Polynomial Identities for Binomial Sums of Harmonic Numbers of Higher Order

Abstract

We study the formulas for binomial sums of harmonic numbers of higher order ${\displaystyle \sum _{k=0}^n}{H}_k^{\left(r\right)}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(1-q)}^k{q}^{n-k}=H_n^{\left(r\right)}-{\displaystyle \sum _{j=1}^n}{\mathcal{D}}_r(n,j){\displaystyle \frac{q^j}{j}}.$ Recently, Mneimneh proved that ${\mathcal{D}}_1(n,j)=1$. In this paper, we find several different expressions of ${\mathcal{D}}_r(n,j)$ for $r\ge 1$.

1. Introduction

For a positive integer r, define the n-th harmonic number of order r by

$$H_n^{\left(r\right)}:=\sum _{i=1}^n{\displaystyle \frac{1}{i^r}}.$$

When $r=1$, $H_n=H_n^{\left(1\right)}$ is the original harmonic number. In this paper, we study the formula

$$\sum _{k=0}^n{H}_k^{\left(r\right)}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(1-q)}^k{q}^{n-k}=H_n^{\left(r\right)}-\sum _{j=1}^n{\mathcal{D}}_r(n,j){\displaystyle \frac{q^j}{j}}.$$

(1)

In [1], for a positive integer n and $0\le q\le 1$, it is shown that ${\mathcal{D}}_1(n,j)=1$. Namely,

$$\sum _{k=0}^n{H}_k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(1-q)}^k{q}^{n-k}=H_n-\sum _{j=1}^n{\displaystyle \frac{q^j}{j}}.$$

(2)

This relation is derived by the author from an interesting probabilistic analysis. The identity (2) is a generalization of

$$\sum _{k=0}^n{H}_k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)=2^n\left(H_n-\sum _{j=1}^n{\displaystyle \frac{1}{j{2}^j}}\right),$$

which has been proved in [2] in the field of symbolic computation and in [3] in finite differences. The identity (2) is a special case of a general result of Boyadzhiev [4]:

$$\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)a^k{b}^{n-k}{H}_k={(a+b)}^n{H}_n-\left(b{(a+b)}^{n-1}+{\displaystyle \frac{b^2}{2}}{(a+b)}^{n-2}+\cdots +{\displaystyle \frac{b^n}{n}}\right).$$

(3)

In addition, Boyadzhiev’s main result (3) has been generalized in [5] to multiple harmonic-like numbers.

The main aim of this paper is to show several different expressions of ${\mathcal{D}}_r(n,j)$ as no simple form has been found.

In fact, more different generalizations of (1) or (2) can be considered. For example, recently, in [6], the so-called hyperharmonic number generalizes the harmonic number of the order r in the formula. In [7], more generalized sums and their application to multiple polylogarithms are given. In [8], some expressions of Mneimneh-type binomial sums are established, involving multiple harmonic-type sums in terms of finite sums of Stirling numbers, Bell numbers, and some related variables, and a conjecture is proposed. Then, in [9], this conjecture is resolved and generalized, and the transformation of generalized Mneimneh-like sums is presented. When we generalize too much, a lot of interesting and essential properties may be lost. Therefore, we do not consider further generalized harmonic numbers in this paper.

2. Observation

By using the harmonic numbers to express ${\mathcal{D}}_r(n,j)$, for $1\le r\le 7$, we can manually obtain the following. Some initial trials for $r=1,2,3$ are given below.

$$\begin{array}{cc}\hfill & {\mathfrak{D}}_1(n,j)=1,\hfill \\ \hfill & {\mathfrak{D}}_2(n,j)=H_n-H_{n-j},\hfill \\ \hfill & {\mathfrak{D}}_3(n,j)={\displaystyle \frac{{(H_n-H_{n-j})}^2}{2}}+{\displaystyle \frac{H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)}}{2}},\hfill \\ \hfill & {\mathfrak{D}}_4(n,j)={\displaystyle \frac{{(H_n-H_{n-j})}^3}{6}}+{\displaystyle \frac{(H_n-H_{n-j})(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)})}{2}}+{\displaystyle \frac{H_n^{\left(3\right)}-H_{n-j}^{\left(3\right)}}{3}},\hfill \\ \hfill & {\mathcal{D}}_5(n,j)={\displaystyle \frac{{(H_n-H_{n-j})}^4}{4!}}+{\displaystyle \frac{{(H_n-H_{n-j})}^2(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)})}{4}}\hfill \\ \hfill & +{\displaystyle \frac{(H_n-H_{n-j})(H_n^{\left(3\right)}-H_{n-j}^{\left(3\right)})}{3}}+{\displaystyle \frac{{(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)})}^2}{8}}+{\displaystyle \frac{H_n^{\left(4\right)}-H_{n-j}^{\left(4\right)}}{4}},\hfill \\ \hfill & {\mathcal{D}}_6(n,j)={\displaystyle \frac{{(H_n-H_{n-j})}^5}{5!}}+{\displaystyle \frac{{(H_n-H_{n-j})}^3(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)})}{12}}\hfill \\ \hfill & +{\displaystyle \frac{{(H_n-H_{n-j})}^2(H_n^{\left(3\right)}-H_{n-j}^{\left(3\right)})}{6}}+{\displaystyle \frac{(H_n-H_{n-j}){(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)})}^2}{8}}\hfill \\ \hfill & +{\displaystyle \frac{(H_n-H_{n-j})(H_n^{\left(4\right)}-H_{n-j}^{\left(4\right)}}{4}}+{\displaystyle \frac{(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)})(H_n^{\left(3\right)}-H_{n-j}^{\left(3\right)})}{6}}\hfill \\ \hfill & +{\displaystyle \frac{H_n^{\left(5\right)}-H_{n-j}^{\left(5\right)}}{5}},\hfill \\ \hfill & {\mathcal{D}}_7(n,j)={\displaystyle \frac{{(H_n-H_{n-j})}^6}{6!}}+{\displaystyle \frac{{(H_n-H_{n-j})}^4(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)})}{48}}\hfill \\ \hfill & +{\displaystyle \frac{{(H_n-H_{n-j})}^3(H_n^{\left(3\right)}-H_{n-j}^{\left(3\right)})}{18}}+{\displaystyle \frac{{(H_n-H_{n-j})}^2{(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)})}^2}{16}}\hfill \\ \hfill & +{\displaystyle \frac{{(H_n-H_{n-j})}^2(H_n^{\left(4\right)}-H_{n-j}^{\left(4\right)})}{8}}+{\displaystyle \frac{(H_n-H_{n-j})(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)})(H_n^{\left(3\right)}-H_{n-j}^{\left(3\right)})}{6}}\hfill \\ \hfill & +{\displaystyle \frac{(H_n-H_{n-j})(H_n^{\left(5\right)}-H_{n-j}^{\left(5\right)})}{5}}+{\displaystyle \frac{{(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)})}^3}{48}}\hfill \\ \hfill & +{\displaystyle \frac{(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)})(H_n^{\left(4\right)}-H_{n-j}^{\left(4\right)})}{8}}+{\displaystyle \frac{{(H_n^{\left(3\right)}-H_{n-j}^{\left(3\right)})}^2}{18}}+{\displaystyle \frac{H_n^{\left(6\right)}-H_{n-j}^{\left(6\right)}}{6}}.\hfill \end{array}$$

It is interesting to observe that the number of terms of each of the right-hand sides of ${\mathcal{D}}_r(n,j)$ is equal to the number of partitions of r ($1\le r\le 7$), respectively. In addition, the same terms of generalized harmonic numbers appear in [10,11]

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{H_n}{(n+1)(n+2)}}=1,\hfill \\ \hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(H_n\right)}^2-H_n^{\left(2\right)}}{2(n+1)(n+2)}}=1,\hfill \\ \hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(H_n\right)}^3-3H_n{H}_n^{\left(2\right)}+2H_n^{\left(3\right)}}{3!(n+1)(n+2)}}=1,\hfill \\ \hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(H_n\right)}^4-6{\left(H_n\right)}^2{H}_n^{\left(2\right)}+8H_n{H}_n^{\left(3\right)}+3{\left(H_n^{\left(2\right)}\right)}^2-6H_n^{\left(4\right)}}{4!(n+1)(n+2)}}=1,\hfill \\ \hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{1}{5!(n+1)(n+2)}}({\left(H_n\right)}^5-10{\left(H_n\right)}^3{H}_n^{\left(2\right)}+20{\left(H_n\right)}^2{H}_n^{\left(3\right)}\hfill \\ \hfill & +15H_n{\left(H_n^{\left(2\right)}\right)}^2-30H_n{H}_n^{\left(4\right)}-20H_n^{\left(2\right)}{H}_n^{\left(3\right)}+24H_n^{\left(5\right)})=1,\hfill \\ \hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{1}{6!(n+1)(n+2)}}({\left(H_n\right)}^6-15{\left(H_n\right)}^4{H}_n^{\left(2\right)}+40{\left(H_n\right)}^3{H}_n^{\left(3\right)}\hfill \\ \hfill & +45{\left(H_n\right)}^2{\left(H_n^{\left(2\right)}\right)}^2-90{\left(H_n\right)}^2{H}_n^{\left(4\right)}-120H_n{H}_n^{\left(2\right)}{H}_n^{\left(3\right)}+144H_n{H}_n^{\left(5\right)}\hfill \\ \hfill & -15{\left(H_n^{\left(2\right)}\right)}^3+90H_n^{\left(2\right)}{H}_n^{\left(4\right)}+40{\left(H_n^{\left(3\right)}\right)}^2-120H_n^{\left(5\right)})=1.\hfill \end{array}$$

Some Initial Trials

For observation, we calculate the expressions of ${\mathcal{D}}_r(n,j)$ for small r, one by one, using the method in [12]. Here, we mention the cases for $r=1,2,3$.

When $r=1$, we find the following relation. Thus, by Theorem 1 in the next section, we can obtain ${\mathfrak{D}}_1(n,j)=1$.

Lemma 1. 

$$\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right)=1.$$

(4)

When $r=2$, we find the following relation. Here, ${\left(n\right)}_j=n(n-1)\cdots (n-j+1)$ ($j\ge 1$) is the falling factorial with ${\left(n\right)}_0=1$, and $\left[{\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right]$ denotes the (unsigned) Stirling number of the first kind, arising from the relation ${\left(x\right)}_n={\sum }_{k=0}^n{(-1)}^{n-k}\left[{\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right]x^k$.

Lemma 2. 

$$\begin{array}{l}{\displaystyle \sum _{l=0}^{j-1}}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right){\displaystyle \frac{1}{n-l}}=H_n-H_{n-j}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\displaystyle \frac{1}{{\left(n\right)}_j}}{\displaystyle \sum _{\nu =0}^{j-1}}{(-1)}^{j-\nu -1}(\nu +1)\left[{\displaystyle \genfrac{}{}{0pt}{}{j}{\nu +1}}\right]n^{\nu }.\end{array}$$

(5)

Note that

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right){\displaystyle \frac{1}{n-l}}\ne {\displaystyle \frac{l+1}{{\left(n\right)}_j}}\left[{\displaystyle \genfrac{}{}{0pt}{}{j}{l+1}}\right]n^l.$$

Hence, we have the following formula.

Corollary 1. 

$$\begin{array}{l}{\displaystyle \sum _{k=0}^n}{H}_k^{\left(2\right)}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(1-q)}^k{q}^{n-k}=H_n^{\left(2\right)}-{\displaystyle \sum _{j=1}^n}(H_n-H_{n-j}){\displaystyle \frac{q^j}{j}}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em} \hspace{1em}\hspace{1em}=H_n^{\left(2\right)}-{\displaystyle \sum _{j=1}^n}\left({\displaystyle \frac{1}{{\left(n\right)}_j}}{\displaystyle \sum _{\nu =0}^{j-1}}{(-1)}^{j-\nu -1}(\nu +1)\left[{\displaystyle \genfrac{}{}{0pt}{}{j}{\nu +1}}\right]n^{\nu }\right){\displaystyle \frac{q^j}{j}}.\end{array}$$

(6)

Proof of Lemma 1. 

Put

$$\begin{array}{cc}\hfill A(n,j)& =\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right),\hfill \\ \hfill B(n,j)& =\sum _{l=0}^j{(-1)}^{j-l}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right).\hfill \end{array}$$

Since

$$\begin{array}{cc}\hfill B(n+1,j)& ={\displaystyle \frac{n+1}{n-j+1}}B(n,j)={\displaystyle \frac{(n+1)n}{(n-j+1)(n-j)}}B(n-1,j)\hfill \\ \hfill & =\cdots ={\displaystyle \frac{(n+1)n\cdots (j+1)}{(n-j+1)!}}B(j,j)\hfill \\ \hfill & ={\displaystyle \frac{(n+1)!}{(n-j+1)!j!}}\sum _{l=0}^j{(-1)}^{j-l}\left({\displaystyle \genfrac{}{}{0pt}{}{j}{l}}\right)\hfill \\ \hfill & =\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{j}}\right){(1-1)}^j=0,\hfill \end{array}$$

we have

$$A(n,j+1)-A(n,j)=B(n,j)=0.$$

Hence, we obtain

$$A(n,j)=A(n,j-1)=\cdots =A(n,1)={(-1)}^0\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{n-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{0}}\right)=1.$$

□

Proof of Lemma 2. 

The Formula (5) is yielded from the definition of the Stirling numbers of the first kind:

$$\begin{array}{cc}\hfill {\left(x\right)}_j& =\sum _{k=0}^j{(-1)}^{j-k}\left[{\displaystyle \genfrac{}{}{0pt}{}{j}{k}}\right]x^k\hfill \\ \hfill & =\sum _{\nu =0}^{j-1}{(-1)}^{j-\nu -1}\left[{\displaystyle \genfrac{}{}{0pt}{}{j}{\nu +1}}\right]x^{\nu +1}(\mathrm{if}j\ge 1).\hfill \end{array}$$

Differentiatingboth sides with respect to x gives

$${\left(x\right)}_j\sum _{l=0}^{j-1}{\displaystyle \frac{1}{x-l}}=\sum _{\nu =0}^{j-1}{(-1)}^{j-\nu -1}(\nu +1)\left[{\displaystyle \genfrac{}{}{0pt}{}{j}{\nu +1}}\right]x^{\nu }.$$

Thus, the right-hand side of (5) is equal to

$$\sum _{l=0}^{j-1}{\displaystyle \frac{1}{n-l}}=H_n-H_{n-j}.$$

Put the left-hand side of (5) as

$$C(n,j):=\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right){\displaystyle \frac{1}{n-l}}.$$

Then,

$$\begin{array}{cc}\hfill & C(n,j)-C(n,j-1)\hfill \\ \hfill & =\sum _{l=0}^{j-2}{(-1)}^{j-l-1}\left(\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right)\right){\displaystyle \frac{1}{n-l}}\hfill \\ \hfill & +\left({\displaystyle \genfrac{}{}{0pt}{}{n}{j-1}}\right){\displaystyle \frac{1}{n-j+1}}\hfill \\ \hfill & =\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l}{n-j+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right){\displaystyle \frac{1}{n-l}}\hfill \\ \hfill & =\left({\displaystyle \genfrac{}{}{0pt}{}{n}{j-1}}\right)\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{l}}\right){\displaystyle \frac{1}{n-l}}.\hfill \end{array}$$

Now,

$$\begin{array}{cc}\hfill & \sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{l}}\right){\displaystyle \frac{1}{n-l}}\hfill \\ \hfill & ={\left.\int \sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{l}}\right)x^{n-l-1}dx\right|}_{x=1}\hfill \\ \hfill & ={\left.\int x^{n-1}{\left(1-{\displaystyle \frac{1}{x}}\right)}^{j-1}dx\right|}_{x=1}\hfill \\ \hfill & ={\left.{\left(1-{\displaystyle \frac{1}{x}}\right)}^j{\displaystyle \frac{{}_{2}F_1(-j+1,n-j+1;n-j+2;x)}{{(x-1)}^j(n-j+1)}}\right|}_{x=1}\hfill \\ \hfill & ={\displaystyle \frac{\Gamma \left(j\right)\Gamma (n-j+2)}{(n-j+1)\Gamma (n+1)}}={\displaystyle \frac{(j-1)!(n-j)!}{n!}},\hfill \end{array}$$

where ${}_{2}F_1(a,b;c;z)$ is the Gauss hypergeometric function. Hence,

$$C(n,j)-C(n,j-1)={\displaystyle \frac{1}{n-j+1}}.$$

Therefore,

$$\begin{array}{cc}\hfill C(n,j)& =C(n,j-1)+{\displaystyle \frac{1}{n-j+1}}\hfill \\ \hfill & =C(n,j-2)+{\displaystyle \frac{1}{n-j+2}}+{\displaystyle \frac{1}{n-j+1}}\hfill \\ \hfill & =\cdots \hfill \\ \hfill & =C(n,1)+{\displaystyle \frac{1}{n-1}}+\cdots +{\displaystyle \frac{1}{n-j+2}}+{\displaystyle \frac{1}{n-j+1}}\hfill \\ \hfill & =\sum _{l=0}^{j-1}{\displaystyle \frac{1}{n-l}}.\hfill \end{array}$$

(7)

□

When $r=3$, we have the following.

Lemma 3. 

$$\begin{array}{cc}\hfill & \sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right){\displaystyle \frac{1}{{(n-l)}^2}}\hfill \\ \hfill & ={\displaystyle \frac{{(H_n-H_{n-j})}^2}{2}}+{\displaystyle \frac{H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)}}{2}}.\hfill \end{array}$$

Therefore, we have the following formula.

Corollary 2. 

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{H}_k^{\left(3\right)}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(1-q)}^k{q}^{n-k}\hfill \\ \hfill & =H_n^{\left(3\right)}-\sum _{j=1}^n\left({\displaystyle \frac{{(H_n-H_{n-j})}^2}{2}}+{\displaystyle \frac{H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)}}{2}}\right){\displaystyle \frac{q^j}{j}}.\hfill \end{array}$$

Proof of Lemma 3. 

Put

$$D(n,j):=\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right){\displaystyle \frac{1}{{(n-l)}^2}}.$$

Then,

$$D(n,j)-D(n,j-1)=\left({\displaystyle \genfrac{}{}{0pt}{}{n}{j-1}}\right)\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{l}}\right){\displaystyle \frac{1}{{(n-l)}^2}}.$$

(8)

Weshall prove that

$$D(n,j)-D(n,j-1)={\displaystyle \frac{1}{n-j+1}}(H_n-H_{n-j}).$$

(9)

By (9), we obtain

$$\begin{array}{cc}\hfill D(n,j)& =\left({\displaystyle \frac{H_n}{n-j+1}}-{\displaystyle \frac{H_{n-j}}{n-j+1}}\right)+\left({\displaystyle \frac{H_n}{n-j+2}}-{\displaystyle \frac{H_{n-j+1}}{n-j+2}}\right)\hfill \\ \hfill & +\cdots +\left({\displaystyle \frac{H_n}{n-1}}-{\displaystyle \frac{H_{n-2}}{n-1}}\right)+\left({\displaystyle \frac{H_n}{n}}-{\displaystyle \frac{H_{n-1}}{n}}\right)\hfill \\ \hfill & =H_n\left({\displaystyle \frac{1}{n-j+1}}+{\displaystyle \frac{1}{n-j+2}}+\cdots +{\displaystyle \frac{1}{n-1}}+{\displaystyle \frac{1}{n}}\right)\hfill \\ \hfill & -\left({\displaystyle \frac{H_{n-j}}{n-j+1}}+{\displaystyle \frac{H_{n-j+1}}{n-j+2}}+\cdots +{\displaystyle \frac{H_{n-2}}{n-1}}+{\displaystyle \frac{H_{n-1}}{n}}\right)\hfill \\ \hfill & =H_n(H_n-H_{n-j})-\left({\displaystyle \frac{H_n^2-H_n^{\left(2\right)}}{2}}-{\displaystyle \frac{H_{n-j}^2-H_{n-j}^{\left(2\right)}}{2}}\right)\hfill \\ \hfill & ={\displaystyle \frac{{(H_n-H_{n-j})}^2}{2}}+{\displaystyle \frac{H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)}}{2}}.\hfill \end{array}$$

In order to prove (9), we put

$$E(n,j)=(n-j+1)\left(D(n,j)-D(n,j-1)\right).$$

Then, by (8) and Lemma 1 (4), we have

$$\begin{array}{cc}\hfill & E(n,j)-E(n,j-1)\hfill \\ \hfill & ={\displaystyle \frac{1}{n-j+1}}\sum _{l=0}^{j-2}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right)+{\displaystyle \frac{1}{n-j+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{j-1}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{n-j+1}}\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right)={\displaystyle \frac{1}{n-j+1}}.\hfill \end{array}$$

Hence, by $D(n,1)=1/n^2$, we obtain

$$\begin{array}{cc}\hfill & D(n,j)-D(n,j-1)={\displaystyle \frac{E(n,j)}{n-j+1}}\hfill \\ \hfill & ={\displaystyle \frac{1}{n-j+1}}\left({\displaystyle \frac{1}{n-j+1}}+{\displaystyle \frac{1}{n-j+2}}+\cdots +{\displaystyle \frac{1}{n-1}}+E(n,1)\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{n-j+1}}(H_n-H_{n-j}).\hfill \end{array}$$

which is the right-hand side of (9). □

3. Expressions (Main Results)

Let $n,j,r$ be positive integers.

Theorem 1. 

$${\mathcal{D}}_r(n,j)=\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right){\displaystyle \frac{1}{{(n-l)}^{r-1}}}.$$

Theorem 2. 

For $r\ge 1$,

$${\mathcal{D}}_{r+1}(n,j)=\sum _{j_1=1}^j\sum _{j_2=1}^{j_1}\cdots \sum _{j_r=1}^{j_{r-1}}{\displaystyle \frac{1}{(n-j_1+1)(n-j_2+1)\cdots (n-j_r+1)}}.$$

${\mathcal{D}}_r(n,j)$ ($r\ge 2$) can be expressed in terms of the determinant [13] ([Ch. I S2]). See also [14,15].

Theorem 3. 

$$\begin{array}{c}{\mathcal{D}}_{r+1}(n,j)\hfill \\ \hspace{1em}\hspace{1em}={\displaystyle \frac{1}{r!}}\left|\begin{array}{ccccc}{H}_n-H_{n-j}& -1& 0& \cdots & 0\\ H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)}& H_n-H_{n-j}& -2& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ H_n^{(r-1)}-H_{n-j}^{(r-1)}& H_n^{(r-2)}-H_{n-j}^{(r-2)}& H_n^{(r-3)}-H_{n-j}^{(r-3)}& \cdots & -r+1\\ H_n^{\left(r\right)}-H_{n-j}^{\left(r\right)}& H_n^{(r-1)}-H_{n-j}^{(r-1)}& H_n^{(r-2)}-H_{n-j}^{(r-2)}& \cdots & H_n-H_{n-j}\end{array}\right|.\end{array}$$

Remark 1. 

By using the inversion formula (see, e.g., ([Lemma 1] [16]), ([13] (p. 28)) regarding (12) below, we also have

$$\begin{array}{c}{(-1)}^{r-1}(H_n^{\left(r\right)}-H_{n-j}^{\left(r\right)})\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}=\left|\begin{array}{ccccc}{\mathcal{D}}_2(n,j)& 1& 0& \cdots & 0\\ 2{\mathcal{D}}_3(n,j)& {\mathcal{D}}_2(n,j)& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ (r-1){\mathcal{D}}_r(n,j)& {\mathcal{D}}_{r-1}(n,j)& {\mathcal{D}}_{r-2}(n,j)& \cdots & 1\\ r{\mathcal{D}}_{r+1}(n,j)& {\mathcal{D}}_r(n,j)& {\mathcal{D}}_{r-1}(n,j)& \cdots & {\mathcal{D}}_2(n,j)\end{array}\right|.\end{array}$$

${\mathcal{D}}_r(n,j)$ ($r\ge 2$) can be expressed by a combinatorial sum ([Proposition 1 (17)] [11]):

Theorem 4. 

$$\begin{array}{c}{\mathcal{D}}_{r+1}(n,j)\hfill \\ \hfill =\sum _{i_1+2i_2+3i_3+\cdots =r}{\displaystyle \frac{1}{i_1!i_2!i_3!\cdots }}{\left({\displaystyle \frac{H_n-H_{n-j}}{1}}\right)}^{i_1}{\left({\displaystyle \frac{H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)}}{2}}\right)}^{i_2}{\left({\displaystyle \frac{H_n^{\left(3\right)}-H_{n-j}^{\left(3\right)}}{3}}\right)}^{i_3}\cdots .\end{array}$$

Remember that the (complete exponential) Bell polynomial ${\mathbf{Y}}_n(x_1,x_2,\cdots ,x_n)$ is defined by

$$exp\left(\sum _{m=1}^{\infty }{x}_m{\displaystyle \frac{t^m}{m!}}\right)=1+\sum _{n=1}^{\infty }{\mathbf{Y}}_n(x_1,x_2,\cdots ,x_n){\displaystyle \frac{t^n}{n!}}$$

(see, e.g., ([Ch.3.3] [17])). That is,

$$\begin{array}{cc}\hfill & {\mathbf{Y}}_n(x_1,x_2,\cdots ,x_n)\hfill \\ \hfill & =\sum _{k=1}^n\sum {\displaystyle \frac{n!}{i_1!i_2!\cdots i_{n-k+1}!}}{\left({\displaystyle \frac{x_1}{1!}}\right)}^{i_1}{\left({\displaystyle \frac{x_2}{2!}}\right)}^{i_2}\cdots {\left({\displaystyle \frac{x_{n-k+1}}{(n-k+1)!}}\right)}^{i_{n-k+1}}\hfill \end{array}$$

with ${\mathbf{Y}}_0=1$. Here, the second sum satisfies the following conditions:

$$i_1+2i_2+3i_3+\cdots +(n-k+1)i_{n-k+1}=n,i_1+i_2+i_3+\cdots =k.$$

Theorem 5. 

For $r\ge 1$, we have

$${\mathcal{D}}_{r+1}(n,j)={\displaystyle \frac{1}{r!}}{\mathbf{Y}}_r\left(H_n-H_{n-j},1!(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)}),2!(H_n^{\left(3\right)}-H_{n-j}^{\left(3\right)}),\cdots \right).$$

4. Proof

Proof of Theorem 1. 

We shall show that

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{H}_k^{\left(r\right)}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(1-q)}^k{q}^{n-k}\hfill \\ \hfill & =H_n^{\left(r\right)}-\sum _{j=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{j}}\right)\left(\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{l}}\right){\displaystyle \frac{1}{{(n-l)}^r}}\right)q^j.\hfill \end{array}$$

(10)

We have

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{H}_k^{\left(r\right)}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(1-q)}^k{q}^{n-k}=\sum _{k=0}^n{H}_k^{\left(r\right)}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\sum _{l=0}^k{(-1)}^{k-l}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{l}}\right)q^{n-l}\hfill \\ \hfill & =\sum _{l=0}^n{q}^{n-l}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right)\sum _{k=l}^n{(-1)}^{k-l}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l}{n-k}}\right)H_k^{\left(r\right)}\hfill \\ \hfill & =\sum _{j=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{j}}\right)q^j\sum _{\nu =0}^j{(-1)}^{j-\nu }\left({\displaystyle \genfrac{}{}{0pt}{}{j}{\nu }}\right)H_{n-\nu }^{\left(r\right)}\hfill \\ \hfill & =H_n^{\left(r\right)}-\sum _{j=1}^n{(-1)}^{j-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{j}}\right)q^j\sum _{\nu =0}^j{(-1)}^{\nu }\left({\displaystyle \genfrac{}{}{0pt}{}{j}{\nu }}\right)\sum _{l=0}^{n-1}{\displaystyle \frac{1}{{(n-l)}^r}}.\hfill \end{array}$$

Since

$$\sum _{\nu =0}^l{(-1)}^{\nu }\left({\displaystyle \genfrac{}{}{0pt}{}{j}{\nu }}\right)={(-1)}^l\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{l}}\right)\left(\mathrm{proved}\mathrm{by}\mathrm{induction}\mathrm{on}l(\ge 0)\right)$$

and

$$\sum _{\nu =0}^j{(-1)}^{\nu }\left({\displaystyle \genfrac{}{}{0pt}{}{j}{\nu }}\right)={(1-1)}^j=0,$$

we have

$$\begin{array}{cc}\hfill & \sum _{\nu =0}^j{(-1)}^{\nu }\left({\displaystyle \genfrac{}{}{0pt}{}{j}{\nu }}\right)\sum _{l=0}^{n-1}{\displaystyle \frac{1}{{(n-l)}^r}}\hfill \\ \hfill & =\sum _{l=0}^{j-1}\left(\sum _{\nu =0}^l{(-1)}^{\nu }\left({\displaystyle \genfrac{}{}{0pt}{}{j}{\nu }}\right)\right){\displaystyle \frac{1}{{(n-l)}^r}}+\sum _{l=j}^{n-1}\left(\sum _{\nu =0}^j{(-1)}^{\nu }\left({\displaystyle \genfrac{}{}{0pt}{}{j}{\nu }}\right)\right){\displaystyle \frac{1}{{(n-l)}^r}}\hfill \\ \hfill & =\sum _{l=0}^{j-1}{(-1)}^l\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{l}}\right){\displaystyle \frac{1}{{(n-l)}^r}}.\hfill \end{array}$$

By (10),

$$\begin{array}{cc}\hfill {\mathcal{D}}_r(n,j)& =j!\left({\displaystyle \genfrac{}{}{0pt}{}{n}{j}}\right)\left(\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{l}}\right){\displaystyle \frac{1}{{(n-l)}^r}}\right)\hfill \\ \hfill & =\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right){\displaystyle \frac{1}{{(n-l)}^{r-1}}}.\hfill \end{array}$$

□

Proof of Theorem 2. 

By Theorem 1,

$$\begin{array}{cc}\hfill & {\mathcal{D}}_r(n,j)-{\mathcal{D}}_r(n,j-1)\hfill \\ \hfill & =\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left(\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j+1}}\right)\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right){\displaystyle \frac{1}{{(n-l)}^{r-1}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{n-j+1}}\sum _{l=0}^{j-1}{(-1)}^{j-l-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-l-1}{n-j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{l}}\right){\displaystyle \frac{1}{{(n-l)}^{r-2}}}\hfill \\ \hfill & ={\displaystyle \frac{{\mathcal{D}}_{r-1}(n,j)}{n-j+1}}.\hfill \end{array}$$

Hence, by ${\mathcal{D}}_r(n,0)=0$, we have

$$\begin{array}{cc}\hfill {\mathcal{D}}_{r+1}(n,j)& ={\mathcal{D}}_{r+1}(n,j-1)+{\displaystyle \frac{{\mathcal{D}}_r(n,j)}{n-j+1}}\hfill \\ \hfill & ={\mathcal{D}}_{r+1}(n,j-2)+{\displaystyle \frac{{\mathcal{D}}_r(n,j-1)}{n-j+2}}+{\displaystyle \frac{{\mathcal{D}}_r(n,j)}{n-j+1}}\hfill \\ \hfill & =\cdots =\sum _{j_1=1}^j{\displaystyle \frac{{\mathcal{D}}_r(n,j_1)}{n-j_1+1}}\hfill \\ \hfill & =\sum _{j_1=1}^j{\displaystyle \frac{1}{n-j_1+1}}\sum _{j_2=1}^{j_1}{\displaystyle \frac{{\mathcal{D}}_{r-1}(n,j_2)}{n-j_2+1}}\hfill \\ \hfill & =\sum _{j_1=1}^j{\displaystyle \frac{1}{n-j_1+1}}\sum _{j_2=1}^{j_1}{\displaystyle \frac{1}{n-j_2+1}}\sum _{j_3=1}^{j_2}{\displaystyle \frac{{\mathcal{D}}_{r-2}(n,j_3)}{n-j_3+1}}\hfill \\ \hfill & =\cdots \hfill \\ \hfill & =\sum _{j_1=1}^j{\displaystyle \frac{1}{n-j_1+1}}\sum _{j_2=1}^{j_1}{\displaystyle \frac{1}{n-j_2+1}}\cdots \sum _{j_r=1}^{j_{r-1}}{\displaystyle \frac{{\mathcal{D}}_1(n,j_r)}{n-j_r+1}}\hfill \\ \hfill & =\sum _{j_1=1}^j{\displaystyle \frac{1}{n-j_1+1}}\sum _{j_2=1}^{j_1}{\displaystyle \frac{1}{n-j_2+1}}\cdots \sum _{j_r=1}^{j_{r-1}}{\displaystyle \frac{1}{n-j_r+1}}.\hfill \end{array}$$

□

In order to prove Theorem 4 and Theorem 3, we need the following relations.

Lemma 4. 

For the sequences ${\left\{p_n\right\}}_{n\ge 1}$ and ${\left\{h_n\right\}}_{n\ge 1}$, we have

$$\begin{array}{cc}\hfill & {(-1)}^{n-1}{p}_n=\left|\begin{array}{ccccc}{h}_1& 1& 0& \cdots & 0\\ 2h_2& h_1& 1& & ⋮\\ 3h_3& h_2& & \ddots & \\ ⋮& ⋮& & & 1\\ n{h}_n& h_{n-1}& \cdots & h_2& h_1\end{array}\right|\hfill \\ \hfill & ⟺n!h_n=\left|\begin{array}{ccccc}{p}_1& -1& 0& \cdots & 0\\ p_2& p_1& -2& & ⋮\\ p_3& p_2& & \ddots & 0\\ ⋮& ⋮& & & -n+1\\ p_n& p_{n-1}& \cdots & p_2& p_1\end{array}\right|\hfill \\ \hfill & =\sum _{{\displaystyle \genfrac{}{}{0pt}{}{i_1+2i_2+\cdots +n{i}_n}{i_1,i_2,\cdots ,i_n\ge 0}}}{\displaystyle \frac{n!}{i_1!i_2!\cdots i_n!}}{\left({\displaystyle \frac{p_1}{1}}\right)}^{i_1}{\left({\displaystyle \frac{p_2}{2}}\right)}^{i_2}\cdots {\left({\displaystyle \frac{p_n}{n}}\right)}^{i_n}.\hfill \end{array}$$

Proof. 

The last identity is a simple modification of Trudi’s formula (([18] (Volume 3, p. 214)), [19]):

$$\begin{array}{c}\left|\begin{array}{ccccc}{a}_1& a_0& \cdots & & 0\\ a_2& a_1& \cdots & & \\ ⋮& ⋮& \ddots & & ⋮\\ a_{n-1}& a_{n-2}& \cdots & a_1& a_0\\ a_n& a_{n-1}& \cdots & a_2& a_1\end{array}\right|\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}=\sum _{i_1+2i_2+\cdots +n{t}_n=n}{\displaystyle \frac{(i_1+\cdots +i_n)!}{i_1!\cdots i_n!}}{(-a_0)}^{n-i_1-\cdots -i_n}{a}_1^{i_1}{a}_2^{i_2}\cdots a_n^{i_n}.\end{array}$$

Notice that the expansion of the second determinant is equivalent to the following relation:

$$n{h}_n=\sum _{i=1}^n{p}_i{h}_{n-1}\mathrm{with}{h}_0=1.$$

(11)

Byapplying the inversion formula (see, e.g., ([Lemma 1] [16])), we can obtain the first identity. □

Proof of Theorem 3. 

The determinant in Theorem 3 is equivalent to the recurrence relation:

$${\mathcal{D}}_{r+1}(n,j)={\displaystyle \frac{1}{r}}\sum _{i=1}^r(H_n^{(r-i+1)}-H_{n-j}^{(r-i+1)}){\mathcal{D}}_i(n,j).$$

(12)

Byapplying the relation (11) in the first identity of the second part of Lemma 4 to (12), we can obtain the desired determinant identity. The identity of this remark can be given from the first part of Lemma 4. □

Proof of Theorem 4. 

The result follows from the second part of Lemma 4 by setting $h_r={\mathcal{D}}_{r+1}(n,j)$ and $p_i=H_n^{\left(i\right)}-H_{n-j}^{\left(i\right)}$, satisfying (12). □

Proof of Theorem 5. 

Since Bell polynomials satisfy the following recurrence relation:

$${\mathbf{Y}}_r(x_1,x_2,\cdots ,x_r)=\sum _{i=1}^r\left({\displaystyle \genfrac{}{}{0pt}{}{r-1}{i-1}}\right)x_{r-i+1}{\mathbf{Y}}_{i-1}(x_1,x_2,\cdots ,x_{i-1})$$

(see, e.g., [17]), by setting $x_{\ell }=(\ell -1)!(H_n^{(\ell )}-H_{n-j}^{(\ell )})$, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{r!}}{\mathbf{Y}}_r\left(H_n-H_{n-j},1!(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)}),2!(H_n^{\left(3\right)}-H_{n-j}^{\left(3\right)}),\cdots \right)\hfill \\ \hfill & ={\displaystyle \frac{1}{r}}\sum _{i=1}^r(H_n^{(r-i+1)}-H_{n-j}^{(r-i+1)})\hfill \\ \hfill & \times {\displaystyle \frac{{\mathbf{Y}}_{i-1}\left(H_n-H_{n-j},1!(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)}),2!(H_n^{\left(3\right)}-H_{n-j}^{\left(3\right)}),\cdots \right)}{(i-1)!}}.\hfill \end{array}$$

Since

$$\begin{array}{cc}\hfill & {\mathcal{D}}_2(n,j)=H_n-H_{n-j}\hfill \\ \hfill & ={\mathbf{Y}}_1\left(H_n-H_{n-j},1!(H_n^{\left(2\right)}-H_{n-j}^{\left(2\right)}),2!(H_n^{\left(3\right)}-H_{n-j}^{\left(3\right)}),\cdots \right),\hfill \end{array}$$

for $r\ge 1$, we can write the form in Theorem 5. □