Series involving degenerate harmonic numbers and degenerate Stirling numbers

ABSTRACT

Recently, degenerate harmonic numbers and degenerate hyperharmonic numbers are introduced by Kim-Kim. In this paper, we study the series involving the degenerate harmonic numbers and degenerate Stirling numbers and investigate those properties.

KEYWORDS:

• Harmonic numbers
• degenerate harmonic numbers
• degenerate hyperharmonic numbers
• degenerate Stirling numbers
• degenerate r-Stirling numbers

MATHEMATICS SUBJECT CLASSIFICATIONS:

• 11B73
• 11B83
• 11B75

1. Introduction

Many mathematicians introduced the special polynomials and numbers and their degenerate versions, and obtained some combinatorial identities and properties associated with them [1–12]. Kim-Kim [13] derived some identities involving the degenerate harmonic numbers and certain finite sums of general nature by using certain degenerate differential and degenerate difference operators. Kim et al. [10] introduced the degenerate harmonic numbers and degenerate hyperharmonic numbers, and obtained some identities by using the umbral calculus. Kim-Kim [11] also defined the discrete harmonic numbers and derived some identities related them. Kim et al. [9] derived some identities involving generalized harmonic numbers and generalized harmonic functions. Kim-Kim [14] introduced the degenerate harmonic and degenerate higher-order harmonic numbers and derived several infinite sums involving the degenerate higher-order harmonic numbers can be represented by using the degenerate zeta functions.

Dolgy et al. [15] derived some properties and identities involving the degenerate harmonic numbers and degenerate hyperharmonic numbers in connection with degenerate Stirling numbers of the first kind. L$\acute{o}$pez-Bonilla and L$\acute{o}$pez-V$\acute{a}$zquez [8] obtained some identities of the harmonic numbers in terms of Stirling numbers of the second kind. Wang et al. [12] investigated some identities involving generalized harmonic numbers with degenerate Changhee-Genocchi sequences and two kinds of degenerate Stirling numbers and other special numbers. Kim-Kim [5, 6] introduced the degenerate harmonic numbers, degenerate hyperharmonic numbers and derived some explicit expressions of the degenerate hyperharmonic numbers in terms of degenerate harmonic numbers and degenerate r-Stirling numbers of the second kind and other polynomials and numbers.

In particular, we are interested in degenerate harmonic numbers and degenerate Stirling numbers. In this paper, we introduce some series of the degenerate harmonic numbers and degenerate Stirling numbers. At the first,we give some definitions and identities needed in this paper.

For $\lambda \in \mathbb{R}$, the degenerate exponential function is defined by (1) ${\mathrm{e}}_{\lambda }^x\left(t\right)=(1+\lambda t{)}^{\frac{x}{\lambda }}=\sum _{n=0}^{\mathrm{\infty }}(x{)}_{n,\lambda }\frac{t^n}{n!},$(1) (see [5,6,16–18]). Where $(x{)}_{0,\lambda }=1$, $(x{)}_{n,\lambda }=x(x-\lambda )\dots (x-(n-1)\lambda ),(n\ge 1)$, (see [2–6, 16–18]).

Note that $\underset{\lambda \to 0}{lim}{\mathrm{e}}_{\lambda }^x\left(t\right)={\mathrm{e}}^{xt}.$When x = 1, $e_{\lambda }\left(t\right)={\mathrm{e}}_{\lambda }^1\left(t\right)$.

Let ${\log }_{\lambda }\left(t\right)$ be the inverse function of $e_{\lambda }\left(t\right)$, so we have ${\log }_{\lambda }\left(e_{\lambda }\right(t\left)\right)=e_{\lambda }\left({\log }_{\lambda }\right(t\left)\right)=t$. Then, we easily see that (2) ${\log }_{\lambda }\left(t\right)=\frac{1}{\lambda }(t^{\lambda }-1),$(2) (see [19]). From (2(2) ${\log }_{\lambda }\left(t\right)=\frac{1}{\lambda }(t^{\lambda }-1),$(2) ), we note that (3) ${\log }_{\lambda }(1+t)=\frac{1}{\lambda }\left(\right(1+t{)}^{\lambda }-1)=\sum _{n=1}^{\mathrm{\infty }}\frac{{\lambda }^{n-1}(1{)}_{n,1/\lambda }}{n!}{t}^n,$(3) (see [6,17,19]). The harmonic numbers are defined by (4) $H_0=0,H_n=1+\frac{1}{2}+\cdots +\frac{1}{n},(n\ge 1),$(4) (see [6–10,20–22]). From (4(4) $H_0=0,H_n=1+\frac{1}{2}+\cdots +\frac{1}{n},(n\ge 1),$(4) ), we note that the generating function of harmonic numbers are given by (5) $-\frac{1}{1-t}\log (1-t)=\sum _{n=1}^{\mathrm{\infty }}{H}_n{t}^n,(|t|<1),$(5) (see [6,8,9,20,21,23]). Recently, the degenerate harmonic numbers are defined by (6) $-\frac{{\log }_{\lambda }(1-t)}{1-t}=\sum _{n=1}^{\mathrm{\infty }}{H}_{n,\lambda }{t}^n,$(6) (see [5,6,13,24]). By (6(6) $-\frac{{\log }_{\lambda }(1-t)}{1-t}=\sum _{n=1}^{\mathrm{\infty }}{H}_{n,\lambda }{t}^n,$(6) ), we have (7) $H_{n,\lambda }=\sum _{k=1}^n\frac{(1{)}_{k,1/\lambda }{\lambda }^{k-1}}{k!}(-1{)}^{k-1},H_{0,\lambda }=0,$(7) (see [5,6,13]). The degenerate Stirling numbers of the second kind are defined by (8) $(x{)}_{n,\lambda }=\sum _{n=0}^{\mathrm{\infty }}{S}_{2,\lambda }(n,k\left)\right(x{)}_n,(n\ge 0),$(8) (see [17,19,25]). Where $(x{)}_0=1,(x{)}_n=x(x-1)\dots (x-n+1),(n\ge 1)$.

From (8(8) $(x{)}_{n,\lambda }=\sum _{n=0}^{\mathrm{\infty }}{S}_{2,\lambda }(n,k\left)\right(x{)}_n,(n\ge 0),$(8) ), we note that (9) $\frac{1}{k!}\left(e_{\lambda }\right(t)-1{)}^k=\sum _{n=k}^{\mathrm{\infty }}{S}_{2,\lambda }(n,k)\frac{t^n}{n!},(k\ge 0),$(9) (see [17,19,25]). On the other hand, we get (10) $\begin{array}{rl}\frac{1}{k!}\left(e_{\lambda }\right(t)-1{)}^k& =\frac{1}{k!}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})(-1{)}^{k-l}{\mathrm{e}}_{\lambda }^l(t)\\ & =\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{k!}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})\right(-1{)}^{k-l}\left(l{)}_{n,\lambda }\right)\frac{t^n}{n!}.\end{array}$(10) Thus, by (9(9) $\frac{1}{k!}\left(e_{\lambda }\right(t)-1{)}^k=\sum _{n=k}^{\mathrm{\infty }}{S}_{2,\lambda }(n,k)\frac{t^n}{n!},(k\ge 0),$(9) ) and (10(10) $\begin{array}{rl}\frac{1}{k!}\left(e_{\lambda }\right(t)-1{)}^k& =\frac{1}{k!}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})(-1{)}^{k-l}{\mathrm{e}}_{\lambda }^l(t)\\ & =\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{k!}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})\right(-1{)}^{k-l}\left(l{)}_{n,\lambda }\right)\frac{t^n}{n!}.\end{array}$(10) ), we have (11) $\frac{1}{k!}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})(-1{)}^{k-l}(l{)}_{n,\lambda }=\{\begin{array}{ll}{S}_{2,\lambda }(n,k),& \text{if}n\ge k,\\ 0,& \text{if}n<k,\end{array}$(11) (see [5,6,19,24]). For $n\ge 0,r\ge 1$, the degenerate hyperharmonic numbers are introduced by Kim-Kim to be (12) $H_{n,\lambda }^{\left(1\right)}=H_{n,\lambda },H_{0,\lambda }^{\left(r\right)}=0(r\ge 1),H_{n,\lambda }^{\left(r\right)}=\sum _{k=1}^n{H}_{k,\lambda }^{(r-1)},(r\ge 2,n\ge 1),$(12) (see [10]). From (12(12) $H_{n,\lambda }^{\left(1\right)}=H_{n,\lambda },H_{0,\lambda }^{\left(r\right)}=0(r\ge 1),H_{n,\lambda }^{\left(r\right)}=\sum _{k=1}^n{H}_{k,\lambda }^{(r-1)},(r\ge 2,n\ge 1),$(12) ), we have (13) $H_{n,\lambda }^{(r+1)}=\frac{(\genfrac{}{}{0ex}{}{n+r}{r})}{(\genfrac{}{}{0ex}{}{k-\lambda }{r})}(H_{n+r,\lambda }-H_{r,\lambda }),(n,r\in \mathbb{N}),$(13) (see [5,6]). By (12(12) $H_{n,\lambda }^{\left(1\right)}=H_{n,\lambda },H_{0,\lambda }^{\left(r\right)}=0(r\ge 1),H_{n,\lambda }^{\left(r\right)}=\sum _{k=1}^n{H}_{k,\lambda }^{(r-1)},(r\ge 2,n\ge 1),$(12) ) and (13(13) $H_{n,\lambda }^{(r+1)}=\frac{(\genfrac{}{}{0ex}{}{n+r}{r})}{(\genfrac{}{}{0ex}{}{k-\lambda }{r})}(H_{n+r,\lambda }-H_{r,\lambda }),(n,r\in \mathbb{N}),$(13) ), we get (14) $-\frac{1}{(1-t{)}^r}{\log }_{\lambda }(1-t)=\sum _{n=1}^{\mathrm{\infty }}{H}_{n,\lambda }^{\left(r\right)}{t}^n,$(14) (see [10,13]). In this paper, we study the series involving degenerate harmonic and hyperharmonic numbers and degenerate Stirling numbers of the second kind. In addition, we give some new identities related to degenerate harmonic and hyperharmonic numbers.

2. Series involving degenerate harmonic numbers and degenerate stirling numbers

In this section, we investigate some identities and properties involving degenerate harmonic numbers and degenerate hyperharmonic numbers by the degenerate logarithmic functions. Meanwhile, we also derive some series associated with degenerate Stirling numbers and degenerate r-Stirling numbers.

By binomial transformation, we get (15) $a_n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(-1{)}^{n-k}{b}_k\iff b_n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a_k,(n\ge 0).$(15)

Theorem 2.1

For $n\in \mathbb{N}$, we have $H_{n,-\lambda }=\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})\frac{{\lambda }^{k-1}}{k!}(1{)}_{k,1/\lambda },$and $\frac{{\lambda }^{n-1}}{n!}\left(1{)}_{n,1/\lambda }\right(-1{)}^{n-1}=\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})H_{k,-\lambda }(-1{)}^{k-1}.$

Proof.

From (3(3) ${\log }_{\lambda }(1+t)=\frac{1}{\lambda }\left(\right(1+t{)}^{\lambda }-1)=\sum _{n=1}^{\mathrm{\infty }}\frac{{\lambda }^{n-1}(1{)}_{n,1/\lambda }}{n!}{t}^n,$(3) ) and (6(6) $-\frac{{\log }_{\lambda }(1-t)}{1-t}=\sum _{n=1}^{\mathrm{\infty }}{H}_{n,\lambda }{t}^n,$(6) ), we note that (16) $\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{{\lambda }^{n-1}(1{)}_{n,1/\lambda }}{n!}(-1{)}^{n-1}{t}^n& =-{\log }_{\lambda }(1-t)={\log }_{-\lambda }(1+\frac{t}{1-t})\\ & =-\frac{1}{(1-t)(1+\frac{t}{1-t})}(-{\log }_{-\lambda }(1+\frac{t}{1-t}))\\ & =\sum _{k=1}^{\mathrm{\infty }}{H}_{k,-\lambda }(-1{)}^{k-1}{t}^k{\left(\frac{1}{1-t}\right)}^{k+1}\\ & =\sum _{k=1}^{\mathrm{\infty }}{H}_{k,-\lambda }(-1{)}^{k-1}{t}^k\sum _{l=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{k+l}{k})t^l\\ & =\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})H_{k,-\lambda }\right(-1{)}^{k-1})t^n.\end{array}$(16) Thus, by comparing the coefficients on the both sides of the (16(16) $\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{{\lambda }^{n-1}(1{)}_{n,1/\lambda }}{n!}(-1{)}^{n-1}{t}^n& =-{\log }_{\lambda }(1-t)={\log }_{-\lambda }(1+\frac{t}{1-t})\\ & =-\frac{1}{(1-t)(1+\frac{t}{1-t})}(-{\log }_{-\lambda }(1+\frac{t}{1-t}))\\ & =\sum _{k=1}^{\mathrm{\infty }}{H}_{k,-\lambda }(-1{)}^{k-1}{t}^k{\left(\frac{1}{1-t}\right)}^{k+1}\\ & =\sum _{k=1}^{\mathrm{\infty }}{H}_{k,-\lambda }(-1{)}^{k-1}{t}^k\sum _{l=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{k+l}{k})t^l\\ & =\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})H_{k,-\lambda }\right(-1{)}^{k-1})t^n.\end{array}$(16) ), we get (17) $\frac{{\lambda }^{n-1}(1{)}_{n,1/\lambda }}{n!}(-1{)}^{n-1}=\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})H_{k,-\lambda }(-1{)}^{k-1},(n\ge 1).$(17) From (15(15) $a_n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(-1{)}^{n-k}{b}_k\iff b_n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a_k,(n\ge 0).$(15) ) and (17(17) $\frac{{\lambda }^{n-1}(1{)}_{n,1/\lambda }}{n!}(-1{)}^{n-1}=\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})H_{k,-\lambda }(-1{)}^{k-1},(n\ge 1).$(17) ), we get (18) $H_{n,-\lambda }=\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})\frac{{\lambda }^{k-1}}{k!}(1{)}_{k,1/\lambda },(n\ge 1).$(18) Therefore, by (17(17) $\frac{{\lambda }^{n-1}(1{)}_{n,1/\lambda }}{n!}(-1{)}^{n-1}=\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})H_{k,-\lambda }(-1{)}^{k-1},(n\ge 1).$(17) ) and (18(18) $H_{n,-\lambda }=\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})\frac{{\lambda }^{k-1}}{k!}(1{)}_{k,1/\lambda },(n\ge 1).$(18) ), we obtain the Theorem 2.1.

Theorem 2.2

For $n\in \mathbb{N}$, we have $\frac{{\lambda }^{n-1}}{n!}\left(1{)}_{n,1/\lambda }\right(-1{)}^{n-1}=\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})H_{k,-\lambda }^{(r+1)}(-1{)}^{k-1}.$

Proof.

By (3(3) ${\log }_{\lambda }(1+t)=\frac{1}{\lambda }\left(\right(1+t{)}^{\lambda }-1)=\sum _{n=1}^{\mathrm{\infty }}\frac{{\lambda }^{n-1}(1{)}_{n,1/\lambda }}{n!}{t}^n,$(3) ) and (14(14) $-\frac{1}{(1-t{)}^r}{\log }_{\lambda }(1-t)=\sum _{n=1}^{\mathrm{\infty }}{H}_{n,\lambda }^{\left(r\right)}{t}^n,$(14) ), we get (19) $\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{{\lambda }^{n-1}(1{)}_{n,1/\lambda }}{n!}(-1{)}^{n-1}{t}^n& =-{\log }_{\lambda }(1-t)={\log }_{-\lambda }(1+\frac{t}{1-t})\\ & =-\frac{-{\log }_{-\lambda }(1+\frac{t}{1-t})}{(1-t{)}^{r+1}{(1+\frac{t}{1-t})}^{r+1}}\\ & =\sum _{k=1}^{\mathrm{\infty }}{H}_{k,-\lambda }^{(r+1)}(-1{)}^{k-1}{t}^k{\left(\frac{1}{1-t}\right)}^{k+r+1}\\ & =\sum _{k=1}^{\mathrm{\infty }}{H}_{k,-\lambda }^{(r+1)}(-1{)}^{k-1}{t}^k\sum _{l=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{k+r+l}{k+r})t^l\\ & =\sum _{n=1}^{\mathrm{\infty }}\left(\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})H_{k,-\lambda }^{(r+1)}\right(-1{)}^{k-1})t^n.\end{array}$(19) Therefore, by (19(19) $\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{{\lambda }^{n-1}(1{)}_{n,1/\lambda }}{n!}(-1{)}^{n-1}{t}^n& =-{\log }_{\lambda }(1-t)={\log }_{-\lambda }(1+\frac{t}{1-t})\\ & =-\frac{-{\log }_{-\lambda }(1+\frac{t}{1-t})}{(1-t{)}^{r+1}{(1+\frac{t}{1-t})}^{r+1}}\\ & =\sum _{k=1}^{\mathrm{\infty }}{H}_{k,-\lambda }^{(r+1)}(-1{)}^{k-1}{t}^k{\left(\frac{1}{1-t}\right)}^{k+r+1}\\ & =\sum _{k=1}^{\mathrm{\infty }}{H}_{k,-\lambda }^{(r+1)}(-1{)}^{k-1}{t}^k\sum _{l=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{k+r+l}{k+r})t^l\\ & =\sum _{n=1}^{\mathrm{\infty }}\left(\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})H_{k,-\lambda }^{(r+1)}\right(-1{)}^{k-1})t^n.\end{array}$(19) ), we obtain the Theorem 2.2.

Theorem 2.3

For $n\ge 1$, we have $H_{n,-\lambda }^{(r+1)}=\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})\frac{{\lambda }^{k-1}}{k!}(1{)}_{k,1/\lambda }.$

Proof.

Let (20) $a_n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})(-1{)}^{k-1}{b}_k.$(20) By (20(20) $a_n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})(-1{)}^{k-1}{b}_k.$(20) ), we have (21) $\begin{array}{rl}\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})(-1{)}^{k-1}{a}_k& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})(-1{)}^{k-1}\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k+r}{j+r})(-1{)}^{j-1}{b}_j\\ & =\sum _{j=0}^n{b}_j(-1{)}^{j-1}\sum _{k=j}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})(\genfrac{}{}{0ex}{}{k+r}{j+r})(-1{)}^{k-1}\\ & =\sum _{j=0}^n{b}_j(-1{)}^{j-1}\sum _{k=0}^{n-j}(\genfrac{}{}{0ex}{}{n+r}{k+j+r})(\genfrac{}{}{0ex}{}{k+j+r}{j+r})(-1{)}^{k-1+j}\\ & =\sum _{j=0}^n{b}_j(\genfrac{}{}{0ex}{}{n+r}{j+r})\sum _{k=0}^{n-j}(\genfrac{}{}{0ex}{}{n-j}{k})(-1{)}^k\\ & =b_n.\end{array}$(21) Thus, by Theorem 2.2 and (21(21) $\begin{array}{rl}\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})(-1{)}^{k-1}{a}_k& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})(-1{)}^{k-1}\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k+r}{j+r})(-1{)}^{j-1}{b}_j\\ & =\sum _{j=0}^n{b}_j(-1{)}^{j-1}\sum _{k=j}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})(\genfrac{}{}{0ex}{}{k+r}{j+r})(-1{)}^{k-1}\\ & =\sum _{j=0}^n{b}_j(-1{)}^{j-1}\sum _{k=0}^{n-j}(\genfrac{}{}{0ex}{}{n+r}{k+j+r})(\genfrac{}{}{0ex}{}{k+j+r}{j+r})(-1{)}^{k-1+j}\\ & =\sum _{j=0}^n{b}_j(\genfrac{}{}{0ex}{}{n+r}{j+r})\sum _{k=0}^{n-j}(\genfrac{}{}{0ex}{}{n-j}{k})(-1{)}^k\\ & =b_n.\end{array}$(21) ), we obtain the Theorem 2.3.

Proposition 2.4

For $\alpha \in \mathbb{R}$, we have $\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a_k=\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(-1{)}^k{a}_k{x}^k.$

Proof.

For $\alpha \in \mathbb{R}$, we have (22) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a_k\\ & =\sum _{k=0}^n{a}_k\sum _{n=k}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})(\genfrac{}{}{0ex}{}{n}{k})\\ & =\sum _{k=0}^n{a}_k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^{n+k}(-1{)}^{n+k}(\genfrac{}{}{0ex}{}{\alpha }{n+k})(\genfrac{}{}{0ex}{}{n+k}{k})\\ & =\sum _{k=0}^n{a}_k{\left(\frac{x}{1+x}\right)}^k(-1{)}^k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{k})(\genfrac{}{}{0ex}{}{\alpha -k}{n})\\ & =\sum _{k=0}^n{a}_k(-1{)}^k{\left(\frac{x}{1+x}\right)}^k(\genfrac{}{}{0ex}{}{\alpha }{k}){(1-\frac{x}{1+x})}^{\alpha -k}\\ & =\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(-1{)}^k{x}^k{a}_k.\end{array}$(22) Therefore, by (22(22) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a_k\\ & =\sum _{k=0}^n{a}_k\sum _{n=k}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})(\genfrac{}{}{0ex}{}{n}{k})\\ & =\sum _{k=0}^n{a}_k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^{n+k}(-1{)}^{n+k}(\genfrac{}{}{0ex}{}{\alpha }{n+k})(\genfrac{}{}{0ex}{}{n+k}{k})\\ & =\sum _{k=0}^n{a}_k{\left(\frac{x}{1+x}\right)}^k(-1{)}^k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{k})(\genfrac{}{}{0ex}{}{\alpha -k}{n})\\ & =\sum _{k=0}^n{a}_k(-1{)}^k{\left(\frac{x}{1+x}\right)}^k(\genfrac{}{}{0ex}{}{\alpha }{k}){(1-\frac{x}{1+x})}^{\alpha -k}\\ & =\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(-1{)}^k{x}^k{a}_k.\end{array}$(22) ), we obtain the Proposition 2.4.

Theorem 2.5

For $\alpha \in \mathbb{R}$, we have $\frac{1}{(1+x{)}^{\alpha }}\sum _{k=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})H_{k,-\lambda }{x}^k=\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\frac{{\lambda }^{n-1}}{n!}(1{)}_{n,1/\lambda }.$

Proof.

Let us take $a_n=(-1{)}^n{H}_{n,-\lambda },(n\ge 1),a_0=0$.

From (22(22) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a_k\\ & =\sum _{k=0}^n{a}_k\sum _{n=k}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})(\genfrac{}{}{0ex}{}{n}{k})\\ & =\sum _{k=0}^n{a}_k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^{n+k}(-1{)}^{n+k}(\genfrac{}{}{0ex}{}{\alpha }{n+k})(\genfrac{}{}{0ex}{}{n+k}{k})\\ & =\sum _{k=0}^n{a}_k{\left(\frac{x}{1+x}\right)}^k(-1{)}^k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{k})(\genfrac{}{}{0ex}{}{\alpha -k}{n})\\ & =\sum _{k=0}^n{a}_k(-1{)}^k{\left(\frac{x}{1+x}\right)}^k(\genfrac{}{}{0ex}{}{\alpha }{k}){(1-\frac{x}{1+x})}^{\alpha -k}\\ & =\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(-1{)}^k{x}^k{a}_k.\end{array}$(22) ), we have (23) $\begin{array}{rl}\frac{1}{(1+x{)}^{\alpha }}\sum _{k=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})H_{k,-\lambda }{x}^k& =\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})(-1{)}^k{H}_{k,-\lambda }\\ & =\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\frac{{\lambda }^{n-1}}{n!}(1{)}_{n,1/\lambda }(-1{)}^n\\ & =\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\frac{{\lambda }^{n-1}}{n!}(1{)}_{n,1/\lambda }.\end{array}$(23) In particular, let $x=1,\alpha =m\in \mathbb{N}$, we have $\frac{1}{2^m}\sum _{k=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m}{k})H_{k,-\lambda }=\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^n(\genfrac{}{}{0ex}{}{m}{n})\frac{{\lambda }^{n-1}}{n!}(1{)}_{n,1/\lambda }.$Therefore, by (23(23) $\begin{array}{rl}\frac{1}{(1+x{)}^{\alpha }}\sum _{k=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})H_{k,-\lambda }{x}^k& =\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})(-1{)}^k{H}_{k,-\lambda }\\ & =\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\frac{{\lambda }^{n-1}}{n!}(1{)}_{n,1/\lambda }(-1{)}^n\\ & =\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\frac{{\lambda }^{n-1}}{n!}(1{)}_{n,1/\lambda }.\end{array}$(23) ), we obtain the Theorem 2.5.

Corollary 2.6

For $\alpha \in \mathbb{R}$, we have ${\left(\frac{1}{1+x}\right)}^{\alpha }\sum _{k=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(-1{)}^k\frac{{\lambda }^{k-1}}{k!}(1{)}_{k,1/\lambda }{x}^k=\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})H_{n,-\lambda }.$

Proof.

From (18(18) $H_{n,-\lambda }=\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})\frac{{\lambda }^{k-1}}{k!}(1{)}_{k,1/\lambda },(n\ge 1).$(18) ), we note that (24) $\begin{array}{rl}& {\left(\frac{1}{1+x}\right)}^{\alpha }\sum _{k=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(-1{)}^k\frac{{\lambda }^{k-1}}{k!}(1{)}_{k,1/\lambda }{x}^k\\ & =\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})\frac{(1{)}_{k,1/\lambda }}{k!}{\lambda }^{k-1}\\ & =\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})H_{n,-\lambda }.\end{array}$(24) Therefore, by (24(24) $\begin{array}{rl}& {\left(\frac{1}{1+x}\right)}^{\alpha }\sum _{k=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(-1{)}^k\frac{{\lambda }^{k-1}}{k!}(1{)}_{k,1/\lambda }{x}^k\\ & =\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})\frac{(1{)}_{k,1/\lambda }}{k!}{\lambda }^{k-1}\\ & =\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})H_{n,-\lambda }.\end{array}$(24) ), we obtain the Corollary 2.6.

Theorem 2.7

For $\alpha \in \mathbb{R},r\in \mathbb{N}$, we have $\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha +r}{n+r})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})a_k=\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})(-1{)}^k{a}_k{x}^k.$

Proof.

Now, we observe that (25) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha +r}{n+r})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})a_k\\ & =\sum _{k=0}^{\mathrm{\infty }}{a}_k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^{n+k}(-1{)}^{n+k}(\genfrac{}{}{0ex}{}{\alpha +r}{n+k+r})(\genfrac{}{}{0ex}{}{n+r+k}{k+r})\\ & =\sum _{k=0}^{\mathrm{\infty }}{a}_k{\left(\frac{x}{1+x}\right)}^k(-1{)}^k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha +r}{n+k+r})(\genfrac{}{}{0ex}{}{n+r+k}{k+r})\\ & =\sum _{k=0}^{\mathrm{\infty }}{a}_k{\left(\frac{x}{1+x}\right)}^k(-1{)}^k(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha -k}{n})\\ & =\sum _{k=0}^{\mathrm{\infty }}{a}_k{\left(\frac{x}{1+x}\right)}^k(-1{)}^k{(1-\frac{x}{1+x})}^{\alpha -k}(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})\\ & =\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})(-1{)}^k{a}_k{x}^k.\end{array}$(25) where $\alpha \in \mathbb{R}$ and $r\in \mathbb{N}$.

Therefore, by (25(25) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha +r}{n+r})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})a_k\\ & =\sum _{k=0}^{\mathrm{\infty }}{a}_k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^{n+k}(-1{)}^{n+k}(\genfrac{}{}{0ex}{}{\alpha +r}{n+k+r})(\genfrac{}{}{0ex}{}{n+r+k}{k+r})\\ & =\sum _{k=0}^{\mathrm{\infty }}{a}_k{\left(\frac{x}{1+x}\right)}^k(-1{)}^k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha +r}{n+k+r})(\genfrac{}{}{0ex}{}{n+r+k}{k+r})\\ & =\sum _{k=0}^{\mathrm{\infty }}{a}_k{\left(\frac{x}{1+x}\right)}^k(-1{)}^k(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha -k}{n})\\ & =\sum _{k=0}^{\mathrm{\infty }}{a}_k{\left(\frac{x}{1+x}\right)}^k(-1{)}^k{(1-\frac{x}{1+x})}^{\alpha -k}(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})\\ & =\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})(-1{)}^k{a}_k{x}^k.\end{array}$(25) ), we obtain the Theorem 2.7.

Theorem 2.8

For $\alpha \in \mathbb{R}$ and $r\in \mathbb{N}$, we have $\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})H_{k,-\lambda }^{(r+1)}{x}^k=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})\frac{{\lambda }^{n-1}}{n!}(1{)}_{n,1/\lambda }.$

Proof.

Let us take $a_k=H_{k,-\lambda }^{(r+1)}(-1{)}^k,a_0=0$.

Then, by Theorem 2.7, we get (26) $\begin{array}{rl}& \frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})H_{k,-\lambda }^{(r+1)}{x}^k\\ & =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})(-1{)}^k{H}_{k,-\lambda }^{(r+1)}\\ & =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})\frac{{\lambda }^{n-1}}{n!}(1{)}_{n,1/\lambda }(-1{)}^n\\ & =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})\frac{{\lambda }^{n-1}}{n!}(1{)}_{n,1/\lambda }.\end{array}$(26) Therefore, by (26(26) $\begin{array}{rl}& \frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})H_{k,-\lambda }^{(r+1)}{x}^k\\ & =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+r}{k+r})(-1{)}^k{H}_{k,-\lambda }^{(r+1)}\\ & =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})\frac{{\lambda }^{n-1}}{n!}(1{)}_{n,1/\lambda }(-1{)}^n\\ & =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha +r}{k+r})\frac{{\lambda }^{n-1}}{n!}(1{)}_{n,1/\lambda }.\end{array}$(26) ), we obtain the Theorem 2.8.

Theorem 2.9

For $\alpha \in \mathbb{R}$, we have $\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(k{)}_{m,\lambda }{x}^k=\sum _{n=0}^m{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})n!S_{2,\lambda }(m,n),(|x|<1).$

Proof.

By (15(15) $a_n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(-1{)}^{n-k}{b}_k\iff b_n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a_k,(n\ge 0).$(15) ) and let $a_k=(-1{)}^k(k{)}_{m,\lambda },(m\ge 0)$ in (22(22) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a_k\\ & =\sum _{k=0}^n{a}_k\sum _{n=k}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})(\genfrac{}{}{0ex}{}{n}{k})\\ & =\sum _{k=0}^n{a}_k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^{n+k}(-1{)}^{n+k}(\genfrac{}{}{0ex}{}{\alpha }{n+k})(\genfrac{}{}{0ex}{}{n+k}{k})\\ & =\sum _{k=0}^n{a}_k{\left(\frac{x}{1+x}\right)}^k(-1{)}^k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{k})(\genfrac{}{}{0ex}{}{\alpha -k}{n})\\ & =\sum _{k=0}^n{a}_k(-1{)}^k{\left(\frac{x}{1+x}\right)}^k(\genfrac{}{}{0ex}{}{\alpha }{k}){(1-\frac{x}{1+x})}^{\alpha -k}\\ & =\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(-1{)}^k{x}^k{a}_k.\end{array}$(22) ). Then we have (27) $\begin{array}{rl}\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(k{)}_{m,\lambda }{x}^k& =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(-1{)}^{n-k}(k{)}_{m,\lambda }\\ & =\sum _{n=0}^m{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})n!S_{2,\lambda }(m,n).\end{array}$(27) Therefore, by (27(27) $\begin{array}{rl}\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(k{)}_{m,\lambda }{x}^k& =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(-1{)}^{n-k}(k{)}_{m,\lambda }\\ & =\sum _{n=0}^m{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})n!S_{2,\lambda }(m,n).\end{array}$(27) ), we obtain the Theorem 2.9.

Let us take $\alpha =-1/2$ and $x=1$ in Theorem 2.9. Then we have $\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{2k}{k})(-1{)}^k\frac{(k{)}_{m,\lambda }}{2^{2k}}=\frac{1}{\sqrt{2}}\sum _{n=0}^m{\left(\frac{1}{2}\right)}^{3n}(\genfrac{}{}{0ex}{}{2n}{n})(-1{)}^{n}n!S_{2,\lambda }(m,n).$Recently, Kim-Kim introduced r-Stirling numbers of the second kind to be (28) $(x+r{)}_{n,\lambda }=\sum _{k=0}^n{\{\genfrac{}{}{0ex}{}{n+r}{k+r}\}}_{r,\lambda }(x{)}_k,(n\ge 0).$(28)

Theorem 2.10

For $r\in \mathbb{N},\alpha \in \mathbb{R}$, we have $\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(k+r{)}_{m,\lambda }{x}^k=\sum _{n=0}^m{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})n!{\{\genfrac{}{}{0ex}{}{m+r}{n+r}\}}_{r,\lambda }.$

Proof.

For (28(28) $(x+r{)}_{n,\lambda }=\sum _{k=0}^n{\{\genfrac{}{}{0ex}{}{n+r}{k+r}\}}_{r,\lambda }(x{)}_k,(n\ge 0).$(28) ), we note that (29) $\frac{1}{k!}\left(e_{\lambda }\right(t)-1{)}^k{e}_{\lambda }^r(t)=\sum _{n=k}^{\mathrm{\infty }}{\{\genfrac{}{}{0ex}{}{n+r}{k+r}\}}_{r,\lambda }\frac{t^n}{n!}.$(29) On the other hand, we get (30) $\begin{array}{rl}\frac{1}{k!}\left(e_{\lambda }\right(t)-1{)}^k{e}_{\lambda }^r(t)& =\frac{1}{k!}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})(-1{)}^{k-l}{\mathrm{e}}_{\lambda }^{l+r}(t)\\ & =\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{k!}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})\right(-1{)}^{k-l}(l+r{)}_{n,\lambda })\frac{t^n}{n!}.\end{array}$(30) By (29(29) $\frac{1}{k!}\left(e_{\lambda }\right(t)-1{)}^k{e}_{\lambda }^r(t)=\sum _{n=k}^{\mathrm{\infty }}{\{\genfrac{}{}{0ex}{}{n+r}{k+r}\}}_{r,\lambda }\frac{t^n}{n!}.$(29) ) and (30(30) $\begin{array}{rl}\frac{1}{k!}\left(e_{\lambda }\right(t)-1{)}^k{e}_{\lambda }^r(t)& =\frac{1}{k!}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})(-1{)}^{k-l}{\mathrm{e}}_{\lambda }^{l+r}(t)\\ & =\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{k!}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})\right(-1{)}^{k-l}(l+r{)}_{n,\lambda })\frac{t^n}{n!}.\end{array}$(30) ), we get (31) $\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})(-1{)}^{k-l}(l+r{)}_{n,\lambda }=\{\begin{array}{ll}k!{\{\genfrac{}{}{0ex}{}{n+r}{k+r}\}}_{r,\lambda },& \text{if}n\ge k,\\ 0,& \text{if}n<k.\end{array}$(31) For $r\in \mathbb{N}$, let us take $a_k=(-1{)}^k(k+r{)}_{m,\lambda },(m\ge 0)$ in (22(22) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a_k\\ & =\sum _{k=0}^n{a}_k\sum _{n=k}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})(\genfrac{}{}{0ex}{}{n}{k})\\ & =\sum _{k=0}^n{a}_k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^{n+k}(-1{)}^{n+k}(\genfrac{}{}{0ex}{}{\alpha }{n+k})(\genfrac{}{}{0ex}{}{n+k}{k})\\ & =\sum _{k=0}^n{a}_k{\left(\frac{x}{1+x}\right)}^k(-1{)}^k\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(-1{)}^n(\genfrac{}{}{0ex}{}{\alpha }{k})(\genfrac{}{}{0ex}{}{\alpha -k}{n})\\ & =\sum _{k=0}^n{a}_k(-1{)}^k{\left(\frac{x}{1+x}\right)}^k(\genfrac{}{}{0ex}{}{\alpha }{k}){(1-\frac{x}{1+x})}^{\alpha -k}\\ & =\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(-1{)}^k{x}^k{a}_k.\end{array}$(22) ). Then, by (31(31) $\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})(-1{)}^{k-l}(l+r{)}_{n,\lambda }=\{\begin{array}{ll}k!{\{\genfrac{}{}{0ex}{}{n+r}{k+r}\}}_{r,\lambda },& \text{if}n\ge k,\\ 0,& \text{if}n<k.\end{array}$(31) ), we get (32) $\begin{array}{rl}\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(k+r{)}_{m,\lambda }{x}^k& =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=0}^m(\genfrac{}{}{0ex}{}{n}{k})(-1{)}^{n-k}(k+r{)}_{m,\lambda }\\ & =\sum _{n=0}^m{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})n!{\{\genfrac{}{}{0ex}{}{m+r}{n+r}\}}_{r,\lambda }.\end{array}$(32) Therefore, by (32(32) $\begin{array}{rl}\frac{1}{(1+x{)}^{\alpha }}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{k})(k+r{)}_{m,\lambda }{x}^k& =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})\sum _{k=0}^m(\genfrac{}{}{0ex}{}{n}{k})(-1{)}^{n-k}(k+r{)}_{m,\lambda }\\ & =\sum _{n=0}^m{\left(\frac{x}{1+x}\right)}^n(\genfrac{}{}{0ex}{}{\alpha }{n})n!{\{\genfrac{}{}{0ex}{}{m+r}{n+r}\}}_{r,\lambda }.\end{array}$(32) ), we obtain the Theorem 2.10.

3. Conclusion

In this paper, we introduce the degenerate harmonic numbers and degenerate Stirling numbers, which are the degenerate versions of the harmonic numbers and Stirling numbers. We obtain some series involving degenerate harmonic numbers, degenerate hyperharmonic numbers and degenerate falling factorial in Theorems 2.1, 2.2, 2.3 and 2.5. Meanwhile, some identities related to degenerate harmonic numbers, degenerate hyperharmonic numbers and combinatorial numbers are obtained in Theorems 2.7 and 2.8. In addition, we investigate the relations between the degenerate falling factorial and degenerate Stirling numbers of the second kind and degenerate r-Stirling numbers of the second kind in Theorems 2.9 and 2.10.

We will continue to study various degenerate versions of many special polynomials and numbers and their various applications in physics, science and other fields.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Additional information

Funding

This research was funded by the National Natural Science Foundation of China (No. 12271320), Key Research and Development Program of Shaanxi (No. 2023-ZDLGY-02).