A closed formula for the generating function of p-Bernoulli numbers

TL;DR

This paper derives a closed-form generating function for p-Bernoulli numbers using geometric polynomials and applies it to obtain formulas for sums involving Bernoulli and harmonic numbers with Stirling numbers.

Contribution

Introduces a new generating function for p-Bernoulli numbers and derives related closed formulas for summations involving Bernoulli and harmonic numbers.

Findings

Closed-form generating function for p-Bernoulli numbers.

Explicit formulas for finite sums of Bernoulli and harmonic numbers.

Connections between p-Bernoulli numbers and Stirling numbers.

Abstract

In this paper, using geometric polynomials, we obtain a generating function of p-Bernoulli numbers. As a consequences this generating function, we derive closed formulas for the finite summation of Bernoulli and harmonic numbers involving Stirling numbers of the second kind.

Equations

$$B_{n+1,p}=p{B}_{n,p}-\frac{{\left(p+1\right)}^2}{p+2}{B}_{n,p+1}.$$

$$B_{n,p}=\sum _{k=0}^n{\left(-1\right)}^k\left\{\genfrac{}{}{0ex}{}{n}{k}\right\}{\left(\genfrac{}{}{0ex}{}{k+p+1}{k}\right)}^{-1}k!,$$

$$B_{n,p}=\frac{p+1}{p!}\sum _{j=0}^p{\left(-1\right)}^j\left[\genfrac{}{}{0ex}{}{p}{j}\right]B_{n+j},$$

$$\sum _{n=0}^{\infty }{B}_{n,1}\frac{t^n}{n!}$$

$$\sum _{n=0}^{\infty }{B}_{n,2}\frac{t^n}{n!}$$

$$\sum _{n=0}^{\infty }{B}_{n,p}\frac{t^n}{n!}=\frac{\left(p+1\right)\left(t-H_p\right)e^{pt}}{{\left(e^t-1\right)}^{p+1}}+\left(p+1\right)\sum _{k=1}^p\left(\genfrac{}{}{0ex}{}{p}{k}\right)\frac{H_k}{{\left(e^t-1\right)}^{k+1}},$$

$$H_n=\sum _{j=1}^n\frac{1}{j}.$$

$$\frac{1}{1-x\left(e^t-1\right)}=\sum _{n=0}^{\infty }{w}_n\left(x\right)\frac{t^n}{n!}\text{,}$$

$$w_n\left(x\right)=\sum _{k=0}^n\left\{\genfrac{}{}{0ex}{}{n}{k}\right\}k!x^k,$$

$$\sum _{n=0}^{\infty }\left\{\genfrac{}{}{0ex}{}{n}{k}\right\}\frac{t^n}{n!}=\frac{{\left(e^t-1\right)}^k}{k!}.$$

$$\sum _{n=0}^{\infty }{B}_{n,p}\frac{t^n}{n!}=\frac{\left(p+1\right)\left(t-H_p\right)e^{pt}}{{\left(e^t-1\right)}^{p+1}}+\left(p+1\right)\sum _{k=1}^p\left(\genfrac{}{}{0ex}{}{p}{k}\right)\frac{H_k}{{\left(e^t-1\right)}^{k+1}}.$$

$$\frac{{\left(-1\right)}^p}{\left(p+1\right)!}{B}_{n,p}=\underset{p+1\text{ times}}{\underbrace{\underset{-1}{\overset{0}{\int }}\underset{0}{\overset{x_p}{\int }}\dots \underset{0}{\overset{x_3}{\int }}\underset{0}{\overset{x_2}{\int }}}}{w}_n\left(x_1\right)d{x}_{1}d{x}_2\dots d{x}_{p-1}d{x}_p.$$

$$\underset{0}{\overset{x_2}{\int }}{w}_n\left(x_1\right)d{x}_1=\sum _{k=0}^n\left\{\genfrac{}{}{0ex}{}{n}{k}\right\}k!\frac{x_2^{k+1}}{k+1}.$$

$$\underset{0}{\overset{x_3}{\int }}\underset{0}{\overset{x_2}{\int }}{w}_n\left(x_1\right)d{x}_{1}d{x}_2=\sum _{k=0}^n\left\{\genfrac{}{}{0ex}{}{n}{k}\right\}k!\frac{x_3^{k+2}}{\left(k+1\right)\left(k+2\right)}.$$

$$\underset{0}{\overset{x_p}{\int }}\dots \underset{0}{\overset{x_3}{\int }}\underset{0}{\overset{x_2}{\int }}{w}_n\left(x_1\right)d{x}_{1}d{x}_2\dots d{x}_{p-1}=\sum _{k=0}^n\left\{\genfrac{}{}{0ex}{}{n}{k}\right\}k!\frac{x_p^{k+p}}{\left(k+1\right)\cdots \left(k+p\right)}.$$

$$\frac{{\left(-1\right)}^p}{\left(p+1\right)!}\sum _{n=0}^{\infty }{B}_{n,p}\frac{t^n}{n!}$$

$$=\underset{-1}{\overset{0}{\int }}\underset{0}{\overset{x_p}{\int }}\dots \underset{0}{\overset{x_3}{\int }}\left[\underset{0}{\overset{x_2}{\int }}\frac{1}{1-x_1\left(e^t-1\right)}d{x}_1\right]d{x}_2\dots d{x}_{p-1}d{x}_p.$$

$$\underset{0}{\overset{x_2}{\int }}\frac{1}{1-x_1\left(e^t-1\right)}d{x}_1=\frac{-1}{e^t-1}\ln \left(1-x_2\left(e^t-1\right)\right).$$

$$\frac{-1}{e^t-1}\underset{0}{\overset{x_3}{\int }}\ln \left(1-x_2\left(e^t-1\right)\right)d{x}_2$$

$$=\frac{1}{{\left(e^t-1\right)}^2}\left[\left(1-x_3\left(e^t-1\right)\right)\ln \left(1-x_3\left(e^t-1\right)\right)-\left(1-x_3\left(e^t-1\right)\right)+1\right].$$

$$\underset{0}{\overset{x_p}{\int }}\dots \underset{0}{\overset{x_3}{\int }}\underset{0}{\overset{x_2}{\int }}{w}_n\left(x_1\right)d{x}_{1}d{x}_2\dots d{x}_{p-1}$$

$$=\frac{{\left(-1\right)}^p}{{\left(e^t-1\right)}^p\left(p-1\right)!}{\left(1-x_p\left(e^t-1\right)\right)}^{p-1}\ln \left(1-x_p\left(e^t-1\right)\right)$$

$$-\frac{{\left(-1\right)}^p}{{\left(e^t-1\right)}^p\left(p-1\right)!}\left[H_{p-1}{\left(1-x_p\left(e^t-1\right)\right)}^{p-1}+H_{p-1}\right]$$

$$+\sum _{k=1}^{p-2}\frac{{\left(-1\right)}^{p-k}{H}_{p-1-k}{x}_p^k}{{\left(e^t-1\right)}^{p-k}\left(p-1-k\right)!k!}.$$

$$\underset{0}{\overset{y}{\int }}\underset{0}{\overset{x_p}{\int }}\dots \underset{0}{\overset{x_3}{\int }}\underset{0}{\overset{x_2}{\int }}{w}_n\left(x_1\right)d{x}_{1}d{x}_2\dots d{x}_{p-1}d{x}_p$$

$$=\frac{{\left(-1\right)}^p}{{\left(e^t-1\right)}^p\left(p-1\right)!}\underset{0}{\overset{y}{\int }}{\left(1-x_p\left(e^t-1\right)\right)}^{p-1}\ln \left(1-x_p\left(e^t-1\right)\right)d{x}_p$$

$$-\frac{{\left(-1\right)}^p{H}_{p-1}}{{\left(e^t-1\right)}^p\left(p-1\right)!}\underset{0}{\overset{y}{\int }}{\left(1-x_p\left(e^t-1\right)\right)}^{p-1}d{x}_p+\frac{{\left(-1\right)}^p{H}_{p-1}}{{\left(e^t-1\right)}^p\left(p-1\right)!}\underset{0}{\overset{y}{\int }}d{x}_p$$

$$+\sum _{k=1}^{p-2}\frac{{\left(-1\right)}^{p-k}{H}_{p-1-k}}{{\left(e^t-1\right)}^{p-k}\left(p-1-k\right)!k!}\underset{0}{\overset{y}{\int }}{x}_p^{k}d{x}_p.$$

$$\frac{{\left(-1\right)}^p}{{\left(e^t-1\right)}^p\left(p-1\right)!}\underset{0}{\overset{y}{\int }}{\left(1-x_p\left(e^t-1\right)\right)}^{p-1}\ln \left(1-x_p\left(e^t-1\right)\right)d{x}_p$$

$$=\frac{{\left(-1\right)}^{p+1}}{{\left(e^t-1\right)}^{p+1}p!}\left[{\left(1-y\left(e^t-1\right)\right)}^p\ln \left(1-y\left(e^t-1\right)\right)-\frac{{\left(1-y\left(e^t-1\right)\right)}^p}{p}+\frac{1}{p}\right].$$

$$\frac{{\left(-1\right)}^p{H}_{p-1}}{{\left(e^t-1\right)}^p\left(p-1\right)!}\underset{0}{\overset{y}{\int }}{\left(1-x_p\left(e^t-1\right)\right)}^{p-1}d{x}_p$$

$$=\frac{{\left(-1\right)}^{p+1}{H}_{p-1}}{{\left(e^t-1\right)}^{p+1}p!}{\left(1-y\left(e^t-1\right)\right)}^p-\frac{{\left(-1\right)}^{p+1}{H}_{p-1}}{{\left(e^t-1\right)}^{p+1}p!}.$$

$$\frac{{\left(-1\right)}^p{H}_{p-1}}{{\left(e^t-1\right)}^p\left(p-1\right)!}\underset{0}{\overset{y}{\int }}d{x}_p=\frac{{\left(-1\right)}^{p+1}{H}_{p-1}}{{\left(e^t-1\right)}^p\left(p-1\right)!}y$$

$$\sum _{k=1}^{p-2}\frac{{\left(-1\right)}^{p-k}{H}_{p-1-k}}{{\left(e^t-1\right)}^{p-k}\left(p-1-k\right)!k!}\underset{0}{\overset{y}{\int }}{x}_p^{k}d{x}_p=\sum _{k=2}^{p-1}\frac{{\left(-1\right)}^{p+1-k}{H}_{p-k}{y}^k}{{\left(e^t-1\right)}^{p+1-k}\left(p-k\right)!k!},$$

$$\underset{0}{\overset{y}{\int }}\underset{0}{\overset{x_p}{\int }}\dots \underset{0}{\overset{x_3}{\int }}\underset{0}{\overset{x_2}{\int }}{w}_n\left(x_1\right)d{x}_{1}d{x}_2\dots d{x}_{p-1}d{x}_p$$

$$=\frac{{\left(-1\right)}^{p+1}}{{\left(e^t-1\right)}^{p+1}p!}{\left(1-y\left(e^t-1\right)\right)}^p\ln \left(1-y\left(e^t-1\right)\right)$$

Full text

**2000 Mathematics Subject Classification: **11B68, 11B83

Key words: $p$-Bernoulli number, geometric polynomial, harmonic number.

1 Introduction

Rahmani [9] introduced $p$-Bernoulli numbers by constructing an infinite matrix as follows:

The first row of the matrix $B_{0,p}=1$ and each entry $B_{n,p}$ is given recursively by

[TABLE]

The first column of the matrix $B_{n,0}=B_{n}$. Here, $B_{n}$ is the $n$th Bernoulli number.

For every integer $p\geq-1$, these numbers have an explicit formula

[TABLE]

and are closely related to Bernoulli numbers by the following formula

[TABLE]

where $\genfrac{[}{]}{0.0pt}{}{n}{k}$ is the Stirling number of the first kind [6]. The first few generating functions for $B_{n,p}$ ($p=1,2$) are

[TABLE]

The main purpose of this study is to give a close form of the above results as

[TABLE]

where $H_{n}$ is the harmonic numbers, defined by [6, p. 258]

[TABLE]

As a consequences of (3), we have closed formulas for the finite summation of Bernoulli and harmonic numbers.

For the proof of (3), we use some properties of geometric polynomials. The geometric polynomials are defined by means of the following generating function [10]

[TABLE]

and have the explicit formula

[TABLE]

where $\genfrac{\{}{\}}{0.0pt}{}{n}{k}$ is the Stirling number of the second kind [6]. The Stirling numbers of the second kind are defined by means of the following generating function

[TABLE]

Some other properties of geometric polynomials can be found in [1, 2, 3, 4, 5, 7].

2 A new generating function for $p$-Bernoulli numbers

In this section, the main theorem and its applications are given.

Now, we give the main theorem of this paper.

Theorem 1

For $p\geq 0$, the following generating function holds true:

[TABLE]

For the proof of main theorem, we first need the following proposition.

Proposition 2

For $n>p\geq 0$, we have

[TABLE]

Proof. If we integrate both sides of (5) with respect to $x_{1}$ from [math] to $x_{2}$, we have

[TABLE]

Integrating both sides of the above equation with respect to $x_{2}$ from [math] to $x_{3}$, we obtain

[TABLE]

Applying the same procedure for $p$ times yields

[TABLE]

Finally, integrating both sides of the above equation with respect to $x_{p}$ from $-1$ to [math] and using (1) gives the desired equation.

We note that taking $p=0$ in (8) gives [8, Theorem 1.2].

Now, we are ready to give the proof of the main theorem.

Proof of Theorem 1. Multiplying both sides of (8) with $\frac{t^{n}}{n!}$ and summing over $n$ from [math] to infinitive, we have

[TABLE]

If we evaluate the first integral, we obtain

[TABLE]

For the second time, we evaluate

[TABLE]

By induction on $p$, let us assume that the following equation holds

[TABLE]

Now, we want to prove that (9) holds for the case $p+1.$ Let us integrate both sides of (9) with respect to $x_{p}$ from [math] to $y.$ Then we have

[TABLE]

The first integral in the right hand-side equals

[TABLE]

For the second integral in the right hand-side, we obtain

[TABLE]

For the third and fourth integrals, we find

[TABLE]

and

[TABLE]

respectively. Combining (10), (11), (12) and (13), we achieve that

[TABLE]

Finally, setting $y=-1$ in the above equation and using (8), we arrive at the desired equation.

As an application of Theorem 1, we give the following theorem.

Theorem 3

For $n>p\geq 0$, we have

[TABLE]

Proof. Multiplying both sides of (7) with $\frac{\left(e^{t}-1\right)^{p+1}}{\left(p+1\right)!}$ and using (6), the left hand side of (7) becomes

[TABLE]

For the right hand side of (7), we obtain

[TABLE]

Finally, comparing the coefficients of $\frac{t^{n}}{n!}$ in (15) and (16) completes the proof.

Using (2) in Theorem 3 gives the following corollary.

Corollary 4

For $n>p\geq 0$,

[TABLE]

As a consequences of Corollary 4, the following sums are obtained $\left(p=1,2\right):$

[TABLE]

Setting $n=p+1$ in Theorem 3 and using $B_{0,p}=1$, we arrive at the following corollary.

Corollary 5

For $p\geq 1$, we obtain a new closed formula for the finite summation of harmonic numbers

[TABLE]