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\begin{center}
\vskip 1cm{\LARGE\bf New Series Identities with Cauchy, Stirling, \\ 
and Harmonic Numbers, \\
\vskip .08in
and Laguerre Polynomials}
\vskip 1cm
\Large Khristo N. Boyadzhiev\\
Department of Mathematics \\
Ohio Northern University\\
Ada, Ohio 45810\\
USA \\
\href{mailto:k-boyadzhiev@onu.edu}{\tt k-boyadzhiev@onu.edu} \\
 \end{center}

\vskip .2in


\begin{abstract}
We use an interplay between Newton series and binomial
formulas to generate a number of series identities involving
Cauchy numbers (also known as Gregory coefficients or Bernoulli numbers
of the second kind), harmonic numbers, Laguerre polynomials, and Stirling
numbers of the first kind.
\end{abstract}


\section{Introduction} \label{sec1}

The Newton interpolation series is a classic tool in analysis with important applications \cite{12, 17, 18, 19}. In this paper, we use the Newton series in order to construct two summation formulas: the first involving Cauchy numbers,
and the second involving Stirling numbers of the first kind. We obtain a number of series identities---closed form evaluations of several series involving Cauchy numbers together with harmonic numbers. We also present similar series with harmonic numbers and Stirling numbers of the first kind. The results are close to some classical results about such series. 

We often refer to sequence numbers in the {\it On-Line Encyclopedia of
Integer Sequences} (OEIS) \cite{sloane}.
The Cauchy numbers $c_{n} $ (OEIS sequence \seqnum{A006232} and \seqnum{A006233}) are defined by the generating function
$$\frac{x}{\ln (x+1)} =\sum _{n=0}^{\infty }\frac{c_{n} }{n!}  \, x^{n} \quad (|x|\; &lt;1)$$ 
and they can also be defined as an integral of the falling factorial
(see Eq.~\eqref{2} below). These numbers are called Cauchy numbers of
the first kind by Comtet \cite[p.\ 294]{11}. The numbers $c_{n} /n!$
are  called Bernoulli numbers of the second kind by some authors (see
the comments in \cite{3}). In Milne-Thomson's book the Cauchy numbers
appear as $B_{n}^{(n)} \eqref{1}$  \cite[p.\ 135]{17}. Blagouchine in
his recent papers \cite{2, 3, 4} used the notation $G_{n} =c_{n} /n!$,
and the name Gregory coefficients for $G_{n}$.

A detailed study of the Cauchy numbers $c_{n} $ and several interesting identities involving these numbers were presented by Merlini et al.\  \cite{16}. Cauchy numbers were also studied by Liu et al.\  \cite{15} and Zhao \cite{25}.

The Cauchy numbers appear in some important formulas (such as the Laplace summation formula in \cite[p.\ 181]{17}), where they act like a counterpart to the Bernoulli numbers in the Euler-Maclaurin formula. 

The Stirling numbers of the first kind $s(n,k)$ (OEIS \seqnum{A048994} and \seqnum{A008275}) play a major role in this paper. They appeared for the first time in the works of James Stirling together with the Stirling numbers of the second kind $S(n,k)$ \cite{6, 22}. They are related to the Cauchy numbers by the formula \cite[p.\ 294]{11}
$$c_{n} =\sum _{k=0}^{n}\frac{s(n,k)}{k+1}   .$$ 
A detailed study of the Stirling numbers can be found in the books of Jordan \cite{14}, Comtet \cite{11}, and Graham et al.\  \cite{13} (where the unsigned Stirling numbers of the first kind $$(-1)^{n-k} s(n,k)$$ are considered). Informative comments and interesting series containing $s(n,k)$ can be found in Adamchik \cite{1} and Blagouchine \cite[Sec.\ 2]{2}. Various results for series with Stirling numbers and harmonic numbers $H_{n} $ were recently obtained by Choi \cite{10} and Wang and Lyu \cite{23}.

To give the reader an idea of our identities, here are three samples:
\begin{align*}
\ln (1+x)&amp;=\sum _{n=0}^{\infty }\frac{(-1)^{n} c_{n} x^{n+1} }{(1+x)(1+2x)\cdots(1+nx)} \\
2\zeta (3)-1&amp;=\sum _{n=1}^{\infty }\frac{(-1)^{n-1} c_{n} (H_{n}^{2} +H_{n}^{(2)} )}{n!n} \\
\sum _{j=1}^{k}\zeta (j+1)&amp;=\sum _{n=k}^{\infty }\frac{(-1)^{n-k} s(n,k)H_{n} }{(n+1)!}  . \\
\end{align*}
Starting from the Newton interpolation series with simple nodes $n=0,
1, 2,\ldots$,  we derive two special representations (Proposition
\ref{pr1}  and Proposition \ref{pr8}). Using these two propositions,
with the help of several binomial identities we generate a number
of expansions involving Cauchy, harmonic, and Stirling numbers. Our
results are presented in the form of examples. In Section \ref{sec3}
we work mostly with Cauchy numbers and in Section \ref{sec4} we present
series with Stirling numbers of the first kind and harmonic numbers.
Section \ref{sec5} contains a formula for series with Stirling numbers
and Laguerre polynomials. Particular cases of this formula provide series
with Laguerre polynomials and harmonic numbers.



\section{The Newton interpolation series}\label{sec2}

We use finite differences. Given an appropriate function $f(x)$, by definition we have $\Delta f(x)=f(x+1)-f(x),\; \, \Delta ^{n} f(x)=\Delta (\Delta ^{n-1} f(x))$ for $n=1,2,\ldots$ and $\Delta ^{0} f(x)=f(x)$.    

The Newton interpolation series for a given function $f(x)$ is the representation of this function in terms of the polynomials 
$$(z)_{n} =z(z-1)\cdots(z-n+1),\;  (z)_{0} =1$$ 
 \cite{6,12,17,18}. Namely,
\begin{equation} \label{1} 
f(z)=\sum _{n=0}^{\infty }\frac{\Delta ^{n} f(0)}{n!}  \, z(z-1)\cdots(z-n+1). 
\end{equation} 
This is the case of simple nodes $0,1,2\ldots$ . The Newton series expansion is not as popular as the Taylor series. One reason for this is the 
difficulty of studying the remainder and determining
the region of convergence. There are other difficulties, too. For instance, if we apply \eqref{1} to the function $f(x)=\sin(2\pi x)$, then $\Delta ^{n} f(0)=0$ for $n\geq 0$ and the series is identically zero.

Alexander Gelfond has done a thorough study of Newton interpolation
series in \cite[Ch.\ 2]{12}. Another study can be found in  Norlund
\cite[Ch. 5]{18} (also \cite{19} and \cite[pp.\ 302--304]{17}). The
sufficient conditions for a function $f(z)$ to be represented by such
a series are too elaborate to list here; however, we can say that if
the function is analytic in a region of the form $\Re(z)&gt;\lambda $
for some $\lambda $ and has moderate growth in that region, then the
representation \eqref{1} holds for $\Re(z)&gt;\lambda $. Moreover, for any
two numbers $\varepsilon ,R&gt;0$ the series is uniformly convergent for
$\Re(z)\geq \lambda +\varepsilon$ and $  |z-\lambda |\; &lt;R$. Boas and Buck
showed \cite[p.\ 34]{5} that entire functions of exponential type less
than $\ln 2$ have such representation.

The polynomials $(z)_{n} $ often appear in mathematics. For example, they are the ordinary generating functions for the Stirling numbers of the first kind $s(n,k)$ \cite[p.\ 50]{11}:
$$z(z-1)\cdots(z-n+1)=\sum _{k=0}^{n}s(n,k)\, z^{k}  .$$ 
They also appear in the definition of the Cauchy numbers $c_{n} ,\; n=0,1,\ldots$ \cite[p.\ 294]{11}:
\begin{equation} \label{2} 
c_{n} =\int _{0}^{1}z(z-1)\cdots(z-n+1)\, dz  ,
\end{equation} 
with $c_{0} =1,\; c_{1} =\frac{1}{2} ,\; c_{2} =-\frac{1}{6} ,\; c_{3} =\frac{1}{4} ,\; \ldots$ etc. For more details on these numbers see Merlini et al.\  \cite{16} and also Liu et al.\  \cite{15}.  

Integrating the series \eqref{1} term by term and using Eq.~\eqref{2} we come to the most interesting formula,
\begin{equation} \label{3} 
\int _{0}^{1}f(x)\, dx=\sum _{n=0}^{\infty }\, \frac{c_{n} }{n!}   \, \Delta ^{n} f(0), 
\end{equation} 
which clearly demonstrates the importance of the Cauchy numbers. This representation appears in Jordan \cite[Eq.\ (1), p.\ 277]{14}. Note that in his book Jordan works with the numbers
$$b_{n} =\int _{0}^{1}{x\choose n}\; dx=\frac{c_{n} }{n!} .$$ 
(A simpler version was used by Gregory in the 17$^{\text{th}}$ century.)

It is easy to compute that
\begin{equation} \label{4} 
\Delta ^{n} f(x)=(-1)^{n} \sum _{k=0}^{n}{n\choose k} (-1)^{k} f(x+k) 
\end{equation} 
and in particular
$$\Delta ^{n} f(0)=(-1)^{n} \sum _{k=0}^{n}{n\choose k} (-1)^{k} f(k).$$ 
Replacing the finite differences by the binomial expression in \eqref{3} we come to the following proposition.

\begin{proposition} \label{pr1} For appropriate functions $f(x)$ we have the representation
\begin{equation} \label{5} 
\int _{0}^{1}f(x)\, dx=\sum _{n=0}^{\infty }\, \frac{(-1)^{n} c_{n} }{n!}   \, \left(\sum _{k=0}^{n}{n\choose k} (-1)^{k} f(k)\right) . 
\end{equation} 
\end{proposition}
As we shall see, this formula is a valuable tool for obtaining identities with Cauchy numbers.

\begin{definition} The \textit{binomial transform} of one sequence $a_{0} ,\, \, a_{1} \, ,\ldots$  is the new sequence defined by 

 $$b_{n} =\sum _{k=0}^{n}{n\choose k} (-1)^{k} a_{k} $$  with inversion $$a_{n} =\sum _{k=0}^{n}{n\choose k} (-1)^{k} b_{k}. $$
\end{definition}
The binomial transform of the sequence $\{ f(k)\} $ is usually easier to 
compute, compared to the computation of $\Delta ^{n} f(0)$, so that formula \eqref{5} often works better than \eqref{3}.  A table of binomial transform formulas can be found in \cite{7}. Using such formulas we generate a number of examples in the next section.

Here is one simple demonstration of how \eqref{5} works. Taking the function $f(z)=z^{p} $, where $p\geq 0$ is an integer, we find by \eqref{5}  that
$$\frac{1}{p+1} =\sum _{n=0}^{\infty }\, \frac{(-1)^{n} c_{n} }{n!}  \left( \sum _{k=0}^{n}{n\choose k} (-1)^{k} k^{p} \right )\, $$ 
and we recall that the Stirling numbers of the second kind $S(p,n)$ can be represented as
$$S(p,n)=(-1)^{n} n!\sum _{k=0}^{n}{n\choose k} (-1)^{k} k^{p}. $$ 
Thus we get the identity
$$\frac{1}{p+1} =\sum _{n=0}^{p}c_{n}  \, S(p,n).$$ 
(The infinite series terminates, as $S(p,n)=0$ for $n&gt;p$.) This is the Stirling inverse of the representation of the Cauchy numbers in terms of Stirling numbers of the first kind \cite[p.\ 294]{11}, and also \cite[p.\ 267]{14})
$$c_{n} =\sum _{p=0}^{n}\frac{s(n,p)}{p+1}  . $$ 


\section{Series with Cauchy numbers and harmonic numbers}\label{sec3}

In this section we use Proposition \ref{pr1}, together with various binomial identities, to generate series identities involving Cauchy numbers and harmonic numbers.

\begin{example}\label{ex3} Let $y&gt;0$ be arbitrary. We use the function
$$f(z)=\frac{y}{y+z} ,\; \; \Re(z)+y&gt;0$$ 
together with the binomial identity \cite[Entry (8.28)]{7}
$$\sum _{k=0}^{n}{n\choose k} (-1)^{k} \frac{y}{y+k} ={n+y\choose n}^{-1} .$$ 
Equation \eqref{5} implies
$$y\ln \left(1+\frac{1}{y} \right)=\sum _{n=0}^{\infty }\frac{(-1)^{n} c_{n} }{n!}  {n+y\choose n}^{-1} =\sum _{n=0}^{\infty }\frac{(-1)^{n} c_{n} }{(y+1)(y+2)\cdots(y+n)}  $$ 
or
$$\ln \left(1+\frac{1}{y} \right)=\sum _{n=0}^{\infty }\frac{(-1)^{n} c_{n} }{y(y+1)(y+2)\cdots(y+n)}  $$ 
(cf.\  \cite{24} and for a similar representation see \cite[p.\  244]{19}). 
Replacing $1/y$ by $x$ we find the following remarkable expansion of
$\ln (1+x)$:
\begin{equation} \label{6}
\ln (1+x)=\sum _{n=0}^{\infty }\frac{(-1)^{n} c_{n} x^{n+1} }{(1+x)(1+2x)\cdots(1+nx)}   . 
\end{equation} 
When $x=1$, this gives
$$\ln 2=\sum _{n=0}^{\infty }\frac{(-1)^{n} c_{n} }{(n+1)!}  .$$ 
Using the 
asymptotic behavior of the Cauchy numbers at infinity \cite[p.\  294]{11}, namely
$$\frac{c_{n} }{n!} \; \; \sim \frac{(-1)^{n+1} }{n\ln ^{2} n} \quad (n\to \infty ),$$ 
it is easy to verify (by the ratio test, for instance) that the series in \eqref{6} is convergent for all $x&gt;0$.

Before continuing further we recall that the harmonic numbers $H_{n} $ are defined for $n=0,1,\ldots$ by
$$H_{n} =1+\frac{1}{2} +\cdots+\frac{1}{n} \quad (n&gt;0),\; H_{0} =0$$ 
and it is often convenient to express them in the form
$$H_{n} =\psi (n+1)+\gamma $$ 
where $\psi (z)=\frac{d}{dz} \ln \Gamma (z)$ is the digamma function and $\gamma =-\psi \eqref{1}$ is Euler's constant. This formula provides an extension of the harmonic numbers as a holomorphic function $H_{z} $ on $\Re(z)&gt;-1$. We also
use the generalized harmonic numbers $H{\kern 1pt} _{n}^{(p)} $, where $H{\kern 1pt} _{0}^{(p)} =0$ and for $n&gt;0$  we have
$$H_{n}^{{\kern 1pt} (p)} =1+\frac{1}{2^{p} } +\cdots+\frac{1}{n^{p} } =\zeta (p)+\frac{(-1)^{p-1} }{(p-1)!} \psi ^{(p-1)} (n+1)\quad (p&gt;1).$$ 
\end{example}
\begin{example}\label{ex4} Let $m&gt;0$ be an integer. We take the function $f(z)=\frac{1}{(z+m)^{2} } $ and use the binomial identity
$$\sum _{k=0}^{n}{n\choose k} (-1)^{k} \frac{1}{(k+m)^{2} } =\frac{(m-1)!n!(H_{n+m} -H_{m-1} )}{(n+m)!} $$ 
 \cite[Entry 8.38]{7}. Equation \eqref{5} implies
\begin{equation} \label{7} 
\frac{1}{(m+1)!} =\sum _{n=0}^{\infty }\frac{(-1)^{n} c_{n} (H_{n+m} -H_{m-1} )}{(n+m)!}  .  
\end{equation} 
For $m=1,\; m=2,\; {\rm and}\; m=3$ we find correspondingly 
\begin{align*}
\frac{1}{2} &amp;=\sum _{n=0}^{\infty }\frac{(-1)^{n} c_{n} H_{n+1} }{(n+1)!}  \\
\frac{1}{6}&amp; =\sum _{n=0}^{\infty }\frac{(-1)^{n} c_{n} (H_{n+2} -1)}{(n+2)!}  \\
\frac{1}{12} &amp;=\sum _{n=0}^{\infty }\frac{(-1)^{n} c_{n} (2H_{n+3} -3)}{(n+3)!}  , \\
\end{align*}
etc.
\end{example}
\begin{example} \label{ex5}Here we exploit the binomial formula  \cite[Entry 9.16]{7}
\begin{equation} \label{8} 
\sum _{k=0}^{n}{n\choose k} (-1)^{k} H_{k}^{ (3)} =-\frac{1}{2n} (H_{n}^{2} +H_{n}^{(2)} ) 
\end{equation} 
($n=1,2\ldots$ ) where 
$$H_{k}^{ (3)} =1+\frac{1}{2^{3} } +\cdots\frac{1}{k^{3} } =\zeta (3)+\frac{1}{2} \psi '' (k+1). $$ 
We apply \eqref{5} to the function $f(z)=\zeta (3) +\frac{1}{2} \psi ''\, (z+1)$. In this case
$$\int _{0}^{1}\zeta (3)+\frac{1}{2} \psi ''\, (z+1)\, d{\kern 1pt} z =\zeta (3)+\frac{1}{2} \left. \psi '{\kern 1pt} (z+1)\right|_{{\kern 1pt} 0}^{{\kern 1pt} 1} =\zeta (3)+\frac{1}{2} (\psi '(2)-\psi '(1))=\zeta (3)-\frac{1}{2} $$ 
and therefore,
$$2\zeta (3)-1=\sum _{n=0}^{\infty }\, \frac{(-1)^{n} c_{n} }{n!} \left( \sum _{k=0}^{n}{n\choose k} (-1)^{k} H_{k}^{(3)} \right) $$ 
$$=\sum _{n=1}^{\infty }\, \frac{(-1)^{n} c_{n} }{n!} \left(\sum _{k=0}^{n}{n\choose k} (-1)^{k} H_{k}^{(3)} \right) =\sum _{n=1}^{\infty }\frac{(-1)^{n-1} c_{n} (H_{n}^{2} +H_{n}^{(2)} )}{n!n}  $$ 
(the term for $n=0$ in the first sum is zero and we can start the summation from $n=1$). Thus
\begin{equation} \label{9} 
\; 2\zeta (3)-1=\sum _{n=1}^{\infty }\frac{(-1)^{n-1} c_{n} (H_{n}^{2} +H_{n}^{(2)} )}{n!n}  .  
\end{equation} 
It can be seen from the digamma definition of the harmonic numbers that
$$\mathop{\lim }\limits_{p\to 0} \ \frac{1}{p} (H_{p}^{2} +H_{p}^{(2)} )=2\zeta (3)$$ 
which can be considered the zero term in the sum. With this agreement, starting the summation from $n=0$ we can write
$$-1=\sum _{n=0}^{\infty }\frac{(-1)^{n-1} c_{n} (H_{n}^{2} +H_{n}^{(2)} )}{n!n}  .$$ 
\end{example}
\begin{example} \label{ex6}For completeness we now reprove two known identities. First, the classical representation
\begin{equation} \label{10} 
\gamma =\sum _{n=0}^{\infty }\frac{(-1)^{n-1} c_{n} }{n!n}   
\end{equation} 
obtained by Lorenzo Mascheroni in 1790. Informative historical notes on this series can be found in Blagouchine \cite[p.\ 406]{2}.  

We take $f(z)=\psi (z+1)+\gamma ,\; f(k)=H_{k} $ and use the simple identity \cite[Identity (9.3a)]{7}
$$\sum _{k=0}^{n}{n\choose k} (-1)^{k} H_{k} =-\frac{1}{n}  .$$ 
Clearly
$$\int _{0}^{1}\psi (z+1)+\gamma dz= \left. \ln \Gamma (z+1)\right|_{0}^{1} +\gamma =\gamma $$ 
and \eqref{5} implies \eqref{10} immediately. 

Next, we prove the representation
\begin{equation} \label{11} 
\frac{\pi ^{2} }{6} -1=\sum _{n=1}^{\infty }\frac{(-1)^{n-1} c_{n} H_{n} }{n!n}  , 
\end{equation} 
which can be found in \cite{16}, and also \cite[Eq.\ 44, p.\ 413]{2}. We
take $f(z)=\zeta(2)-\psi '(z+1)$, so that $f(k)=H_{k}^{{\kern 1pt}
(2)} =1+\frac{1}{2^{2} } +\cdots+\frac{1}{k^{2} } ,\; f(0)=0$.
We also need the identity  \cite[9.4b]{7}
$$\sum _{k=0}^{n}{n\choose k} (-1)^{k} H_{k}^{{\kern 1pt} (2)} =-\frac{H_{n} }{n} . $$ 
Then we compute
$$\int _{0}^{1}\zeta (2)-\psi '(z+1)\, dz=\;  \zeta (2)-\left. \psi (z+1)\right|_{0}^{1} =\zeta (2)-\psi (2)+\psi (1)=\zeta (2)-1. $$ 
and \eqref{11} follows from \eqref{5}.

As in the previous example, we notice that 
\begin{equation} \label{12} 
\mathop{\lim }\limits_{p\to 0} \frac{H_{p} }{p} =\mathop{\lim }\limits_{p\to 0} \frac{\psi (p+1)+\gamma }{p} =\zeta (2)=\frac{\pi ^{2} }{6} , 
\end{equation} 
and if we agree to write 
$$\frac{\pi ^{2} }{6} =\left. \frac{H_{n} }{n} \right|_{n=0}, $$ 
then Eq.~\eqref{11} becomes (with summation from $n=0$) 
$$-1=\sum _{n=0}^{\infty }\frac{(-1)^{n-1} c_{n} H_{n} }{n!n}  .$$ 
Merlini, Sprugnoli, and Verri \cite{16} found \eqref{10} and \eqref{11} by using a symbolic Laplace summation formula. Blagouchine derived \eqref{11} from a special series transformation formula  \cite[p.\  412]{2}. With the help of his formula, he also obtained several more interesting representations of this kind. Two finite sums connecting Cauchy numbers and harmonic numbers can be found in \cite{25}.
\end{example}
\begin{example}\label{ex7} In this example we use the constant
$$M_{1} =\int _{1}^{2}\frac{\psi (1+t)+\gamma }{t} \;  dt\approx 0.86062,$$ 
and we prove that 
\begin{equation} \label{13} 
M_{1} =\frac{1}{2} +\sum _{n=0}^{\infty }\frac{(-1)^{n-1} c_{n} H_{n} }{(n+1)!}  . 
\end{equation} 
For this purpose we take the function $f(z)=\dfrac{\psi (z+1)+\gamma }{z+1} $ and apply the binomial formula
\begin{equation} \label{14} 
\sum _{k=0}^{n}{n\choose k} (-1)^{k} \frac{H_{k} }{k+1} =-\frac{H_{n} }{n+1}  
\end{equation} 
\cite[Entry 9.32]{7}. The left hand side in \eqref{5} becomes
$$\int _{0}^{1}\frac{\psi (z+1)+\gamma }{z+1} dz= \int _{1}^{2}\frac{\psi (x)+\gamma }{x} dx= \int _{1}^{2}\frac{\psi (x+1)-1/x+\gamma }{x} dx $$ 
$$=M_{1} -\int _{1}^{2}\frac{1}{x^{2} } dx= M_{1} -\frac{1}{2} $$ 
and therefore,
$$M_{1} -\frac{1}{2} =\sum _{n=0}^{\infty }\frac{(-1)^{n-1} c_{n} H_{n} }{(n+1)!}  .$$ 
The constant $M_{1} $ is important, because it represents the value of some interesting series. It was shown in \cite{8} that 
$$M_{1} =\sum _{n=1}^{\infty }\frac{1}{n}  \ln \left(1+\frac{1}{n+1} \right)$$ 
$$M_{1} =\sum _{n=1}^{\infty }\frac{(-1)^{n-1} }{n} \left(n-\zeta (2)-\zeta (3)-\cdots-\zeta (n)\right) $$ 
the first term in this series is $1$. Also
$$M_{1} =\sum _{n=1}^{\infty }H_{n}^{-}  \left(\zeta (n+1)-1\right)$$ 
where 
$$H_{n}^{{\kern 1pt} -} =1-\frac{1}{2} +\frac{1}{3} +\ldots+\frac{(-1)^{n-1} }{n} $$ 
are the skew harmonic numbers. Now we have one more series in this list.
\end{example}


\section{ Series with Stirling numbers of the first kind and harmonic numbers} \label{sec4}

In this section we assume that $f(z)$ is a function analytic in some
half plane $\Re(z)&gt;\lambda $, where $\lambda &lt;0$, and has moderate
growth. Being analytic in a neighborhood of the origin, $f(z)$ is
represented by a Taylor series centered at $z=0$.

Now we manipulate the representation \eqref{1} in a different way. We replace $(z)_{n} $ by 
$$z(z-1)\cdots(z-n+1)=\sum _{m=0}^{n}s(n,m)\, z^{m}  $$ 
and change the order of summation to get
$$f(z)=\sum _{m=0}^{\infty }z^{m}  \left(\sum _{n=0}^{\infty }\frac{s(n,m)}{n!} \Delta ^{n} f(0) \right). $$ 
Comparing this to the Taylor series 
$$f(z)=\sum _{m=0}^{\infty }z^{m}  \frac{f^{(m)} (0)}{m!} $$ 
we conclude that 
$$\frac{f^{(m)} (0)}{m!} =\sum _{n=0}^{\infty }\frac{s(n,m)}{n!} \Delta ^{n} f(0) .$$ 
Applying Eq.~\eqref{4}, we now come to an important identity:
\begin{proposition}\label{pr8} Under the initial assumption on the function $f(z)$, for every $m\geq 0$ we have the representation
\begin{equation} \label{15} 
\frac{f^{(m)} (0)}{m!} =\sum _{n=0}^{\infty }\frac{(-1)^{n} s(n,m)}{n!} \left(\sum _{k=0}^{n}{n\choose k}(-1)^{k} f(k) \right) . 
\end{equation} 
\end{proposition}
\noindent
(The summation, in fact, starts from $n=m$ since $s(n,m)=0$ for $n&lt;m$.)

When $m=1$ we have $s(n,1)=(-1)^{n-1} (n-1)!$ and the formula takes the form
$$f'{\kern 1pt} {\kern 1pt} (0)=\sum _{n=1}^{\infty }\, \frac{1}{n} \, \left(\sum _{k=0}^{n}{n\choose k}(-1)^{k-1} f(k) \right).$$ 
This evaluation was obtained by a different method in \cite{9}.
Formula \eqref{15} is an efficient mechanism for producing various series with Stirling numbers of the first kind together with special numbers.  It resembles formula \eqref{5}, where the Cauchy numbers are replaced by Stirling numbers and on the left hand side we have now derivatives.  Most examples from the previous section can be repeated with the same function $f(z)$ and the same binomial identity, but now using formula \eqref{15} instead of \eqref{5}.

\begin{example} \label{ex9}First we prove a classical result. Taking $f(z)=\psi (z+1)+\gamma $, $f(k)=H_{k} $ we have
$$\sum _{k=0}^{n}{n\choose k}(-1)^{k} f(k) =-\frac{1}{n} $$ 
(see Example \ref{ex6}). The Taylor series of  $\psi (z+1)+\gamma $ is well-known 
\begin{equation} \label{16} 
\psi (z+1)+\gamma =\sum _{k=1}^{\infty }(-1)^{k-1}  \zeta (k+1)\, z^{k} ,\quad |z|&lt;1 
\end{equation} 
and \eqref{15} produces the representation ($k\geq 1$) 
$$\zeta (k+1)=\sum _{n=k}^{\infty }\frac{(-1)^{n-k} s(n,k)}{n!n}  .$$ 
This is almost a folklore mathematical result. In a more general form it can be found in Jordan \cite[p.\ 166]{14}. Comments and extensions can be found in Adamchik \cite[Section 5]{1}, and also in Blagouchine \cite[p.\  412]{2}.

Note that the numbers 
$$\left[\begin{array}{c} {n} \\ {k} \end{array}\right]=(-1)^{n-k} s(n,k)$$ 
are called the {\it unsigned Stirling numbers of the first kind}. They are thoroughly studied in the classical book of Graham, Knuth, and Patashnik \cite[Chapter 6]{13}. These numbers appear in the examples that follow.
\end{example}
\begin{example} \label{ex10} Working with the function $f(z)=\zeta (2)-\psi '(z+1)$, $f(k)=H_{k}^{(2)} $ we have for $n\geq 1$   
$$\sum _{k=0}^{n}{n\choose k}(-1)^{k} f(k) =-\frac{H_{n} }{n} $$ 
and also from \eqref{16}
\begin{align*}
\zeta (2)-\psi '(z+1)&amp;=\zeta (2)+\sum _{k=0}^{\infty }(-1)^{k+1}  (k+1)\zeta (k+2)\, z^{k} \\
&amp;=\sum _{k=1}^{\infty }(-1)^{k+1}  (k+1)\zeta (k+2)\, z^{k} .\\
\end{align*}
This way for $k\geq 1$ we have
\begin{equation} \label{17} 
\zeta (k+2)=\frac{1}{k+1} \sum _{n=k}^{\infty }\frac{(-1)^{n-k} s(n,k)H_{n} }{n!n}  . 
\end{equation} 
The equality also holds for $k=0$ under the agreement \eqref{12}.

Continuing further in this direction we can use identity \eqref{8} from Example \ref{ex5} with the same function $f(z)=\zeta (3)+\frac{1}{2} \psi ''(z+1),f(k)=H_{k}^{{\kern 1pt} (3)} $. Again from \eqref{16}
$$f(z)=\frac{1}{2} \sum _{k=1}^{\infty }(-1)^{k-1}  (k+1)(k+2)\zeta (k+3)z^{k} $$
and \eqref{15} implies 
\begin{equation} \label{18} 
\zeta (k+3)=\frac{1}{(k+1)(k+2)} \sum _{n=k}^{\infty }\frac{(-1)^{n-k} s(n,k)}{n!n}  \left(H_{n}^{2} +H_{n}^{(2)} \right) 
\end{equation} 
for any $k\geq 1$. 

The identities \eqref{17} and \eqref{18} were found independently by a different method by Wang and Lyu in \cite[p.\  171]{23}. 
\end{example}
\begin{example} \label{ex11} 
Again let $k\ge 1$. Applying \eqref{15} to the function
$$\, f(z)=\frac{\psi (z+1)+\gamma }{z+1} ,\; \Re(z)&gt;-1$$ 
and the binomial identity \eqref{14}, we find \cite[p.\  28]{21}
\begin{equation} \label{19} 
\sum _{j=1}^{k}\zeta (j+1)=\sum _{n=k}^{\infty }\frac{(-1)^{n-k} s(n,k)H_{n} }{(n+1)!}. 
\end{equation} 
To prove this,
we use the Taylor series \eqref{16} and notice that for $|z|\; &lt;1$  we have
$$\frac{\psi (z+1)+\gamma }{z+1} =\frac{1}{z+1} \sum _{k=1}^{\infty }(-1)^{k-1}  \zeta (k+1){\kern 1pt} {\kern 1pt} z^{k} =\sum _{k=1}^{\infty }\left((-1)^{k-1} \sum _{j=1}^{k}\zeta (j+1) \right)z^{k}.  $$ 
Then \eqref{15} implies \eqref{19} immediately.
\end{example}
\begin{example} \label{ex12}Here we use the identity
$$\sum _{k=0}^{n}{n\choose k}(-1)^{k} \frac{(-1)^{k} }{(k+1)^{2} }  =\frac{H_{n+1} }{n+1} ,$$ 
together with the function $f(z)=(z+1)^{-2} ,\; \Re(z)&gt;-1$. For $|z|\; &lt;1$ we have the Taylor series 
$$\frac{1}{(z+1)^{2} } =\sum _{k=0}^{\infty }(-1)^{k}  (k+1)z^{k} .$$ 
Therefore, from \eqref{15} we find the curious companion to Eq.~\eqref{19}
\begin{equation} \label{20} 
k+1=\sum _{n=k}^{\infty }\frac{(-1)^{n-k} s(n,k)H_{n+1} }{(n+1)!}  
\end{equation} 
(see  \cite[p.\ 29]{21}).  More generally, using the identity from Example \ref{ex4}
$$\sum _{k=0}^{n}{n\choose k} (-1)^{k} \frac{1}{(k+m)^{2} } =\frac{(m-1)!n!(H_{n+m} -H_{m-1} )}{(n+m)!} $$ 
with the function 
$$f(z)=(z+m)^{-2} =\sum _{k=0}^{\infty }\frac{(-1)^{k} (k+1)z^{k} }{m^{k+2} }  $$ 
we find
\begin{equation} \label{21} 
\frac{k+1}{m^{k+2} (m-1)!} =\sum _{n=k}^{\infty }\frac{(-1)^{n-k} s(n,k)(H_{n+m} -H_{m-1} )}{(n+m)!}  ,
\end{equation} 
where $m=1$ gives Eq.~\eqref{20}. In the same line, using the identity 
$$\sum _{k=0}^{n}{n\choose k} (-1)^{k} \frac{1}{(k+m)^{3} } =\frac{(m-1)!n!}{2(n+m)!} \left((H_{n+m} -H_{m-1} )^{2} +H_{n+m}^{(2)} -H_{m-1}^{(2)} \right)$$ 
 \cite[8.42]{7} together with the function 
$$f(z)=(z+m)^{-3} =\sum _{k=1}^{\infty }\frac{(-1)^{k} (k+1)(k+2)z^{k} }{m^{k+3} }  $$ 
we come to the series
\begin{equation} \label{22} 
\frac{(k+1)(k+2)}{m^{k+3} (m-1)!} =\sum _{n=k}^{\infty }\frac{(-1)^{n-k} s(n,k)}{2(n+m)!}  \left((H_{n+m} -H_{m-1} )^{2} +H_{n+m}^{(2)} -H_{m-1}^{(2)} \right). 
\end{equation} 
\end{example}
\begin{example} \label{ex13}Here we use the identity 
$$\sum _{k=0}^{n}{n\choose k}(-1)^{k} H_{k+1}  =-\frac{1}{n(n+1)} $$ 
together with the function $f(z)=\psi (z+2)+\gamma ,\; \; f(k)=H_{k+1} $. We write 
$$f(z)=\psi (z+2)+\gamma =\psi (z+1+1)+\gamma =\frac{1}{1+z} +\psi (z+1)+\gamma $$ 
for $\Re(z)&gt;-1$.  From \eqref{16}
$$f(z)=\sum _{k=1}^{\infty }(-1)^{k-1}  (\zeta (k+1)-1)\, z^{k} ,\; \; |z|\, &lt;1$$ 
and \eqref{15} produces the representation (see  \cite[p.\ 339]{14})
\begin{equation} \label{23} 
\zeta (k+1)-1=\sum _{n=k}^{\infty }\frac{(-1)^{n-k} s(n,k)}{(n+1)!n}  \quad (k\geq 1). 
\end{equation} 
\end{example}
\begin{example}\label{ex14} Now we present a series connecting the Stirling numbers $s(n,k)$ and the central binomial coefficients ${2n\choose n}$. Starting from the identity in \cite[Entry 8.43]{7}.
$$\sum _{k=0}^{n}{n\choose k}(-1)^{k} \frac{1}{2k+1}  =\frac{4^{n} }{2n+1} {2n\choose n}^{-1} $$ 
we introduce the function $f(z)=(2z+1)^{-1} ,\; \Re(z)&gt;-\frac{1}{2} $  with Taylor series for $|z|\; &lt;1/2$ 
$$\frac{1}{2z+1} =\sum _{k=0}^{\infty }(-1)^{k}  2^{k} z^{k} . $$ 
From \eqref{15} we have the strange representation
\begin{equation} \label{24} 
2^{k} =\sum _{n=k}^{\infty }\frac{(-1)^{n-k} 4^{n} s(n,k)}{n!(2n+1)}  {2n\choose n}^{-1}  . 
\end{equation} 
All these series with positive terms are very slowly convergent.

\end{example}
\section{Series with Laguerre polynomials and harmonic numbers} \label{sec5}

In this section we present series with the classical Laguerre polynomials $L_{{\kern 1pt} n} (x)$, which have the representation
$$L_{n} (x)=\sum _{k=0}^{n}{n\choose k} \, (-1)^{k} \frac{x^{{\kern 1pt} k} }{k!} .$$ 
Let $x&gt;0$ and consider the function $f(z)=\frac{x^{{\kern 1pt} z} }{\Gamma (z+1)} $ so that $f(k)=\frac{x^{k} }{k!} $. Proposition \ref{pr8} implies the formula
\begin{equation} \label{25} 
\left(\frac{d}{d{\kern 1pt} z} \right)^{m} \left. \frac{x^{{\kern 1pt} z} }{\Gamma (z+1)} \, \right|_{{\kern 1pt} z=0} =\sum _{n=m}^{\infty }\frac{(-1)^{n} s(n,m)}{n!} L_{{\kern 1pt} n}  (x). 
\end{equation} 
This way for every $m=1,\, \, 2,\ldots$, we obtain
different series with Laguerre polynomials. 

\begin{example} \label{ex15} When $m=1$, $(-1)^{n} s(n,1)=-(n-1)!$ and we come to the known series
\begin{equation} \label{26} 
-\gamma -\ln x=\sum _{n=1}^{\infty }\frac{1}{n}  \, L_{n} (x). 
\end{equation} 
This is an entry in the reference book \cite[Entry 5.11.1(3)]{20}. Interestingly, this equation resembles \cite[Eq.\ 101]{4}.

For $m=2$ we have $(-1)^{n} s(n,2)=(n-1)!H_{n-1} $ and this provides the new series
\begin{equation} \label{27} 
\frac{\gamma ^{2} }{2} +\gamma \ln x-\frac{\pi ^{2} }{12} +\frac{\ln ^{2} x}{2} =\sum _{n=1}^{\infty }\frac{1}{n}  \, H_{n-1} \, L_{n} (x). 
\end{equation} 
For $m&gt;2$ the derivatives in \eqref{25} become very long. For $m=3$ we just write
\begin{equation} \label{28} 
\left(\frac{d}{d{\kern 1pt} z} \right)^{3} \left. \frac{x^{{\kern 1pt} z} }{\Gamma (z+1)} \, \right|_{{\kern 1pt} z=0} =\frac{1}{2} \sum _{n=k}^{\infty }\, \frac{1}{n} \left(H_{n-1}^{{\kern 1pt} 2} -H_{n-1}^{{\kern 1pt} (2)} \right)L_{{\kern 1pt} n}  (x) 
\end{equation} 
etc.
\end{example}


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\end{thebibliography}
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\noindent 2010 {\it Mathematics Subject Classification}: Primary 11B65; Secondary, 05A19, 33B99, 30B10.

\noindent \emph{Keywords:} Cauchy number, harmonic number, Stirling number, Newton series, binomial identity, Riemann's zeta function, Laguerre polynomial, series with special numbers.

\bigskip
\hrule
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\noindent (Concerned with sequences
\seqnum{A000108},
\seqnum{A006232},
\seqnum{A006233},
\seqnum{A008275}, and
\seqnum{A048994}.)

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received  November 8 2019;
revised versions received  April 6 2020; October 19 2020; December 5 2020.
Published in {\it Journal of Integer Sequences}, December 6 2020.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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