of Bernoulli polynomials to settle (1.3). In particular, we provide a new proof of (1.2).

The main aim of this paper is to present a new generalization of (1.2). We define

$$\begin{aligned} S(m,n;k;z)=(-1)^m \sum _{\nu =0}^m (\nu +n+1){m+1\atopwithdelims ()\nu } {\nu +n\atopwithdelims ()k} B_{\nu +n-k}(z). \end{aligned}$$

The following reciprocity formula holds. (As usual, if $p<0$, then $B_p(z)=0$.)

Theorem 2

Let m, n and k be nonnegative integers. Then, for $z\in {\mathbb {C}}$,

$$\begin{aligned} S(m,n;k;z) + (-1)^{k} S(n,m;k;-z) & =(-1)^m (m+n+1)(m+n+2)\nonumber \\ & \quad \ \times {m+n\atopwithdelims ()k} z^{m+n-k}. \end{aligned}$$

(1.4)

Remark 1

The special case $k=0$ gives (1.2).

If we set $z=0$, then (1.4) yields an extension of Momiyama’s identity (1.1).

Corollary 1

Let m, n, k be nonnegative integers. Then

$$\begin{aligned} S^*(m,n;k) = (-1)^{k+1} S^*(n,m;k), \end{aligned}$$

(1.5)

where

$$\begin{aligned} S^*(m,n;k)= (-1)^m \sum _{\nu =0}^m (\nu +n+1) {m+1\atopwithdelims ()\nu }{\nu +n\atopwithdelims ()k} B_{\nu +n-k}. \end{aligned}$$

An application of Theorem 2 leads to an identity for the alternating sum

$$\begin{aligned} A(m,n;k)= \sum _{\nu =0}^m (-1)^{\nu } (\nu +n+1){m+1\atopwithdelims ()\nu } {\nu +n\atopwithdelims ()k}B_{\nu +n-k}. \end{aligned}$$

We obtain the following companion to (1.5).

Corollary 2

Let k, m, n be nonnegative integers with $k+2\le \min (m,n)$. Then

$$\begin{aligned} A(m,n;k)=(-1)^{k+1} A(n,m;k). \end{aligned}$$

(1.6)

A second application of Theorem 2 gives a reciprocity formula for the polynomial

$$\begin{aligned} P(m,n;k;z)=\sum _{\nu =0}^{m} {m\atopwithdelims ()\nu } {\nu +n \atopwithdelims ()k} z^{\nu +n-k}. \end{aligned}$$

Corollary 3

Let k, m, n be nonnegative integers. Then, for $z\in {\mathbb {C}}$,

$$\begin{aligned} P(m,n;k;z)=(-1)^{m+n+k} P(n,m;k;-z-1). \end{aligned}$$

(1.7)

Remark 2

From (1.7) with $z=-1/2$ we get

$$\begin{aligned} (-1)^m \sum _{\nu =0}^m (-2)^{\nu } {m+n-\nu \atopwithdelims ()k}{m\atopwithdelims ()\nu } =(-1)^{k+n} \sum _{\nu =0}^n (-2)^{\nu } {m+n-\nu \atopwithdelims ()k}{n\atopwithdelims ()\nu }. \end{aligned}$$

By using differentiation or integration certain summation formulas involving Bernoulli polynomials lead to interesting new identities. Hereby, the recurrence relation $B'_n(x)=nB_{n-1}(x)$ plays an important role. We show that by applying an integral formula and (1.3) we obtain the following counterpart of

$$\begin{aligned} (-1)^m \sum _{\nu =0}^m {m\atopwithdelims ()\nu } B_{\nu +n}= (-1)^n\sum _{\nu =0}^n {n\atopwithdelims ()\nu } B_{\nu +m} \end{aligned}$$

(1.8)

which is due to Gessel [3].

Corollary 4

Let m and n be nonnegative integers. Then

$$\begin{aligned} & (-1)^n \sum _{\nu =0}^m {m\atopwithdelims ()\nu } (2^{-\nu -n}-1) B_{\nu +n} - (-1)^m \sum _{\nu =0}^n {n\atopwithdelims ()\nu } (2^{-\nu -m}-1) B_{\nu +m}\nonumber \\ & \qquad =\frac{m-n}{2^{m+n}}. \end{aligned}$$

(1.9)

Remark 3

Combining (1.8) and (1.9) gives

$$\begin{aligned} (-1)^n \sum _{\nu =0}^m{m\atopwithdelims ()\nu } 2^{m-\nu } B_{\nu +n} - (-1)^m \sum _{\nu =0}^n{n\atopwithdelims ()\nu } 2^{n-\nu } B_{\nu +m} = m-n. \end{aligned}$$

2 Proofs

I. We need the following formulas:

$$\begin{aligned} B_n(x+y)= & \sum _{\nu =0}^n {n\atopwithdelims ()\nu } B_{\nu }(x) y^{n-\nu } \quad (n=0,1,\ldots ), \end{aligned}$$

(2.1)

$$\begin{aligned} B_n(x+1)= & (-1)^n B_{n}(-x) =B_n(x)+n x^{n-1} \quad (n=0,1,\ldots ). \end{aligned}$$

(2.2)

These formulas can be found in Abramowitz and Stegun [1, (23.1.7), (23.1.8), (23.1.9)]. The integral formula

$$\begin{aligned} \int _0^{1/2} B_n(x)dx= \frac{1-2^{n+1}}{(n+1) 2^n} B_{n+1} \quad (n=0,1,\ldots ) \end{aligned}$$

(2.3)

is given in Moll and Vignat [6]. Applying (2.2) and (2.3) leads to

$$\begin{aligned} \int _0^{1/2} B_n(-x)dx= \Bigl (\frac{-1}{2}\Bigr )^n \Bigl (\frac{1-2^{n+1}}{n+1} B_{n+1}+1\Bigr ) \quad (n=1,2,\ldots ). \end{aligned}$$

(2.4)

The next two identities can be found in Gould [4, (1.13), (3.49)]:

$$\begin{aligned} & \sum _{\nu =0}^{m+1} (-1)^{\nu } {m+1\atopwithdelims ()\nu } {\nu }^p =0 \quad (0\le p\le m), \end{aligned}$$

(2.5)

$$\begin{aligned} & \sum _{\nu =0}^k (-1)^{\nu } {k\atopwithdelims ()\nu } {x-\nu \atopwithdelims ()r} ={x-k\atopwithdelims ()r-k} \quad (k=0,1,\ldots ). \end{aligned}$$

(2.6)

Applying the binomial theorem gives

$$\begin{aligned} \sum _{\nu =0}^r(\nu +s){r\atopwithdelims ()\nu } t^{\nu } =(t+1)^{r-1} \bigl ( (r+s)t+s\bigr ) \quad (r=0,1,\ldots ). \end{aligned}$$

(2.7)

II. We show that the identities (1.2) and (1.3) are equivalent. Let

$$\begin{aligned} T(m,n;z)=\sum _{\nu =0}^{m+1} a_{\nu }(m,n;z) \quad \text{ and }\quad T^*(m,n;z)= \sum _{\nu =0}^{m} a_{\nu }(m,n;z) \end{aligned}$$

(2.8)

with

$$\begin{aligned} a_{\nu }(m,n;z)=(-1)^m (\nu +n+1){m+1\atopwithdelims ()\nu } B_{\nu +n}(z). \end{aligned}$$

Using (2.2) gives

$$\begin{aligned} T(m,n;z)+T(n,m;-z)= & T^*(m,n;z)+a_{m+1}(m,n;z)+T^*(n,m;-z)\\ & \qquad +a_{n+1}(n,m;-z) \\ & \quad = T^*(m,n;z) +T^*(n,m;-z)\\ & \quad -(-1)^m (m+n+1)(m+n+2) z^{m+n}. \end{aligned}$$

This implies that (1.2) and (1.3) are equivalent.

III. Now, we prove the theorems and corollaries stated in Sect. 1.

Proof of Theorem 1

Using (2.8), (2.2), (2.1) with $x=-z$, $y=1$ and the elementary formula

$$\begin{aligned} (r+1){r\atopwithdelims ()j} =(j+1){r+1\atopwithdelims ()j+1} \end{aligned}$$

with $r=\nu + n$ gives

$$\begin{aligned} T(m,n;z)= & (-1)^{m+n} \sum _{\nu =0}^{m+1} (-1)^{\nu } (\nu +n+1) {m+1\atopwithdelims ()\nu } B_{\nu +n}(1-z) \\= & (-1)^{m+n} \sum _{\nu =0}^{m+1} (-1)^{\nu } (\nu +n+1) {m+1\atopwithdelims ()\nu } \sum _{j=0}^{\nu +n} {\nu +n\atopwithdelims ()j} B_{j}(-z) \\= & (-1)^{m+n} \sum _{\nu =0}^{m+1} (-1)^{\nu } {m+1\atopwithdelims ()\nu } \sum _{j=0}^{\nu +n} (j+1) {\nu +n+1\atopwithdelims ()j+1} B_j(-z) \\= & (-1)^{m+n}\sum _{j=0}^{m+n+1} (j+1) B_j(-z)\sum _{\nu =0}^{m+1} (-1)^{\nu } {m+1\atopwithdelims ()\nu } {\nu +n+1\atopwithdelims ()j+1}.\\ \end{aligned}$$

Applying (2.5) we conclude that the inner sum is equal to zero, if $0\le j\le m-1$. Thus

$$\begin{aligned} T(m,n;z)= & (-1)^{m+n} \sum _{j=m}^{m+n+1} (j+1)B_j(-z) \sum _{\nu =0}^{m+1} (-1)^{\nu } {m+1\atopwithdelims ()\nu } {\nu + n+1\atopwithdelims ()j+1} \nonumber \\= & (-1)^{m+n}\sum _{j=0}^{n+1} (j+m+1)B_{j+m}(-z) \sum _{\nu =0}^{m+1} (-1)^{\nu } {m+1\atopwithdelims ()\nu } {\nu + n+1\atopwithdelims ()j+m+1}. \nonumber \\ \end{aligned}$$

(2.9)

Using (2.6) with $k=m+1$, $r=j+m+1$, $x=m+n+2$ we get

$$\begin{aligned} \sum _{\nu =0}^{m+1} (-1)^{\nu } {m+1\atopwithdelims ()\nu } {\nu +n+1\atopwithdelims ()j+m+1}= & (-1)^{m+1}\sum _{\nu =0}^{m+1} (-1)^{\nu } {m+1\atopwithdelims ()\nu } {m+n+2-\nu \atopwithdelims ()j+m+1} \nonumber \\= & (-1)^{m+1}{n+1\atopwithdelims ()j}. \end{aligned}$$

(2.10)

From (2.9) and (2.10) we obtain

$$\begin{aligned} T(m,n;z)=(-1)^{n+1} \sum _{j=0}^{n+1}(j+m+1){n+1\atopwithdelims ()j} B_{j+m}(-z)=-T(n,m;-z). \end{aligned}$$

This settles (1.3). $\square$

Proof of Theorem 2

Using (2.1) gives

$$\begin{aligned} & (-1)^m \sum _{\nu =0}^m (\nu +n+1){m+1\atopwithdelims ()\nu } B_{\nu +n}(x+z)\nonumber \\ & \qquad =(-1)^m \sum _{\nu =0}^m (\nu +n+1){m+1\atopwithdelims ()\nu } \sum _{k=0}^\infty {\nu +n\atopwithdelims ()k} B_{\nu +n-k}(z) x^k \nonumber \\ & \qquad =(-1)^m \sum _{k=0}^\infty \sum _{\nu =0}^m (\nu +n+1){m+1\atopwithdelims ()\nu } {\nu +n\atopwithdelims ()k} B_{\nu +n-k}(z) x^k.\nonumber \\ \end{aligned}$$

(2.11)

Next, we exchange m and n and we replace x and z by $-x$ and $-z$, respectively. Then we obtain from (2.11):

$$\begin{aligned} & (-1)^n \sum _{\nu =0}^n (\nu +m+1){n+1\atopwithdelims ()\nu } B_{\nu +m}(-(x+z) )\nonumber \\ & \qquad =(-1)^n \sum _{k=0}^\infty \sum _{\nu =0}^n (\nu +m+1){n+1\atopwithdelims ()\nu } {\nu +m\atopwithdelims ()k} B_{\nu +m-k}(-z) (-1)^k x^k.\nonumber \\ \end{aligned}$$

(2.12)

Moreover, we have

$$\begin{aligned} & (-1)^m (m+n+1) (m+n+2) (x+z)^{m+n}\nonumber \\ & \quad =(-1)^m (m+n+1)(m+m+2) \sum _{k=0}^{\infty }{m+n\atopwithdelims ()k} z^{m+n-k} x^k. \end{aligned}$$

(2.13)

We apply (1.2) with $x+z$ instead of z and (2.11), (2.12), (2.13). A comparison of the coefficients yields

$$\begin{aligned} & (-1)^m \sum _{\nu =0}^m (\nu +n+1) {m+1\atopwithdelims ()\nu }{\nu +n\atopwithdelims ()k} B_{\nu +n-k}(z)\\ & \qquad \quad +(-1)^{k+n} \sum _{\nu =0}^n (\nu +m+1) {n+1\atopwithdelims ()\nu }{\nu +m\atopwithdelims ()k} B_{\nu +m-k}(-z)\\ & \qquad = (-1)^m (m+n+1) (m+n+2) {m+n\atopwithdelims ()k} z^{m+n-k}, \end{aligned}$$

as claimed. $\square$

Proof of Corollary 2

We define

$$\begin{aligned} C(m,n;k)= \sum _{\nu =0}^m (-1)^{\nu } (\nu +n+1)(\nu +n-k) {m+1\atopwithdelims ()\nu } {\nu +n\atopwithdelims ()k} \end{aligned}$$

and

$$\begin{aligned} D(m,n;k)= \sum _{\nu =0}^m (-1)^{\nu } {m+1\atopwithdelims ()\nu } {\nu +n+1\atopwithdelims ()k}. \end{aligned}$$

Then

$$\begin{aligned} C(m,n;k)=(k+1)(k+2)D(m,n;k+2). \end{aligned}$$

Using

$$\begin{aligned} B_n(1)=(-1)^n B_n \quad \text{ and }\quad B_n(-1)=B_n +(-1)^n n \quad (n=0,1,\ldots ) \end{aligned}$$

gives

$$\begin{aligned} S(m,n;k;1) & =(-1)^m \sum _{\nu =0}^m (\nu +n+1){m+1\atopwithdelims ()\nu } {\nu +n\atopwithdelims ()k}B_{\nu +n-k}(1)\nonumber \\ & = (-1)^{m+n+k} A(m,n;k) \end{aligned}$$

(2.14)

and

$$\begin{aligned} S(n,m;k;-1)= & (-1)^n \sum _{\nu =0}^n (\nu +m+1){n+1\atopwithdelims ()\nu } {\nu +m\atopwithdelims ()k}B_{\nu +m-k}(-1) \nonumber \\= & S^*(n,m;k) +(-1)^{m+n+k} C(n,m;k). \end{aligned}$$

(2.15)

We apply Theorem 2 with $z=1$ and (2.14), (2.15). This yields

$$\begin{aligned} & (-1)^{m+k} (m+n+1)(m+n+2){m+n\atopwithdelims ()k} \nonumber \\ & \qquad = (-1)^k S(m,n;k;1)+S(n,m;k;-1)\nonumber \\ & \qquad = (-1)^{m+n} A(m,n;k) +S^*(n,m;k) +(-1)^{m+n+k} C(n,m;k).\nonumber \\ \end{aligned}$$

(2.16)

Next, we exchange m and n, multiply by $(-1)^{k+1}$, and use (1.5). Then

$$\begin{aligned} & (-1)^{n+1} (m+n+1)(m+n+2){m+n\atopwithdelims ()k}\nonumber \\ & \qquad = (-1)^{m+n+k+1}A(n,m;k) +(-1)^{k+1} S^*(m,n;k)\nonumber \\ & \qquad \quad \ +(-1)^{m+n+1} C(m,n;k)\nonumber \\ & \qquad = (-1)^{m+n+k+1} A(n,m;k) +S^*(n,m;k) +(-1)^{m+n+1} C(m,n;k).\nonumber \\ \end{aligned}$$

(2.17)

From (2.16) and (2.17) we obtain

$$\begin{aligned} A(m,n;k)+(-1)^k A(n,m;k)=F(m,n;k) \end{aligned}$$

(2.18)

with

$$\begin{aligned} F(m,n;k)= & \bigl ( (-1)^{n+k} +(-1)^m \bigr ) (m+n+1)(m+n+2){m+n\atopwithdelims ()k}\nonumber \\ & -(-1)^k C(n,m;k)-C(m,n;k) \nonumber \\ & = (-1)^k (k+1)(k+2) \Bigg [ (-1)^n {m+n+2\atopwithdelims ()k+2} -D(n,m;k+2) \Bigg ]\nonumber \\ & + (k+1)(k+2) \Bigg [ (-1)^m {m+n+2\atopwithdelims ()k+2} -D(m,n;k+2) \Bigg ].\nonumber \\ \end{aligned}$$

(2.19)

If $0\le p\le m$, then we conclude from (2.5) that

$$\begin{aligned} D(m,n;p)= & \sum _{\nu =0}^{m+1} (-1)^{\nu }{m+1\atopwithdelims ()\nu }{ \nu +n+1\atopwithdelims ()p}+(-1)^m {m+n+2\atopwithdelims ()p}\\= & (-1)^m {m+n+2\atopwithdelims ()p}. \end{aligned}$$

Since $k+2\le \min (m,n)$, we obtain from (2.19) that $F(m,n;k)=0$, so that (2.18) implies (1.6). $\square$

Proof of Corollary 3

We present two proofs. First proof. A short calculation gives that (1.7) holds for $k=0$ and $k=1$. Moreover, since

$$\begin{aligned} (-1)^{n+k} P(n,0;k;-z-1)= & (-1)^{n+k} \sum _{\nu =0}^n {\nu \atopwithdelims ()k}{n\atopwithdelims ()\nu }(-z-1)^{\nu -k} \\= & (-1)^{n+k} {n\atopwithdelims ()k} \sum _{\nu =0}^n {n-k\atopwithdelims ()\nu -k}(-z-1)^{\nu -k} \\= & {n\atopwithdelims ()k} z^{n-k} =P(0,n;k;z), \end{aligned}$$

we conclude that (1.7) is valid for $m=0$ and $n=0$. Let $m\ge 1$, $n\ge 1$, $k\ge 2$. Using (2.2) gives

$$\begin{aligned} & S(m-1,n-1;k-2;z+1)-S(m-1,n-1;k-2;z) \nonumber \\ & \quad = (-1)^{m-1} \sum _{\nu =0}^{m-1} (\nu +n){m\atopwithdelims ()\nu } {\nu +n-1\atopwithdelims ()k-2} \bigl ( B_{\nu +n+1-k}(z+1)-B_{\nu +n+1-k}(z) \bigr ) \nonumber \\ & \quad = (-1)^{m-1} \sum _{\nu =0}^{m-1} (\nu +n){m\atopwithdelims ()\nu } {\nu +n-1\atopwithdelims ()k-2} (\nu +n+1-k) z^{\nu +n-k} \nonumber \\ & \quad = (k-1)k (-1)^{m-1} \sum _{\nu =0}^{m-1} {m\atopwithdelims ()\nu } {\nu +n\atopwithdelims ()k} z^{\nu +n-k} \nonumber \\ & \quad = (k-1)k \Bigg [ (-1)^{m-1}P(m,n;k;z)-(-1)^{m-1} {m+n\atopwithdelims ()k}z^{m+n-k}\Bigg ] \end{aligned}$$

(2.20)

and

$$\begin{aligned} & (-1)^{k} \bigl ( S(n-1,m-1;k-2;-z-1)-S(n-1,m-1;k-2;-z) \bigr ) \nonumber \\ & \quad = (-1)^{k+n} \sum _{\nu =0}^{n-1} (\nu +m) {n\atopwithdelims ()\nu } {\nu +m-1\atopwithdelims ()k-2} \bigl ( B_{\nu +m+1-k}(-z)-B_{\nu +m+1-k}(-z-1)\bigr ) \nonumber \\ & \quad = (-1)^{m+n} \sum _{\nu =0}^{n-1} (-1)^{\nu } (\nu +m) (\nu +m+1-k) {\nu +m-1\atopwithdelims ()k-2}{n\atopwithdelims ()\nu } (z+1)^{\nu +m-k} \nonumber \\ & \quad = (k-1)k (-1)^{m+n} \sum _{\nu =0}^{n-1} (-1)^{\nu } {\nu +m\atopwithdelims ()k}{n\atopwithdelims ()\nu } (z+1)^{\nu +m-k} \nonumber \\ & \quad = (k-1)k \Bigg [ (-1)^{n +k} P(n,m;k;-z-1)+(-1)^{m-1}{m+n\atopwithdelims ()k} (z+1)^{m+n-k}\Bigg ].\nonumber \\ \end{aligned}$$

(2.21)

From Theorem 2 we obtain

$$\begin{aligned} & S(m-1,n-1;k-2;z+1)-S(m-1,n-1;k-2;z) \nonumber \\ & \qquad +(-1)^{k} S(n-1,m-1;k-2;-z-1)-(-1)^{k} S(n-1,m-1;k-2;-z) \nonumber \\ & \quad = (-1)^{m+1} (m+n-1)(m+n){m+n-2\atopwithdelims ()k-2} \bigl ( (z+1)^{m+n-k} -z^{m+n-k} \bigr ) \nonumber \\ & \quad = (k-1)k (-1)^{m-1} {m+n\atopwithdelims ()k}\bigl ( (z+1)^{m+n-k} -z^{m+n-k} \bigr ). \end{aligned}$$

(2.22)

Combining (2.20), (2.21) and (2.22) leads to (1.7).$\square$

Second proof. We have

$$\begin{aligned} P(m,n;k;z)= & \sum _{\nu =0}^m {m\atopwithdelims ()\nu } {\nu +n\atopwithdelims ()k}z^{\nu +n-k}= \frac{1}{k!} \sum _{\nu =0}^m{m\atopwithdelims ()\nu } (z^{\nu +n})^{(k)}\\= & \frac{1}{k!} \bigl ( z^n (z+1)^m\Bigr )^{(k)} \end{aligned}$$

and

$$\begin{aligned} (-1)^{m+n+k} P(n,m;k;-z-1)= & \sum _{\nu =0}^n (-1)^{n-\nu } {n\atopwithdelims ()\nu } {\nu +m\atopwithdelims ()k} (z+1)^{\nu +m-k} \\= & \frac{1}{k!} \sum _{\nu =0}^n (-1)^{n-\nu } {n\atopwithdelims ()\nu } \bigl ( (z+1)^{\nu +m} \bigr )^{(k)} \\= & \frac{1}{k!} \bigl ( z^n (z+1)^m\Bigr )^{(k)}. \end{aligned}$$

$\square$

Proof of Corollary 4

We assume that $m>n\ge 0$. Since, for $m\ge 1$,

$$\begin{aligned} \sum _{\nu =0}^m {m\atopwithdelims ()\nu } (2^{-\nu }-1)B_{\nu } =2^{-m} B_m(2)-B_m(1)= (-1)^m ( 2^{-m}-1) B_m +m 2^{-m}, \end{aligned}$$

we conclude that (1.9) holds for $n=0$. Next, let $m>n\ge 1$. Using (2.3) gives

$$\begin{aligned} & \frac{1}{2}\int _0^{1/2} (-1)^{m-1} \sum _{\nu =0}^{m} (\nu +n){m\atopwithdelims ()\nu } B_{\nu +n-1}(z) dz \nonumber \\ & \qquad =(-1)^{m-1} \sum _{\nu =0}^{m} {m\atopwithdelims ()\nu } (2^{-\nu -n}-1)B_{\nu +n} \end{aligned}$$

(2.23)

and from (2.4) and (2.7) with $r=n$, $s=m$, $t=-1/2$ we get

$$\begin{aligned} & \frac{1}{2} \int _0^{1/2} (-1)^{n} \sum _{\nu =0}^{n} (\nu +m) {n\atopwithdelims ()\nu } B_{\nu +m-1}(-z)dz \nonumber \\ & \quad = \frac{(-1)^{n}}{2}\sum _{\nu =0}^{n} {n \atopwithdelims ()\nu } \Bigl ( -\frac{1}{2} \Bigr )^{\nu +m-1} \bigl [ (1-2^{\nu +m}) B_{\nu +m}+\nu +m\bigr ] \nonumber \\ & \quad = (-1)^{m+n-1} \Bigg [\sum _{\nu =0}^{n} (-1)^{\nu } {n\atopwithdelims ()\nu } \bigl ( 2^{-\nu -m} -1 \bigr ) B_{\nu +m} +\frac{m-n}{2^{m+n}}\Bigg ]. \end{aligned}$$

(2.24)

Applying (1.3), (2.23), (2.24) and

$$\begin{aligned} \sum _{\nu =0}^n (-1)^{\nu } {n\atopwithdelims ()\nu } (2^{-\nu -m}-1)B_{\nu +m} =(-1)^m \sum _{\nu =0}^n {n\atopwithdelims ()\nu } (2^{-\nu -m}-1)B_{\nu +m} \end{aligned}$$

we obtain (1.9). $\square$