text:
\begin{document}

\title{Euler sums of generalized harmonic numbers and connected extensions}
\author{M\"{u}m\"{u}n CAN, Levent KARGIN, Ayhan D\.IL, G\"{u}ltekin SOYLU\\
Department of Mathematics, Akdeniz University, Antalya, Turkey.}
\date{}
\maketitle

\begin{abstract}
This paper presents the evaluation of the Euler sums of generalized
hyperharmonic numbers $H_{n}^{\left(  p,q\right)  }$
\[
\zeta_{H^{\left(  p,q\right)  }}\left(  r\right)  =\sum\limits_{n=1}^{\infty
}\dfrac{H_{n}^{\left(  p,q\right)  }}{n^{r}}
\]
in terms of the famous Euler sums of generalized harmonic numbers. Moreover,
several infinite series, whose terms consist of certain harmonic numbers and
reciprocal binomial coefficients, are evaluated in terms of Riemann zeta
values.

\end{abstract}

\section{Introduction}

The classical Euler sum $\zeta_{H}\left(  r\right)  $ is the following
Dirichlet series
\[
\zeta_{H}\left(  r\right)  =\sum\limits_{n=1}^{\infty}\dfrac{H_{n}}{n^{r}},
\]
where $H_{n}$ is the $n$th harmonic number. This series is also known as the
harmonic zeta function. The famous Euler's identity for this sum is
\cite{CS,LE,Ni}
\begin{equation}
2\zeta_{H}\left(  r\right)  =\left(  r+2\right)  \zeta\left(  r+1\right)
-\sum_{j=1}^{r-2}\zeta\left(  r-j\right)  \zeta\left(  j+1\right)  ,\text{
}r\in\mathbb{N}\backslash\left\{  1\right\}  , \label{ES}
\end{equation}
where $\zeta\left(  r\right)  $ is the classical Riemann zeta function (for
more details, see for instance \cite{SrC}). Many generalizations of Euler sums
(the so called Euler-type sums) are given using generalizations of harmonic
numbers (see \cite{AC,BA,BJG,Bo,BK,Br,DiMeCe,KK,S1,S,Xu1,XYS,YW,Y,Y1,WL,ZCB}).
Evaluation of Euler-type sums and construction of closed forms are active
fields of study in analytical number theory. Furthermore \cite{BJG,Bo, Br, C,
C2} are some of the studies that make this area interesting in the sense that
Euler sums have potential applications in quantum field theory and knot
theory, especially in evaluation of Feynman diagrams.

Euler actually considered also the more general form \cite{BB, CB, LE,Ni}
\begin{equation}
\zeta_{H^{\left(  p\right)  }}\left(  m\right)  =\sum_{n=1}^{\infty}
\frac{H_{n}^{\left(  p\right)  }}{n^{m}}, \label{g1}
\end{equation}
where $H_{n}^{\left(  p\right)  }$ defined by
\[
H_{n}^{\left(  p\right)  }=1+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\cdots+\frac
{1}{n^{p}},\text{ }\left(  p\in\mathbb{Z},n\in\mathbb{N}\right)  ,
\]
is the $n$th partial sum of $\zeta\left(  p\right)  $ and is called the $n$th
generalized harmonic number for $p>1$. In particular, $H_{0}^{\left(
p\right)  }=0$ and $H_{n}^{\left(  1\right)  }=H_{n}$, the $n$th harmonic
number. When $p\leq0$ it is called sum of powers of integers.

One of the most important issues here is to write Euler-type sums as
combinations of the Riemann zeta function as in (\ref{ES}). This problem has
remained important for various Euler-type sums from the era of Euler to the
present day. It's shown by Euler himself that, the cases of $p=1$, $p=q$,
$p+q$ odd, and for special pairs $(p,q)\in\{(2,4),(4,2)\}$, the sums of the
form (\ref{g1}) have evaluations in terms of the Riemann zeta function (see
\cite{BB, CB, LE,Ni}). There is a very comprehensive literature on this
subject, both theoretical and numerical (\cite{A, Borwein, Bo,Br,CB,CS,DB,
DiMeCe, MD, S,Xu1,Xu, XYS,XZZ, YW, WL}). One of these results; the Euler
identity (\ref{ES}) was further extended in the works of Borwein et al.
\cite{Borwein} and Huard et al. \cite{HuWY}. For odd weight $N\geq3$ and
$p=1,2,\ldots,N-2,$ we have \cite[Theorem 1]{HuWY} (or \cite[p. 278]
{Borwein})
\begin{align}
\zeta_{H^{\left(  p\right)  }}\left(  N-p\right)   &  =\left(  -1\right)
^{p}\sum_{j=0}^{\left[  \left(  N-p-1\right)  /2\right]  }\binom{N-2j-1}
{p-1}\zeta\left(  N-2j\right)  \zeta\left(  2j\right) \label{GES}\\
&  +\left(  -1\right)  ^{p}\sum_{j=0}^{\left[  p/2\right]  }\binom
{N-2j-1}{N-p-1}\zeta\left(  N-2j\right)  \zeta\left(  2j\right)  -\zeta\left(
0\right)  \zeta\left(  N\right)  .\nonumber
\end{align}
Moreover, these so called "linear Euler sums" satisfy a simple reflection
formula
\begin{equation}
\zeta_{H^{\left(  p\right)  }}\left(  r\right)  +\zeta_{H^{\left(  r\right)
}}\left(  p\right)  =\zeta\left(  p+r\right)  +\zeta\left(  p\right)
\zeta\left(  r\right)  . \label{rp}
\end{equation}

Considering nested partial sums of the harmonic numbers, Conway and Guy
\cite{CG} introduced hyperharmonic numbers for an integer $r>1$ as
\[
h_{n}^{(r)}=\sum_{k=1}^{n}h_{k}^{(r-1)},\text{ }n\in\mathbb{N},
\]
with $h_{n}^{(0)}=1/n,$ $h_{n}^{(1)}=H_{n}$ and $h_{0}^{(r)}=0.$ Hyperharmonic
numbers are also important because they build a step in the transition to the
multiple zeta functions (see \cite{KK, Xu}). Dil and Boyadzhiev \cite{DB}
extended the Euler's identity (\ref{ES}) to the Euler sums of the
hyperharmonic numbers:
\begin{equation}
\zeta_{h^{\left(  q\right)  }}\left(  r\right)  =\sum_{n=1}^{\infty}
\frac{h_{n}^{\left(  q\right)  }}{n^{r}},\text{ }(r>q) ,\label{g2}
\end{equation}
as
\begin{align}
\zeta_{h^{\left(  q\right)  }}\left(  r\right)   & =\frac{1}{\left(
q-1\right)  !}\sum_{k=1}^{q}

\genfrac{[}{]}{0pt}{}{q}{k}

\label{HES}\\
& \times\left\{  \zeta_{H}\left(  r-k+1\right)  -H_{q-1}\zeta\left(
r-k+1\right)  +\sum_{j=1}^{q-1}\mu\left(  r-k+1,j\right)  \right\}  ,
\nonumber
\end{align}
where $

\genfrac{[}{]}{0pt}{}{q}{k}

$ is the Stirling number of the first kind and
\begin{equation}
\mu\left(  r,j\right)  =\sum_{n=1}^{\infty}\frac{1}{n^{r}\left(  n+j\right)
}=\sum_{k=1}^{r-1}\frac{\left(  -1\right)  ^{k-1}}{j^{k}}\zeta\left(
r+1-k\right)  +\left(  -1\right)  ^{r-1}\frac{H_{j}}{j^{r}}. \label{mu}
\end{equation}
Formula (\ref{HES}) was the general form of the results obtained for some
special values of $q$ and $r$ in the study of \cite{MD}.

Studies on evaluating Euler sums (\ref{g1}) and (\ref{g2}) in terms of Riemann
zeta values $\zeta\left(  m\right)  $ have motivated researchers to find
representations harmonic number series of the forms
\[
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p\right)  }}{\left(  n+m\right)
^{r}\dbinom{n+m+l}{l}},\text{ }\sum_{n=1}^{\infty}\frac{h_{n}^{\left(
q\right)  }}{n\dbinom{n+q}{q}}.
\]
It has been shown that some families of these type of series can be evaluated
in terms of Euler sums and Riemann zeta values (see for example for $m=0,$
$p=1,$ $r\in\left\{  0,1\right\}  $ \cite{S2,S4,SC}, for $m=r=0$ \cite{S3},
for $m=0$ \cite{S2,XZZ}$,$ for $m>0,$ $r=1,$ $p\in\left\{  1,2\right\}  $
\cite{S5,SS} and for the series involving hyperharmonic numbers \cite{DB}). We
would like to emphasize that in some studies these type of series have been
expressed in terms of hypergeometric series \cite{Choi,CS,MD,S5,S,SS}.

In this work we mainly concentrate on generalized hyperharmonic numbers
defined as (see \cite{DiMeCe})
\begin{equation}
H_{n}^{\left(  p,r\right)  }=\sum_{k=1}^{n}H_{k}^{\left(  p,r-1\right)
},\text{ }\left(  p\in\mathbb{Z},r\in\mathbb{N}\right)  ,\label{0}
\end{equation}
with $H_{n}^{\left(  p,0\right)  }=1/n^{p}$. These are a unified extension of
generalized harmonic numbers and hyperharmonic numbers:
\[
H_{n}^{\left(  p,1\right)  }=H_{n}^{\left(  p\right)  }\text{ and }
H_{n}^{\left(  1,r\right)  }=h_{n}^{(r)}.
\]

The main objective of this study is the evaluation of Euler sums of
generalized hyperharmonic numbers
\[
\zeta_{H^{\left(  p,q\right)  }}\left(  r\right)  =\sum\limits_{n=1}^{\infty
}\dfrac{H_{n}^{\left(  p,q\right)  }}{n^{r}}.
\]
A step towards the solution of this problem is taken in \cite{DiMeCe}.
However, the recurrence used by the authors did not return a closed formula.
Without an available closed formula, they listed only the following few
special cases
\begin{align*}
\zeta_{H^{\left(  p,2\right)  }}\left(  r\right)   &  =\zeta_{H^{\left(
r-1\right)  }}\left(  p\right)  -\zeta_{H^{\left(  r\right)  }}\left(
p-1\right)  +\zeta_{H^{\left(  r\right)  }}\left(  p\right)  ,\\
2\zeta_{H^{\left(  p,3\right)  }}\left(  r\right)   &  =2\zeta_{H^{\left(
r\right)  }}\left(  p\right)  +3\zeta_{H^{\left(  r-1\right)  }}\left(
p\right)  +\zeta_{H^{\left(  r-2\right)  }}\left(  p\right)  -3\zeta
_{H^{\left(  r\right)  }}\left(  p-1\right) \\
&  \quad+\zeta_{H^{\left(  r\right)  }}\left(  p-2\right)  -2\zeta_{H^{\left(
r-1\right)  }}\left(  p-1\right)  .
\end{align*}
Later, G\"{o}ral and Sertba\c{s} \cite{GS} showed that the Euler sums of
generalized hyperharmonic numbers can be evaluated in terms of the Euler sums
of generalized harmonic numbers and special values of the Riemann zeta
function. However, their method does not determine the coefficients
explicitly. This gap is filled in this study. The following recurrence
relation for $H_{n}^{\left(  p,q\right)  }$ depending on the index $q$,
\[
\left(  q-1\right)  H_{n}^{\left(  p,q\right)  }=\left(  n+q-1\right)
H_{n}^{\left(  p,q-1\right)  }-H_{n}^{\left(  p-1,q-1\right)  }
\]
is obtained. Thanks to this recurrence relation, it is managed to obtain a
closed formula for $H_{n}^{\left(  p,q\right)  }$ in terms of $H_{n}^{\left(
p\right)  }$ in Theorem \ref{teo1}. This enables the evaluation of Euler sums
of generalized hyperharmonic numbers in terms of the Euler sums of generalized
harmonic numbers as
\[
\zeta_{H^{\left(  p,q+1\right)  }}\left(  r\right)  =\frac{1}{q!}\sum
_{m=0}^{q}\sum_{k=0}^{m}\left(  -1\right)  ^{k}

\genfrac{[}{]}{0pt}{}{q+1}{m+1}

\binom{m}{k}\zeta_{H^{\left(  p-k\right)  }}\left(  r+k-m\right)  .
\]
A demonstration of this formula is the following example:
\begin{align*}
2\zeta_{H^{\left(  6,4+1\right)  }}\left(  6\right)   &  =-1925\zeta\left(
11\right)  +\left(  175\pi^{2}-\frac{905}{4}-\frac{3937\pi^{8}}{544\,320}
\right)  \zeta\left(  9\right) \\
&  +\left(  \frac{245\pi^{2}}{12}+\frac{35\pi^{4}}{18}+\frac{31\pi^{10}
}{46\,656}\right)  \zeta\left(  7\right)  +\left(  \frac{\pi^{4}}{4}
+\frac{5\pi^{6}}{1134}+\frac{31\pi^{12}}{6123\,600}\right)  \zeta\left(
5\right) \\
&  -\frac{35}{12}\zeta^{2}\left(  5\right)  -\frac{1}{3}\zeta\left(  3\right)
\zeta\left(  5\right)  +\frac{\pi^{6}}{1134}\zeta\left(  3\right)  +\frac
{\pi^{10}}{29\,160}+\frac{1406\pi^{12}}{638\,512\,875}.
\end{align*}
In addition, a counterpart of the reflection formula (\ref{rp}) is obtained in
the following form:
\[
\zeta_{H^{\left(  p,q+1\right)  }}\left(  r\right)  +\zeta_{H^{\left(
r,q+1\right)  }}\left(  p\right)  .
\]
Section two completes with this formula which serves to calculate sums similar
to the foregoing example with less computational cost.

In the last section we further extend our results. In this direction we
establish new and more general identities for the series whose terms are
generalizations of harmonic numbers and reciprocal binomial coefficients. For
instance,
\[
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p,q\right)  }}{\left(  n+m\right)
\binom{n+m+l}{l}}
\]
is evaluated in terms of Riemann zeta values. This leads to several new
evaluation formulas for particular series involving generalized harmonic and
hyperharmonic numbers. We point out special cases of these formulas which
match with several known results in the literature.

\section{Euler sums of generalized hyperharmonic numbers}

In this section we present an evaluation formula for Euler sums $\zeta
_{H^{\left(  p,q\right)  }}\left(  r\right)  $ under certain conditions. To
state and prove our result we need some preliminaries.

Firstly, recall the polylogarithm defined by
\[
Li_{p}\left(  t\right)  =\sum_{k=1}^{\infty}\frac{t^{k}}{k^{p}},\text{
}(\left\vert t\right\vert \leq1\text{ if }p>1,\text{ and }\left\vert
t\right\vert <1\text{ if }p\leq1)\text{.}
\]
The generating function of the numbers $H_{n}^{\left(  p,q\right)  }$ in terms
of the polylogarithm is \cite{DiMeCe}
\begin{equation}
\sum_{n=0}^{\infty}H_{n}^{\left(  p,q\right)  }t^{n}=\frac{Li_{p}\left(
t\right)  }{\left(  1-t\right)  ^{q}},\text{ }\left\vert t\right\vert
<1,\text{ }p,q\in\mathbb{Z}. \label{ghhgf}
\end{equation}

Our first result presents the following reduction formula for $H_{n}^{\left(
p,q\right)  }$.

\begin{lemma}
Let $p$ and $q$ be integers with $q\geq1.$ Reduction relation for
$H_{n}^{\left(  p,q\right)  }$ in the index $q$\ is
\begin{equation}
\left(  q-1\right)  H_{n}^{\left(  p,q\right)  }=\left(  n+q-1\right)
H_{n}^{\left(  p,q-1\right)  }-H_{n}^{\left(  p-1,q-1\right)  }. \label{red}
\end{equation}

\end{lemma}

\begin{proof}
We define the polynomial $H_{n}^{\left(  p,q\right)  }\left(  z\right)  $ as
\[
H_{n}^{\left(  p,q\right)  }\left(  z\right)  =\sum_{k=0}^{n}H_{k}^{\left(
p,q\right)  }z^{k}.
\]
Considering (\ref{ghhgf}), we obtain the ordinary generating function of
$H_{n}^{\left(  p,q\right)  }\left(  z\right)  $ as
\begin{equation}
\sum_{n=0}^{\infty}H_{n}^{\left(  p,q\right)  }\left(  z\right)  t^{n}
=\frac{Li_{p}\left(  zt\right)  }{\left(  1-t\right)  \left(  1-zt\right)
^{q}}. \label{Hgf}
\end{equation}
From (\ref{Hgf}), it can be seen that
\begin{equation}
z\frac{d}{dz}H_{n}^{\left(  p,q\right)  }\left(  z\right)  =H_{n}^{\left(
p-1,q\right)  }\left(  z\right)  +qzH_{n-1}^{\left(  p,q+1\right)  }\left(
z\right)  . \label{tur1}
\end{equation}
On the other hand, we utilize (\ref{0}) twice to find that
\begin{align*}
H_{n}^{\left(  p,q\right)  }  &  =\sum_{k=1}^{n}H_{k}^{\left(  p,q-1\right)
}=\sum_{k=1}^{n}\sum_{j=1}^{k}H_{j}^{\left(  p,q-2\right)  }\\
&  =\left(  n+1\right)  H_{n}^{\left(  p,q-1\right)  }-\sum_{j=1}^{n}
jH_{j}^{\left(  p,q-2\right)  }\\
&  =\left(  n+1\right)  H_{n}^{\left(  p,q-1\right)  }-\left.  \frac{d}
{dz}H_{n}^{\left(  p,q-2\right)  }\left(  z\right)  \right\vert _{z=1},
\end{align*}
or equivalently
\begin{equation}
\left.  \frac{d}{dz}H_{n}^{\left(  p,q-2\right)  }\left(  z\right)
\right\vert _{z=1}=\left(  n+1\right)  H_{n}^{\left(  p,q-1\right)  }
-H_{n}^{\left(  p,q\right)  }\overset{\text{(\ref{0})}}{=}nH_{n}^{\left(
p,q-1\right)  }-H_{n-1}^{\left(  p,q\right)  }. \label{tur2}
\end{equation}
Therefore, (\ref{tur1}) and (\ref{tur2}) yield the desired formula.
\end{proof}

The objective here is to express $H_{n}^{\left(  p,q\right)  }$ in terms of
$H_{n}^{\left(  p\right)  }.$ In \cite{DiMeCe} this relation is listed for at
most $q=4$ due to the complexity of the process. However, the next result
provides a general solution to this problem where the numbers $H_{n}^{\left(
p,q\right)  }$ are expressed in terms of the numbers $H_{n}^{\left(  p\right)
}$ and $

\genfrac{[}{]}{0pt}{}{q}{j}

_{r}$. Here $

\genfrac{[}{]}{0pt}{}{q}{j}

_{r}$ denotes the $r$-Stirling number of the first kind defined by the
"horizontal" generating function \cite{Broder,Ca,RM}
\begin{equation}
\left(  x+r\right)  \left(  x+r+1\right)  \cdots\left(  x+r+q-1\right)
=\sum\limits_{j=0}^{q}

\genfrac{[}{]}{0pt}{}{q}{j}

_{r}x^{j}. \label{rS1}
\end{equation}
The essence of the theorem's proof is based on the relationship between
$r$-Stirling numbers and symmetric polynomials. The $k$th elementary symmetric
polynomial $e_{k}\left(  X_{1},\ldots,X_{q}\right)  $ in variables
$X_{1},\ldots,X_{q}$ is defined by (see for example \cite{Ma})
\begin{align*}
e_{0}\left(  X_{1},\ldots,X_{q}\right)   &  =1,\\
e_{k}\left(  X_{1},\ldots,X_{q}\right)   &  =\sum_{1\leq j_{1}<j_{2}
<\cdots<j_{k}\leq q}\prod\limits_{i=1}^{k}X_{j_{i}},\text{ }1\leq k\leq q,\\
e_{k}\left(  X_{1},\ldots,X_{q}\right)   &  =0,\text{ }k>q,
\end{align*}
and possess the identity
\begin{equation}
\prod\limits_{j=1}^{q}\left(  x-X_{j}\right)  =\sum\limits_{j=0}^{q}\left(
-1\right)  ^{r-j}e_{q-j}\left(  X_{1},\ldots,X_{q}\right)  x^{j}. \label{**}
\end{equation}
The comparison of (\ref{rS1}) with (\ref{**}) obviously leads to the following
relationship \cite[Theorem 4.1]{RM}
\begin{equation}
e_{q-j}\left(  n+1,n+2,\ldots,n+q\right)  =

\genfrac{[}{]}{0pt}{}{q}{j}

_{n+1}. \label{2}
\end{equation}

\begin{theorem}
\label{teo1}Let $p$ and $q$ be integers with $q\geq0.$ Then,
\begin{equation}
q!H_{n}^{\left(  p,q+1\right)  }=\sum_{k=0}^{q}\left(  -1\right)  ^{k}

\genfrac{[}{]}{0pt}{}{q}{k}

_{n+1}H_{n}^{\left(  p-k\right)  }. \label{4.2}
\end{equation}

\end{theorem}

\begin{proof}
We employ (\ref{red}) on the right-hand side of
\[
\left(  q-1\right)  qH_{n}^{\left(  p,q+1\right)  }=\left(  n+q\right)
\left(  q-1\right)  H_{n}^{\left(  p,q\right)  }-\left(  q-1\right)
H_{n}^{\left(  p-1,q\right)  },
\]
and see that
\begin{align*}
\left(  q-1\right)  qH_{n}^{\left(  p,q+1\right)  }  &  =H_{n}^{\left(
p,q-1\right)  }\left\{  \left(  n+q\right)  \left(  n+q-1\right)  \right\} \\
&  \quad-H_{n}^{\left(  p-1,q-1\right)  }\left\{  \left(  n+q\right)  +\left(
n+q-1\right)  \right\}  +H_{n}^{\left(  p-2,q-1\right)  }\\
&  =\sum_{k=0}^{2}\left(  -1\right)  ^{k}e_{2-k}\left(  n+q-1,n+q\right)
H_{n}^{\left(  p-k,q+1-2\right)  }.
\end{align*}
These initial steps suggest that the following equality should hold:
\begin{align}
& \left(  q+1-r\right)  \left(  q+1-\left(  r-1\right)  \right)  \cdots\left(
q-1\right)  qH_{n}^{\left(  p,q+1\right)  }\label{*}\\
& =\sum_{k=0}^{r}\left(  -1\right)  ^{k}e_{r-k}\left(  n+q-\left(  r-1\right)
,n+q-\left(  r-2\right)  ,\ldots,n+q\right)  H_{n}^{\left(  p-k,q+1-r\right)
}. \nonumber
\end{align}
To prove this by induction we show that it is also true for $r+1\leq q.$ We
multiply (\ref{*}) by $\left(  q-r\right)  $ and then use (\ref{red}). Hence
we find that
\begin{align*}
&  \left(  q-r\right)  \left(  q+1-r\right)  \left(  q+1-\left(  r-1\right)
\right)  \cdots\left(  q-1\right)  qH_{n}^{\left(  p,q+1\right)  }\\
&  \ =\sum_{k=0}^{r}\left(  -1\right)  ^{k}e_{r-k}\left(  n+q-\left(
r-1\right)  ,\ldots,n+q\right)  \left(  n+q-r\right)  H_{n}^{\left(
p-k,q-r\right)  }\\
&  \quad+\sum_{k=0}^{r}\left(  -1\right)  ^{k+1}e_{r-k}\left(  n+q-\left(
r-1\right)  ,\ldots,n+q\right)  H_{n}^{\left(  p-k-1,q-r\right)  }\\
&  \ =e_{r}\left(  n+q-\left(  r-1\right)  ,\ldots,n+q\right)  \left(
n+q-r\right)  H_{n}^{\left(  p,q-r\right)  }\\
&  \quad+\sum_{k=1}^{r}\left(  -1\right)  ^{k}H_{n}^{\left(  p-k,q-r\right)
}\left\{  \left(  n+q-r\right)  e_{r-k}\left(  n+q-\left(  r-1\right)
,\ldots,n+q\right)  \right. \\
&  \qquad\qquad\qquad\qquad\qquad\qquad\qquad\left.  +e_{r+1-k}\left(
n+q-\left(  r-1\right)  ,\ldots,n+q\right)  \right\} \\
&  \quad+\left(  -1\right)  ^{r+1}e_{0}\left(  n+q-r,\ldots,n+q\right)
H_{n}^{\left(  p-\left(  r+1\right)  ,q-r\right)  }\\
&  \ =\sum_{k=0}^{r+1}\left(  -1\right)  ^{k}e_{r+1-k}\left(  n+q-r,\ldots
,n+q\right)  H_{n}^{\left(  p-k,q+1-\left(  r+1\right)  \right)  }.
\end{align*}
The case $r=q$ in (\ref{*}) gives
\[
q!H_{n}^{\left(  p,q+1\right)  }=\sum_{k=0}^{q}\left(  -1\right)  ^{k}
e_{q-k}\left(  n+1,\ldots,n+q\right)  H_{n}^{\left(  p-k\right)  },
\]
which combines with (\ref{2}) to give the statement.
\end{proof}

Now, we are ready to state and prove our evaluation formula for $\zeta
_{H^{\left(  p,q\right)  }}\left(  r\right)  .$ Thanks to this formula the
evaluation of Euler sums of generalized hyperharmonic numbers reduces to the
evaluation of Euler sums of generalized harmonic numbers.

\begin{theorem}
\label{GHHES}For $p,q\geq1$ and $r>q+1,$ we have
\[
\zeta_{H^{\left(  p,q+1\right)  }}\left(  r\right)  =\frac{1}{q!}\sum
_{m=0}^{q}\sum_{k=0}^{m}\left(  -1\right)  ^{k}

\genfrac{[}{]}{0pt}{}{q+1}{m+1}

\binom{m}{k}\zeta_{H^{\left(  p-k\right)  }}\left(  r+k-m\right)  .
\]

\end{theorem}

\begin{proof}
From (\ref{4.2}) and the following identity \cite[p. 1661]{NR}
\[

\genfrac{[}{]}{0pt}{}{n}{k}

_{r+1}=\sum_{m=k}^{n}

\genfrac{[}{]}{0pt}{}{n+1}{m+1}

\binom{m}{k}r^{m-k},
\]
we have
\begin{align*}
H_{n}^{\left(  p,q+1\right)  }  &  =\frac{1}{q!}\sum_{k=0}^{q}\left(
-1\right)  ^{k}

\genfrac{[}{]}{0pt}{}{q}{k}

_{n+1}H_{n}^{\left(  p-k\right)  }\\
&  =\frac{1}{q!}\sum_{k=0}^{q}\sum_{m=k}^{q}\left(  -1\right)  ^{k}

\genfrac{[}{]}{0pt}{}{q+1}{m+1}

\binom{m}{k}n^{m-k}H_{n}^{\left(  p-k\right)  }.
\end{align*}
Multiplying both sides with $n^{-r}$ and summing over $n$ from $1$ to $\infty
$, we deduce the desired result.
\end{proof}

As mentioned introductory the sums $\zeta_{H^{\left(  p,q\right)  }}\left(
r\right)  $ were listed up to $q=3$ in \cite{DiMeCe}. With the help of Theorem
\ref{GHHES} these sums can be evaluated for further choices of $q.$ For
instance for $q=4$ one can obtain:
\begin{align*}
\zeta_{H^{\left(  p,4\right)  }}\left(  r\right)   &  =\zeta_{H^{\left(
p\right)  }}\left(  r\right)  +\frac{11}{6}\zeta_{H^{\left(  p\right)  }
}\left(  r-1\right)  +\zeta_{H^{\left(  p\right)  }}\left(  r-2\right)
+\frac{1}{6}\zeta_{H^{\left(  p\right)  }}\left(  r-3\right) \\
&  \quad-\frac{11}{6}\zeta_{H^{\left(  p-1\right)  }}\left(  r\right)
-2\zeta_{H^{\left(  p-1\right)  }}\left(  r-1\right)  -\frac{1}{2}
\zeta_{H^{\left(  p-1\right)  }}\left(  r-2\right)  +\zeta_{H^{\left(
p-2\right)  }}\left(  r\right) \\
&  \quad+\frac{1}{2}\zeta_{H^{\left(  p-2\right)  }}\left(  r-1\right)
-\frac{1}{6}\zeta_{H^{\left(  p-3\right)  }}\left(  r\right)  .
\end{align*}
Hence, with the use of some values of $\zeta_{H^{\left(  p\right)  }}\left(
r\right)  $ listed in forthcoming Remark \ref{tablo}, a few concrete
expressions of $\zeta_{H^{\left(  p,4\right)  }}\left(  r\right)  $ are:\
\begin{align*}
\bullet\text{ }\zeta_{H^{\left(  1,4\right)  }}\left(  5\right)   &
=\frac{11}{2}\zeta\left(  5\right)  -\left(  1-\frac{11}{36}\pi^{2}\right)
\zeta\left(  3\right)  -\frac{1}{2}\left(  \zeta\left(  3\right)  \right)
^{2}-\frac{11}{216}\pi^{2}-\frac{\pi^{4}}{810}+\frac{\pi^{6}}{540},\\
\bullet\text{ }\zeta_{H^{\left(  2,4\right)  }}\left(  5\right)   &
=-10\zeta\left(  7\right)  +\left(  \frac{5}{6}\pi^{2}-\frac{21}{2}\right)
\zeta\left(  5\right)  +\left(  \frac{\pi^{4}}{45}+\frac{5}{6}\pi^{2}+\frac
{5}{12}\right)  \zeta\left(  3\right) \\
&  \quad+\frac{11}{4}\left(  \zeta\left(  3\right)  \right)  ^{2}+\frac
{7\pi^{4}}{1080}-\frac{55\pi^{6}}{13608},\\
\bullet\text{ }\zeta_{H^{\left(  3,4\right)  }}\left(  5\right)   &
=\zeta_{H^{\left(  3\right)  }}\left(  5\right)  +\frac{154}{3}\zeta\left(
7\right)  +\left(  \frac{14}{3}-\frac{55}{12}\pi^{2}\right)  \zeta\left(
5\right)  -2\left(  \zeta\left(  3\right)  \right)  ^{2}\\
&  \quad-\left(  \frac{7\pi^{2}}{18}+\frac{11\pi^{4}}{270}\right)
\zeta\left(  3\right)  -\frac{\pi^{4}}{540}+\frac{\pi^{6}}{324},\\
\bullet\text{ }\zeta_{H^{\left(  4,4\right)  }}\left(  5\right)   &
=-\frac{11}{6}\zeta_{H^{\left(  3\right)  }}\left(  5\right)  -\frac{125}
{2}\zeta\left(  9\right)  +\left(  \frac{35}{6}\pi^{2}-63\right)  \zeta\left(
7\right) \\
&  \quad+\left(  \frac{35}{6}\pi^{2}+\frac{\pi^{4}}{18}\right)  \zeta\left(
5\right)  +\frac{\pi^{4}}{30}\zeta\left(  3\right)  -\frac{\pi^{6}}
{1944}+\frac{143\pi^{8}}{680400},\\
\bullet\text{ }\zeta_{H^{\left(  5,4\right)  }}\left(  5\right)   &
=231\zeta\left(  9\right)  +\left(  21-\frac{385}{18}\pi^{2}\right)
\zeta\left(  7\right)  -\left(  \frac{11}{60}\pi^{4}+\frac{23}{12}\pi
^{2}\right)  \zeta\left(  5\right) \\
&  \quad+\frac{1}{2}\left(  \zeta\left(  5\right)  \right)  ^{2}+\zeta\left(
3\right)  \zeta\left(  5\right)  -\frac{7\pi^{4}}{540}\zeta\left(  3\right)
-\frac{\pi^{8}}{8100}+\frac{\pi^{10}}{187110}.
\end{align*}
\textbf{\ }

The following corollary gives the reflection formula for Euler sums of
generalized hyperharmonic numbers. Combined with (\ref{GES}), this corollary
shows that $\zeta_{H^{\left(  p,q+1\right)  }}\left(  r\right)  +\zeta
_{H^{\left(  r,q+1\right)  }}\left(  p\right)  $ can be written as a
combination of Riemann zeta values. In this way, particular Euler sums of type
$\zeta_{H^{\left(  p,q\right)  }}\left(  p\right)  $ can be evaluated with
less computation.

\begin{corollary}
\label{cor1}Let $p>q+1,$ $r>q+1$ and $p+r$ be even. Then
\begin{align*}
&  \zeta_{H^{\left(  p,q+1\right)  }}\left(  r\right)  +\zeta_{H^{\left(
r,q+1\right)  }}\left(  p\right) \\
&  =\zeta\left(  p+r\right)  +\frac{2}{q!}\sum_{\substack{m=0\\m\text{ odd}
}}^{q}\sum_{k=0}^{m}\left(  -1\right)  ^{k}

\genfrac{[}{]}{0pt}{}{q+1}{m+1}

\binom{m}{k}\zeta_{H^{\left(  p-k\right)  }}\left(  r+k-m\right) \\
&  +\frac{1}{q!}\sum_{m=0}^{q}\sum_{k=0}^{m}\left(  -1\right)  ^{m+k}

\genfrac{[}{]}{0pt}{}{q+1}{m+1}

\binom{m}{k}\zeta\left(  p-k\right)  \zeta\left(  r+k-m\right)  .
\end{align*}

\end{corollary}

\begin{proof}
Let $\left(  p+r\right)  $ be even. It is obvious from Theorem \ref{GHHES}
that
\begin{align*}
&  \zeta_{H^{\left(  p,q+1\right)  }}\left(  r\right)  +\zeta_{H^{\left(
r,q+1\right)  }}\left(  p\right) \\
&  =\frac{1}{q!}\sum_{m=0}^{q}\sum_{k=0}^{m}\left(  -1\right)  ^{k}

\genfrac{[}{]}{0pt}{}{q+1}{m+1}

\binom{m}{k}\left\{  \sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p-k\right)  }
}{n^{r+k-m}}+\left(  -1\right)  ^{m}\sum_{n=1}^{\infty}\frac{H_{n}^{\left(
r+k-m\right)  }}{n^{p-k}}\right\}  .
\end{align*}
We write the right-hand side as
\begin{align*}
&  \sum_{\substack{m=0\\m\text{ odd}}}^{q}\sum_{k=0}^{m}\left(  -1\right)
^{k}

\genfrac{[}{]}{0pt}{}{q+1}{m+1}

\binom{m}{k}\left\{  \sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p-k\right)  }
}{n^{r+k-m}}-\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  r+k-m\right)  }}{n^{p-k}
}\right\} \\
&  +\sum_{0\leq m\leq q/2}\sum_{k=0}^{2m}\left(  -1\right)  ^{k}

\genfrac{[}{]}{0pt}{}{q+1}{2m+1}

\binom{2m}{k}\left\{  \sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p-k\right)  }
}{n^{r+k-2m}}+\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  r+k-2m\right)  }
}{n^{p-k}}\right\}  .
\end{align*}
By the reflection formula (\ref{rp}) we have
\[
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p-k\right)  }}{n^{r+k-2m}}+\sum
_{n=1}^{\infty}\frac{H_{n}^{\left(  r+k-2m\right)  }}{n^{p-k}}=\zeta\left(
p+r-2m\right)  +\zeta\left(  p-k\right)  \zeta\left(  r+k-2m\right)  .
\]
Moreover, for odd $m,$ it can be seen from (\ref{GES})\ that
\begin{align*}
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p-k\right)  }}{n^{r+k-m}}-\sum
_{n=1}^{\infty}\frac{H_{n}^{\left(  r+k-m\right)  }}{n^{p-k}}  &
=2\zeta_{H^{\left(  p-k\right)  }}\left(  r+k-m\right)  -\zeta\left(
p+r-m\right) \\
&  \quad-\zeta\left(  p-k\right)  \zeta\left(  r+k-m\right)  .
\end{align*}
Hence, we obtain the desired equation.
\end{proof}

\begin{remark}
\label{tablo}For interested readers we would like to list some values of
$\zeta_{H^{\left(  p\right)  }}\left(  r\right)  $, used in the evaluations of
$\zeta_{H^{\left(  p,4\right)  }}\left(  5\right)  ,$ $1\leq p\leq5,$ and
$\zeta_{H^{\left(  6,5\right)  }}\left(  6\right)  .$ These are calculated
with the help of (\ref{ES}), (\ref{GES}) and (\ref{rp}).

\noindent\hspace{-0.09in}
\begin{tabular}
[c]{ll}
$\bullet\text{ }\zeta_{H^{\left(  1\right)  }}\left(  2\right)  =2\zeta\left(
3\right)  ,$ & $\bullet\text{ }\zeta_{H^{\left(  3\right)  }}\left(  6\right)
=\frac{85}{2}\zeta\left(  9\right)  -\frac{7\pi^{2}}{2}\zeta\left(  7\right)
-\frac{\pi^{4}}{15}\zeta\left(  5\right)  ,$\medskip\\
$\bullet\text{ }\zeta_{H^{\left(  1\right)  }}\left(  3\right)  =\frac{\pi
^{4}}{72},$ & $\bullet\text{ }\zeta_{H^{\left(  4\right)  }}\left(  2\right)
=-\zeta^{2}\left(  3\right)  +\frac{37\pi^{6}}{11340},$\medskip\\
$\bullet\text{ }\zeta_{H^{\left(  1\right)  }}\left(  4\right)  =3\zeta\left(
5\right)  -\frac{\pi^{2}}{6}\zeta\left(  3\right)  ,$ & $\bullet\text{ }
\zeta_{H^{\left(  4\right)  }}\left(  3\right)  =-17\zeta\left(  7\right)
+\frac{5\pi^{2}}{3}\zeta\left(  5\right)  +\frac{\pi^{4}}{90}\zeta\left(
3\right)  ,$\medskip\\
$\bullet\text{ }\zeta_{H^{\left(  1\right)  }}\left(  5\right)  =-\frac{1}
{2}\zeta^{2}\left(  3\right)  +\frac{\pi^{6}}{540},$ & $\bullet\text{ }
\zeta_{H^{\left(  4\right)  }}\left(  4\right)  =\frac{13\pi^{8}}{113\,400}
,$\medskip\\
$\bullet\text{ }\zeta_{H^{\left(  2\right)  }}\left(  2\right)  =\frac
{7\pi^{4}}{360},$ & $\bullet\text{ }\zeta_{H^{\left(  4\right)  }}\left(
5\right)  =-\frac{125}{2}\zeta\left(  9\right)  +\frac{35\pi^{2}}{6}
\zeta\left(  7\right)  +\frac{\pi^{4}}{18}\zeta\left(  5\right)  ,$\medskip\\
$\bullet\text{ }\zeta_{H^{\left(  2\right)  }}\left(  3\right)  =-\frac{9}
{2}\zeta\left(  5\right)  +\frac{\pi^{2}}{2}\zeta\left(  3\right)  ,$ &
$\bullet\text{ }\zeta_{H^{\left(  5\right)  }}\left(  2\right)  =11\zeta
\left(  7\right)  -\frac{2\pi^{2}}{3}\zeta\left(  5\right)  -\frac{\pi^{4}
}{45}\zeta\left(  3\right)  ,$\medskip\\
$\bullet\text{ }\zeta_{H^{\left(  2\right)  }}\left(  4\right)  =\zeta
^{2}\left(  3\right)  -\frac{\pi^{6}}{2835},$ & $\bullet\text{ }
\zeta_{H^{\left(  5\right)  }}\left(  4\right)  =\frac{127}{2}\zeta\left(
9\right)  -\frac{35\pi^{2}}{6}\zeta\left(  7\right)  -\frac{2\pi^{4}}{45}
\zeta\left(  5\right)  ,$\medskip\\
$\bullet\text{ }\zeta_{H^{\left(  2\right)  }}\left(  5\right)  =-10\zeta
\left(  7\right)  +\frac{5\pi^{2}}{6}\zeta\left(  5\right)  $ & $\bullet\text{
}\zeta_{H^{\left(  5\right)  }}\left(  5\right)  =\frac{1}{2}\zeta^{2}\left(
5\right)  +\frac{\pi^{10}}{187110},$\medskip\\
$\qquad\qquad\qquad+\frac{\pi^{4}}{45}\zeta\left(  3\right)  ,$ &
$\bullet\text{ }\zeta_{H^{\left(  5\right)  }}\left(  6\right)  =\frac{463}
{2}\zeta\left(  11\right)  -21\pi^{2}\zeta\left(  9\right)  -\frac{7}{30}
\pi^{4}\zeta\left(  7\right)  ,$\medskip\\
$\bullet\text{ }\zeta_{H^{\left(  3\right)  }}\left(  2\right)  =\frac{11}
{2}\zeta\left(  5\right)  -\frac{\pi^{2}}{3}\zeta\left(  3\right)  ,$ &
$\bullet\text{ }\zeta_{H^{\left(  6\right)  }}\left(  3\right)  =\frac
{7\pi^{2}}{2}\zeta\left(  7\right)  -\frac{83}{2}\zeta\left(  9\right)
+\frac{\pi^{4}}{15}\zeta\left(  5\right)  $\\
$\bullet\text{ }\zeta_{H^{\left(  3\right)  }}\left(  3\right)  =\frac{1}
{2}\zeta^{2}\left(  3\right)  +\frac{\pi^{6}}{1890},$ & $\qquad\qquad
\qquad+\frac{\pi^{6}}{945}\zeta\left(  3\right)  ,$\medskip\\
$\bullet\text{ }\zeta_{H^{\left(  3\right)  }}\left(  4\right)  =18\zeta
\left(  7\right)  -\frac{5\pi^{2}}{3}\zeta\left(  5\right)  ,$ &
$\bullet\text{ }\zeta_{H^{\left(  6\right)  }}\left(  5\right)  =21\pi
^{2}\zeta\left(  9\right)  -\frac{461}{2}\zeta\left(  11\right)  +\frac
{7\pi^{4}}{30}\zeta\left(  7\right)  $\\
& $\qquad\qquad\qquad+\frac{\pi^{6}}{945}\zeta\left(  5\right)  .$
\end{tabular}

\end{remark}

\section{\textbf{Series involving harmonic numbers and reciprocal binomial
coefficients}}

In this section, we introduce evaluation formulas for some series involving
the harmonic numbers and their generalizations.

\begin{theorem}
\label{pro1}Let $p\geq1$ and $q,$ $l\geq0$ be integers with $l\geq q$. For
$m\geq1$,
\begin{align}
&  \sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p,q\right)  }}{\left(  n+m\right)
\binom{n+m+l}{l}}\label{L5}\\
&  \ =\sum_{j=0}^{l-q}\binom{l-q}{j}\left\{  \frac{\left(  -1\right)
^{j+p-1}H_{m+j}}{\left(  m+j\right)  ^{p}}+\sum_{k=1}^{p-1}\frac{\left(
-1\right)  ^{j+k-1}}{\left(  m+j\right)  ^{k}}\zeta\left(  p+1-k\right)
\right\}  \nonumber
\end{align}
and
\begin{align}
&  \sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p,q\right)  }}{n\binom{n+l}{l}
}\label{L6}\\
&  \ =\zeta\left(  p+1\right)  -\sum_{j=1}^{l-q}\binom{l-q}{j}\left\{
\frac{\left(  -1\right)  ^{j+p}H_{j}}{j^{p}}+\sum_{k=1}^{p-1}\frac{\left(
-1\right)  ^{j+k}}{j^{k}}\zeta\left(  p+1-k\right)  \right\}  . \nonumber
\end{align}

\end{theorem}

\begin{proof}
Using the formula (see \cite[p.909]{JZ})
\[

{\displaystyle\int\limits_{0}^{1}}

t^{n+m-1}\left(  1-t\right)  ^{l}dt=\frac{1}{\left(  n+m\right)  \binom
{n+m+l}{l}},
\]
we can write
\[
\frac{H_{n}^{\left(  p,q\right)  }}{\left(  n+m\right)  \binom{n+m+l}{l}}=

{\displaystyle\int\limits_{0}^{1}}

H_{n}^{\left(  p,q\right)  }t^{n+m-1}\left(  1-t\right)  ^{l}dt.
\]
With the help of (\ref{ghhgf}), we get
\begin{align*}
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p,q\right)  }}{\left(  n+m\right)
\binom{n+m+l}{l}}  &  =

{\displaystyle\int\limits_{0}^{1}}

t^{m-1}\left(  1-t\right)  ^{l-q}Li_{p}\left(  t\right)  dt\\
&  =\sum_{j=0}^{l-q}\binom{l-q}{j}\left(  -1\right)  ^{j}\sum_{n=1}^{\infty
}\frac{1}{n^{p}\left(  n+m+j\right)  }.
\end{align*}
Then (\ref{L5}) follows from (\ref{mu}). If $m=0,$ then we have
\[
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p,q\right)  }}{n\binom{n+l}{l}}
=\zeta\left(  p+1\right)  +\sum_{j=1}^{l-q}\binom{l-q}{j}\left(  -1\right)
^{j}\sum_{n=1}^{\infty}\frac{1}{n^{p}\left(  n+j\right)  },
\]
which is equivalent to (\ref{L6}).
\end{proof}

Now, we deal with some special cases of Theorem \ref{pro1}. Setting $q=l$
gives
\begin{equation}
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p,q\right)  }}{\left(  n+m\right)
\binom{n+m+q}{q}}=\frac{\left(  -1\right)  ^{p-1}H_{m}}{m^{p}}+\sum
_{i=1}^{p-1}\frac{\left(  -1\right)  ^{i-1}}{m^{i}}\zeta\left(  p+1-i\right)
\label{17}
\end{equation}
and
\begin{equation}
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p,q\right)  }}{n\binom{n+q}{q}}
=\zeta\left(  p+1\right)  . \label{18}
\end{equation}
Note that the variable $q$ does not appear in the right-hand sides and all
these series converge very slowly.

For $p=1,$\textbf{ }(\ref{17}) and (\ref{18}) give Proposition 5 and
Proposition 6 in \cite{DB}
\[
\sum_{n=1}^{\infty}\frac{h_{n}^{\left(  q\right)  }}{n\binom{n+q}{q}}=\frac
{1}{6}\pi^{2} \text{ \ \ and \ \ }\sum_{n=1}^{\infty}\frac{h_{n-1}^{\left(
q\right)  }}{n\binom{n+q}{q}}=1,
\]
respectively. (\ref{17}) also yields \cite[Eq. (2.30)]{XZZ} for $q=1$.
Additionally, when $p=1$ in Theorem \ref{pro1}, we reach that
\[
\sum_{n=1}^{\infty}\frac{h_{n}^{\left(  q\right)  }}{\left(  n+m\right)
\binom{n+m+l}{l}}=\sum_{j=0}^{l-q}\binom{l-q}{j}\left(  -1\right)  ^{j}
\frac{H_{m+j}}{m+j}
\]
and
\[
\sum_{n=1}^{\infty}\frac{h_{n}^{\left(  q\right)  }}{n\binom{n+l}{l}}=\frac
{1}{6}\pi^{2}+\sum_{j=1}^{l-q}\left(  -1\right)  ^{j}\binom{l-q}{j}\frac
{H_{j}}{j}.
\]
Now employing \cite[Eq.(18)]{CHU}
\[
\sum_{k=0}^{m}\left(  -1\right)  ^{k}\binom{m}{k}\frac{H_{n+k}}{n+k}
=\frac{H_{n+m}-H_{m}}{n\binom{n+m}{m}}
\]
and \cite[Eq. (9.4b)]{B2018}
\[
\sum_{j=1}^{n}\left(  -1\right)  ^{j+1}\binom{n}{j}\frac{H_{j}}{j}
=H_{n}^{\left(  2\right)  }
\]
gives the following closed forms for series involving hyperharmonic numbers
with reciprocal binomial coefficients.

\begin{corollary}
Let $q,l\geq0$ be integers with $l\geq q$. For all integers $m\geq1$
\[
\sum_{n=1}^{\infty}\frac{h_{n}^{\left(  q\right)  }}{\left(  n+m\right)
\binom{n+m+l}{l}}=\frac{H_{m+l-q}-H_{l-q}}{m\binom{n+m+l-q}{l-q}}
\]
and
\[
\sum_{n=1}^{\infty}\frac{h_{n}^{\left(  q\right)  }}{n\binom{n+l}{l}}=\frac
{1}{6}\pi^{2}-H_{l-q}^{\left(  2\right)  }.
\]

\end{corollary}

For $p=q=1,$ Theorem \ref{pro1} gives \cite[Eq. (2.31)]{SS}
\[
\sum_{n=1}^{\infty}\frac{H_{n}}{\left(  n+m\right)  \binom{n+m+q}{q}}
=\frac{H_{n+q-1}-H_{q-1}}{n\binom{n+q-1}{q-1}}
\]
and \cite{SC}
\[
\sum_{n=1}^{\infty}\frac{H_{n}}{n\binom{n+l}{l}}=\frac{1}{6}\pi^{2}
-H_{l-1}^{\left(  2\right)  }.
\]
For $q=1,$ (\ref{L6}) becomes
\begin{align}
&  \sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p\right)  }}{n\binom{n+l}{l}
}\label{sofo}\\
&  \ =\zeta\left(  p+1\right)  -\sum_{j=1}^{l-1}\binom{l-1}{j}\left\{
\frac{\left(  -1\right)  ^{j+p}H_{j}}{j^{p}}+\sum_{k=1}^{p-1}\frac{\left(
-1\right)  ^{j+k}}{j^{k}}\zeta\left(  p+1-k\right)  \right\}  , \nonumber
\end{align}
which is also given by Sofo \cite[Theorem 2.2]{S3} in a slightly different form.

Setting $q=1$ in (\ref{L5}) yields the following corollary involving
generalized harmonic numbers.

\begin{corollary}
For all integers $m,p,l\geq1,$
\begin{align}
&  \sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p\right)  }}{\left(  n+m\right)
\binom{n+m+l}{l}}\label{26}\\
&  =\sum_{j=0}^{l-1}\binom{l-1}{j}\left(  -1\right)  ^{j}\left\{
\frac{\left(  -1\right)  ^{p-1}H_{m+j}}{\left(  m+j\right)  ^{p}}+\sum
_{k=1}^{p-1}\frac{\left(  -1\right)  ^{k-1}}{\left(  m+j\right)  ^{k}}
\zeta\left(  p+1-k\right)  \right\}  . \nonumber
\end{align}

\end{corollary}

The following particular cases can be deduced setting $p=2$ and $p=3$:
\begin{align}
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  2\right)  }}{\left(  n+m\right)
\binom{n+m+l}{l}}  &  =\frac{\pi^{2}}{6m\binom{m+l-1}{l-1}}-\sum_{j=0}
^{l-1}\left(  -1\right)  ^{j}\binom{l-1}{j}\frac{H_{m+j}}{\left(  m+j\right)
^{2}},\label{L2}\\
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  3\right)  }}{\left(  n+m\right)
\binom{n+m+l}{l}}  &  =\sum_{j=0}^{l-1}\left(  -1\right)  ^{j}\binom{l-1}
{j}\frac{H_{m+j}}{\left(  m+j\right)  ^{3}}\nonumber\\
&  +\frac{1}{m\binom{m+l-1}{l-1}} \left\{  \zeta\left(  3\right)  -\frac
{\pi^{2}}{6}\left(  H_{m+l-1}-H_{m-1}\right)  \right\}  .\nonumber
\end{align}

It is worth noting that the special case with choices $m=6$ and $l=3$ in
(\ref{L2}) is recorded in \cite[Remark 1]{S5} despite a misprint. The case is
below
\[
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  2\right)  }}{\left(  n+6\right)
\binom{n+9}{3}}=\frac{1}{168}\zeta\left(  2\right)  -\frac{37073}{7902720}.
\]

For our final results, we deal with the special case $q=2$ of Theorem
\ref{pro1}. By aid of (\ref{4.2}), $

\genfrac{[}{]}{0pt}{}{q}{0}

_{r}=r\left(  r+1\right)  \cdots\left(  r+q-1\right)  $ and $

\genfrac{[}{]}{0pt}{}{q}{q}

_{r}=1,$ we have
\[
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p\right)  }}{\binom{n+l}{l}}
=\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p,2\right)  }}{n\binom{n+l}{l}}
+\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p-1\right)  }}{n\binom{n+l}{l}}
-\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p\right)  }}{n\binom{n+l}{l}},
\]
where $l$ is any integer greater than $1$. From (\ref{L6}), (\ref{sofo}) and
some arrangements we obtain
\[
\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p\right)  }}{\binom{n+l}{l}}
=\zeta\left(  p\right)  +\sum_{j=1}^{l-1}\left(  -1\right)  ^{j}\left\{
\binom{l-1}{j}\mu\left(  p-1,j\right)  -\binom{l-2}{j-1}\mu\left(  p,j\right)
\right\}  ,
\]
where $\mu\left(  p,j\right)  $ is given in (\ref{mu}). A slightly different
form of the equation above is given in \cite[Theorem 2.1]{S3}. Similarly
\begin{align*}
\sum_{n=1}^{\infty}\frac{nH_{n}^{\left(  p\right)  }}{\left(  n+m\right)
\binom{n+m+l}{l}}  &  =\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p,2\right)  }
}{\left(  n+m\right)  \binom{n+m+l}{l}}+\sum_{n=1}^{\infty}\frac
{H_{n}^{\left(  p-1\right)  }}{\left(  n+m\right)  \binom{n+m+l}{l}}\\
&  -\sum_{n=1}^{\infty}\frac{H_{n}^{\left(  p\right)  }}{\left(  n+m\right)
\binom{n+m+l}{l}}.
\end{align*}
Then exploiting (\ref{L6}) and (\ref{26}) in the last equation yields the
following corollary.

\begin{corollary}
Let $l>1$ be an integer. Then \textbf{ }
\begin{align*}
&  \sum_{n=1}^{\infty}\frac{nH_{n}^{\left(  p\right)  }}{\left(  n+m\right)
\binom{n+m+l}{l}}\\
&  \ =\mu\left(  p-1,m\right)  +\sum_{j=1}^{l-1}\left(  -1\right)
^{j}\left\{  \binom{l-1}{j}\mu\left(  p-1,m+j\right)  -\binom{l-2}{j-1}
\mu\left(  p,m+j\right)  \right\}  ,
\end{align*}
where $\mu\left(  p,j\right)  $ is given in (\ref{mu}).
\end{corollary}

For\textbf{ }$l=2$\textbf{ }this formula can be read as
\begin{align*}
&  \sum_{n=1}^{\infty}\frac{nH_{n}^{\left(  p\right)  }}{\left(  n+m\right)
\left(  n+m+1\right)  \left(  n+m+2\right)  }\\
&  =\left(  -1\right)  ^{p+1}\frac{m+2}{2\left(  m+1\right)  ^{p}}
H_{m+1}+\left(  -1\right)  ^{p}\frac{H_{m}}{2m^{p-1}}\\
&  \quad+\sum_{k=1}^{p-2}\left(  -1\right)  ^{k-1}\left\{  \frac{1}{2m^{k}
}-\frac{1}{2\left(  m+1\right)  ^{k}}\right\}  \zeta\left(  p-k\right)
+\sum_{k=1}^{p-1}\frac{\left(  -1\right)  ^{k-1}}{2\left(  m+1\right)  ^{k}
}\zeta\left(  p+1-k\right)  .
\end{align*}
The first few cases of this formula are listed below:
\begin{align*}
\sum_{n=1}^{\infty}\frac{nH_{n}}{\left(  n+m\right)  \left(  n+m+1\right)
\left(  n+m+2\right)  }  &  =\frac{1}{2\left(  m+1\right)  }\left(
H_{m+1}+1\right)  ,\\
\sum_{n=1}^{\infty}\frac{nH_{n}^{\left(  2\right)  }}{\left(  n+m\right)
\left(  n+m+1\right)  \left(  n+m+2\right)  }  &  =\frac{\pi^{2}}{12\left(
m+1\right)  }-\frac{\left(  m+2\right)  }{2\left(  m+1\right)  ^{2}}
H_{m+1}+\frac{1}{2m}H_{m},\\
\sum_{n=1}^{\infty}\frac{nH_{n}^{\left(  3\right)  }}{\left(  n+m\right)
\left(  n+m+1\right)  \left(  n+m+2\right)  }  &  =\frac{\zeta\left(
3\right)  }{2\left(  m+1\right)  }+\frac{\pi^{2}}{12m\left(  m+1\right)  ^{2}
}\\
&  \ \ \ +\frac{m+2}{2\left(  m+1\right)  ^{3}}H_{m+1}-\frac{1}{2m^{2}}H_{m}.
\end{align*}


TITLE: Derivative of limited convolution
QUESTION [2 upvotes]: If I take infinite convolution, I know that its derivative is the convolution between one term and other term's derivative.
However, I am interested on derivative of
$$\dfrac {d}{dt}\int_0^t f (t) g (t-a) da$$
i.e. in terms of limited convolution.
Many thanks in advance!
======
EDIT: Please review comments on answer

REPLY [2 votes]: I think @tossimmar 's answer gives some preliminary remarks, but I think this answer will provide the asker with more of what he/she is looking for. First of all, Leibniz's integral rule, in most general form, is
$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\int_{a(t)}^{b(t)}f(t,s)\mathrm{d}s\right)=f\left(t,b(t)\right)b'(t)-f\left(t,a(t)\right)a'(t)+\int_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(t,s)\mathrm{d}s$$
In our case, this gives us
$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\int_{0}^{t}f(t)g(t-s)\mathrm{d}s\right)=f(t)g(0)+\int_0^tf'(t)g(t-s)\mathrm{d}s+\int_0^tf(t)g'(t-s)\mathrm{d}s$$
Of course, we could write this in terms of this partial convolution which I'll denote with an tilded asterisk:
$$\mathrm{D}(f~\tilde{*}~g)=g(0)\cdot f+f'~\tilde{*}~g+f~\tilde{*}~g'$$


TITLE: What happens to atoms inside a black hole?
QUESTION [13 upvotes]: Black holes have very high gravitational force that tends to crush everything. So as we know atoms in a molecule have inter-atomic spacing between them and further electrons also revolve at a certain distance to the nucleus and inside the nucleus there is also some empty space. So does the strong force of a black hole crush them all the way, leaving no space in between them? And, if that happens, then it should give rise to a completely new species of atoms and molecules, right?

REPLY [12 votes]: Even before you have a black hole, you can have a neutron star.  Here, the gravitational force is enough to cause atoms to collapse, as protons absorb electrons in reverse beta decay, making the neutron star essentially a giant nucleus without protons.  If you add mass beyond the maximum stable size of a neutron star, you get a neutron star supernova, and then collapse to a black hole (you can also form black holes directly from a star, of course).
What happens to the atoms on their way to the singularity is not well understood by current research, and the question is in principle unanswerable right now when the volume of the solid becomes smaller than one cubic planck length, as we do not have a universally accepted theory of physics with concrete predictions that is applicable below this length scale.


TITLE: Show that $\{ \frac 12 (1+\cos (x-m\pi))\}$ forms a partition of unity
QUESTION [2 upvotes]: Exercise.$1$(Munkres- pg.$140$) :  let $f:\mathbb R \to \mathbb R$ be defined by the equation
$$f(x)= \left\{ \begin{array}{lcc}
             \frac{(1+\cos x)}{2} &   \text{for }   -\pi \leq x \leq \pi \\
             \\ 0 &    \text{otherwise}.
             \end{array}
   \right.$$
Then $f$ is of class $C^1$. For each integer  $m \geq 0$, set $\phi_{2m+1}(x)=f(x-m\pi)$. For each integer $m \geq1$, set $\phi_{2m}(x)=f(x+m\pi)$. Then the collection {$\phi_i$} forms a partition of unity on $\mathbb R$. The Support $S_{i}$ of $\phi_{i}$ is a closed interval of the form $[k\pi,(k+2)\pi]$,which is compact,and each point of $\mathbb R$ has neighborhood  that intersects at most three of the sets $S_{i}$. We leave it to you to check that $\sum \phi_{i}(x)=1$. thus {$\phi_{i}$} is a partition of unity on $\mathbb R$.
I did the following: if $\phi_{2m+1}(x)=f(x-m\pi)$ then $-\pi\leq x-m\pi\leq \pi$ imply $\pi(m-1)\leq x\leq \pi(m+1)$, the same way for $\phi_{2m}$ we have $-\pi(m+1)\leq x\leq -\pi(m+1)$. The support is adherent set of point of domain where $\phi_i$ is different of zero so I put
$$\operatorname{supp} \phi_i=\overline{[\pi(m-1),\pi(m+1)]\cup [-\pi(m+1),-\pi(m+1)]},$$
but the book wrote
$$\operatorname{supp} \phi_i=[k \pi,(k+2)\pi].$$
Is it a mistake because it has to be $m$ right?
Besides I need a hint to show the last part. I tried to do with $m=0$ and $m=1$, $\phi_1= \frac{1+\cos x}{2}$ and $\phi_2=\frac{1+\cos(x-\pi)}{2}$ and the sum of $\phi_1$ and $\phi_2$ is equal to $1$. But how to generalize this and how to find that just three set of point have intersection in the support?

REPLY [0 votes]: For a function $f : X \to \mathbb R$ let $s(f) = \{ x \in X \mid f(x) \ne 0 \}$. Then the support of $f$,  $\text{supp} f$,  is defined as the closure of $s(f)$ in the space $X$. To answer your question it will be easier to work with $s(-)$ than with $\text{supp}$.
For your function $f : \mathbb R \to \mathbb R$ we have $s(f) = (-\pi,\pi)$. Thus $s(\phi_{2m+1}) = ((m-1)\pi,(m+1)\pi)$ and $s(\phi_{2m}) = ((-m-1)\pi,(-m+1)\pi)$ because $\phi_i$ is obtained from $f$ by a shift of the variable $x$.
This shows that $\text{supp}\phi_i$ always has the form $[k\pi,(k+2)\pi]$ for some $k = k(i)$.
To verify that $\Sigma(x) = \sum \phi_i(x) = 1$ for all $x$, note that only the indices $i$ make a contribution $> 0$ for which $x \in s(\phi_i)$. We have  $s(\phi_1) = (-\pi,\pi)$, $s(\phi_2) = (-2\pi,0)$, $s(\phi_3) = (0,2\pi)$, $s(\phi_4) = (-3\pi,-\pi)$, $s(\phi_5) = (\pi,3\pi)$, etc.
Case 1: $x = r\pi$ with $r \ge 0$. We have $r \pi \in s(\phi_i)$ precisely for $i=2r+1$. This is true because for $i = 2m$ with $m \ge 1$ we have $s(\phi_i) \subset (-\infty,0)$ and for $i = 2m+1$ with $m \ge 0$ we have $r\pi \in s(\phi_i) =  ((m-1)\pi,(m+1)\pi)$ iff $r = m$, i.e. $i = 2r+1$.
Thus $\Sigma(r \pi)= \frac{1 + \cos(r \pi - r \pi)}{2} = 1$.
Case 2: $x = r\pi$ with $r<  0$. Then $x  \in s(\phi_i)$ precisely for $i=-2r$.  This is true because for $i = 2m+1$ with $m \ge 0$ we have $s(\phi_i) \subset (-\pi,\infty)$ and for $i = 2m$ with $m \ge 1$ we have $r\pi \in s(\phi_i) =  ((-m-1)\pi,(-m+1)\pi)$ iff $r = -m$, i.e. $i = -2r$.
Thus $\Sigma(r \pi)= \frac{1 + \cos(r\pi + (-r)\pi)}{2} = 1$.
For the next two case recall that $\cos(y\pm\pi) = -\cos(y)$.
Case 3: $x \in (r\pi, (r+1)\pi)$ with $r \ge 0$. Then $x  \in s(\phi_i)$ precisely for $i=2r+1$ and $r = 2(r+1)+1$. This is true because for $i = 2m$ with $m \ge 1$ we have $s(\phi_i) \subset (-\infty,0)$ and for $i = 2m+1$ with $m \ge 0$ we have
$$(r\pi, (r+1)\pi) \cap s(\phi_i) =  (r\pi, (r+1)\pi) \cap ((m-1)\pi,(m+1)\pi) = \begin{cases} \emptyset & m \ne r, r+1 \\  (r\pi, (r+1)\pi) & m = r, r+1\end{cases}$$
Thus with $y = x + \pi r$ we have $\Sigma(x)= \phi_r(x) + \phi_{r+1}(x) = f(y) + f(y +\pi) = \frac{2 + \cos(y) + \cos(y + \pi)}{2} = 1$.
Case 4: $x \in (r\pi, (r+1)\pi)$ with $r < 0$. This is similar as case 3.


TITLE: Is this hypothesis of a group homomorphism theorem actually a corollary?
QUESTION [2 upvotes]: I'm trying to finally understand Noether's isomorphism theorems in the context of groups, mainly following Hungerford's treatment.
The question is regarding his Theorem 5.6 of Chapter 1, which states:

If $f: G \rightarrow H$ is a homomorphism of groups and $N$ is a normal subgoup of $G$ contained in the kernel of $f$, then there is a unique homomorphism $\bar{f}: G/N \rightarrow H$ such that $\bar{f}(aN) \rightarrow f(a)$ for all $a \in G$. $\text{Im} f = \text{Im} \bar{f}$  and $\ker \bar{f} = \ker f / N$. $\bar{f}$ is an isomorphism if and only if $f$ is an epimorphism and $N = \ker f$.

After proving this on my own, I am not sure why $N \subset \ker f$ is a hypothesis and not a corollary. Please allow me to explain:
First, I am unable to interpret the claim regarding $\bar{f}$ being the unique homomorphism such that $\bar{f}(aN) \rightarrow f(a)$ as anything but a construction of $\bar{f}$. Furthermore, this means $\bar{f}$ can be regarded as a subset of $G/N \times H$, specifically the subset defined by $\{(aN, f(a)) \ | \ a \in G\}$, which is obviously well-defined in the sense of being single-valued, since $f$ is a homomorphism. That makes the terminology unique strange to me, since apparently $\bar{f}$ is unique by construction.
Now we must prove $\bar{f}$ is a homomorphism from $G/N$ to $H$. To show  this, take $aN$ and $bN$ from $G/N$. We have
$$
\bar{f}(aNbN) = \bar{f}(abN) = f(ab) = f(a)f(b) = \bar{f}(aN)\bar{f}(bN).
$$
If this style of manipulating cosets is unfamiliar, consider Van der Waerden's definition of a complex, which is simply an arbitrary subset of the group. If $A$ and $B$ are complexes then the product $AB$ is defined as $\{ab \ | \ a\in A, b\in B\}.$ Van der Waerden asks which properties $A$ and $B$ must satisfy if $AB$ is a subgroup and concludes $AB < G$ iff $AB = BA$. Writing $aB$ for $\{a\}B$, we are granted an interpretation of the set of normal subgroups of $G$ as the center of the free group over $P(G)$. Thus the coset product $aNbN$ may be regarded as the product of complexes $\{a\}N\{b\}N = \{a\}(N\{b\})N =\{a\}(\{b\}N)N =\{a\}\{b\}N = abN.$
Evidently the fact that $\bar{f}$ is a unique homomorphism constructed as given does not rely on $N \subset \ker f$ as a hypothesis. The latter can however be arrived at as a corollary to $\bar{f}$ being a homomorphism. Just compare $\bar{f}(N) = \bar{f}(1_GN)= f(1_G) = 1_H$ with $\bar{f}(N) = \bar{f}(nN) = f(n)$ whenever $n \in N$.
To conclude, I find that the hypothesis $N \subset \ker f$ is not necessary to conclude $\bar{f}$ is a homomorphism, but follows easily from the fact that it is. I feel like there is something I am missing however.

REPLY [2 votes]: The hypothesis $N\subset\ker f$ is necessary to conclude $\overline{f}$ is a well-defined map. Example:  Let $$f:\mathbb{Z}\to\mathbb{Z}_2$$ be the canonical homomorphism. Obviously, $\ker f=2\mathbb{Z}$ and $N=3\mathbb{Z}\not\subset \ker f$. Now we have $1+\mathbb{Z}=4+\mathbb{Z}$ on $\mathbb{Z}/N$ but $1=\overline{f}(1+N)\neq\overline{f}(4+N)=0$ on $\mathbb{Z}_2$. Thus $\overline{f}(x+N)=f(x)$ is not well-defined.


TITLE: power series for $\frac1{x^2+x+1}$
QUESTION [1 upvotes]: So I am trying to find the sequence $(a_n)_{n\geq0}$ such that
$$\frac1{x^2+x+1}=\sum_{n\geq0}a_nx^n$$
My attempts:
I defined $$A(x)=\frac1{x^2+x+1}$$
Hence $$A(x)=\sum_{n\geq0}a_nx^n$$
And so $$(x^2+x+1)\sum_{n\geq0}a_nx^n=1$$
$$\sum_{n\geq0}a_nx^{n+2}+\sum_{n\geq0}a_nx^{n+1}+\sum_{n\geq0}a_nx^n=1$$
$$\sum_{n\geq0}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$a_0(x^2+x+1)+\sum_{n\geq1}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$x=0$:
$$a_0=1$$
So
$$x^2+x+1+a_1(x^3+x^2+x)+\sum_{n\geq2}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$\frac{x^2+x+1+a_1(x^3+x^2+x)+\sum_{n\geq2}a_n(x^{n+2}+x^{n+1}+x^n)}x=\frac1x$$
$$x+1+a_1(x^2+x+1)+\sum_{n\geq2}a_n(x^{n+1}+x^{n}+x^{n-1})=0$$
$x=0$:
$$a_1=-1$$
One more time:
$$x^2+x+1-x^3-x^2-x+a_2(x^4+x^3+x^2)+\sum_{n\geq3}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$-x^3+a_2(x^4+x^3+x^2)+\sum_{n\geq3}a_n(x^{n+2}+x^{n+1}+x^n)=0$$
Divide both sides by $x^2$:
$$-x+a_2(x^2+x+1)+\sum_{n\geq3}a_n(x^{n}+x^{n-1}+x^{n-2})=0$$
$x=0$:
$$a_2=0$$
How do I find $a_n$? Am I doing things right so far? Thanks.

REPLY [2 votes]: You could have continue your way.
Starting from
$$\sum_{n=0}^\infty a_nx^{n+2}+\sum_{n=0}^\infty a_nx^{n+1}+\sum_{n=0}^\infty a_nx^n=1$$ you already found $a_0=1$ and $a_1=-1$.
Now, consider a given degree $m >1$. You then have the recurrence relation
$$a_{m-2}+a_{m-1}+a_m=0$$ and the roots of the characteristic equation
$$r^2+r+1=0 \implies r_\pm=\frac{-1 \pm i \sqrt 3} 2$$ making by the end (I let to you the intermediate steps)
$$a_m=\cos \left(\frac{2 \pi  m}{3}\right)-\frac{1}{\sqrt{3}}\sin \left(\frac{2 \pi
   m}{3}\right)$$ making $a_{3m}=1$, $a_{3m-1}=0$, $a_{3m+1}=-1$.