Polynomial Identities for Binomial Sums of Harmonic Numbers of Higher Order

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 the one

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

which has been proved in [2] in the field of symbolic computation and in [3] in finite differences.

The 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 [4], the so-called hyperharmonic number generalizes harmonic number of order r in the formula. However, when we generalize too much, we often lose the fundamental properties that make us interesting.

2. Observation

By using the harmonic numbers to express ${\mathcal{D}}_r(n,j)$, for $1\le r\le 7$, we can manually get the following1.

$$\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 [6,7]:

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

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 ([8, Ch. I S 2]). See also [9,10].

Theorem 3.

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

Remark 1.

By using the inversion formula (see, e.g., [11, Lemma 1],[12, Theorem 1],[8, p.28],) about (5) below, we also have

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

${\mathcal{D}}_r(n,j)$ ($r\ge 2$) can be expressed by a combinatorial sum ([7, Proposition 1 (17)]):

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

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

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 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

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

(3)

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 (3),

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

□

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 amd Theorem 3, we need the following relations.

Lemma 1.

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 ([13, Vol.3, p.214],[14]):

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

Notice that the expansion of the second determinant is equivalant to the relation

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

(4)

By applying the inversion formula (see, e.g. [11, Lemma 1],[12, Theorem 1]), we can get the first identity. □

Proof of Theorem 3.

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

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

(5)

By applying the relation (4) in the first identity of the second part of Lemma 1 to (5), we can get the desired determinant identity. The identity of Remark can be given from the first part of Lemma 1. □

Proof of Theorem 4.

The result follows from the second part of Lemma 1 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 (5). □

Proof of Theorem 5.

Since Bell polynomials satisfy the 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., [15]), 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. □

5. Some Reductions

In particular, when $r=1$ in Theorem 1, we find the following relation.

Corollary 1.

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

(6)

When $r=2$ in Theorem 1, 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 ${\left(x\right)}_n={\sum }_{k=0}^n{(-1)}^{n-k}\left[{\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right]x^k$.

Corollary 2.

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

(7)

Remark 2.

Note that

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

Proof of Corollary 2.

The formula (7) is yielded from the definition of the Stirling numbers of the first kind:

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

Differentiating both 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 (7) is equal to

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

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