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\begin{center}
\vskip 1cm{\LARGE\bf Ramanujan and a Combinatorial Identity\\
\vskip .1in
Involving Harmonic Numbers
}
\vskip 1cm
\large
Horst Alzer\\
Morsbacher Stra\ss e 10\\
51545 Waldbr\"ol\\
Germany\\
\href{mailto:email}{h.alzer@gmx.de}\\ 
{\ } \\

Man Kam Kwong\\
Department of Applied Mathematics\\
The Hong Kong Polytechnic University\\
Hunghom, Hong Kong\\
\href{mailto:email}{mankamkwong.math@outlook.com}\\
\end{center}

\numberwithin{equation}{section}


\vskip .2 in
\begin{abstract}
We show that trigonometric series and integral formulas given by Ramanujan can be applied to deduce a new  summation identity involving binomial coefficients and the classical harmonic number $H_n=1+1/2+\cdots +1/n$. Moreover, we prove that for $n\geq 2$ the sum $H_2+H_3 + \cdots + H_n$ is not an integer.
\end{abstract}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\section{Introduction and statement of the main result}

The work of  Ramanujan had (and still has) tremendous influence on various fields, such as real analysis and number theory. The aim of this note is to show that certain  trigonometric summation and integral formulas given by Ramanujan can be used to deduce a new combinatorial identity involving the well-known harmonic numbers
$$
H_0=0, \quad
H_n=\sum_{\nu=1}^n \frac{1}{\nu}, \quad n=1,2,3, \ldots .
$$
The properties of these numbers have been studied intensively by many authors. Ramanujan \cite[p.\ 374]{B2} obtained a rapidly convergent series representation for the digamma function $\psi=\Gamma'/\Gamma$,
$$
\psi(z+1)+\gamma= 
\frac{1}{2z} -\frac{1}{2\pi z^2} +\frac{\pi \cot(\pi z)}{e^{2\pi z}-1}
+\sum_{\nu=1}^\infty \frac{z^2}{\nu (\nu^2+z^2)}  +4 \sum_{\nu=1}^\infty \frac{\nu z^2}{(e^{2\pi \nu}-1) (\nu^4 -z^4)},
$$
where $\gamma$ denotes Euler's constant. Let $n\geq 1$ be an integer.  If $z\rightarrow n$, then we obtain 
$$
H_n=\frac{1}{2n} -\frac{1}{2\pi n^2} +\frac{1-(4\pi n +1)e^{2\pi n}}{2n(e^{2\pi n}-1)^2} 
+\sum_{\nu=1}^\infty \frac{n^2}{\nu (\nu^2+n^2)}  +4 \sum_{\nu=1 \atop \nu\neq n}^\infty \frac{\nu n^2}{(e^{2\pi \nu}-1) (\nu^4 -n^4)}.
$$
Lagarias \cite{L} revealed a remarkable connection between harmonic numbers and the classical Riemann hypothesis. He proved that Riemann's hypothesis is equivalent to
$$
\sigma(n) < H_n + e^{H_n}  \log H_n, \quad n=2,3,4, \ldots,
$$
where $\sigma(n)$ denotes the sum of divisors of $n$.

In the literature, we can find numerous interesting sums and series involving $H_n$. We give two examples:
$$
\sum_{\nu=1}^m (-1)^{\nu} {m\choose \nu} \frac{\nu (H_{n+\nu} - H_n)}{{n+\nu\choose \nu}}
 = \frac{m (m^2-m-n^2)}{(m+n)^2 (m+n-1)^2}
 $$
and
 $$
\sum_{\nu=1}^\infty
 \frac{  {2\nu \choose \nu}^2   }{   (2\nu -1)^2 16^{\nu}    } H_{\nu}
= \frac{4}{\pi} \bigl( 3-4\log 2 \bigr);
$$
see 
Choi \cite{Ch},  Chu \cite{C}, Chu and Donno \cite{CD}, and Sofo \cite{S}.


Here is our main result.


\begin{theorem} \label{T1}
{Let $k$ and $n$ be natural numbers with $k\leq n$. Then}
\begin{equation}
\sum_{\nu=1}^n \left\{ {n+k-\nu -1\choose n-1}  - {n+k-\nu -1\choose k-1} \right\}H_{\nu}
={n+k \choose k} \bigl( H_k -H_n \bigr). \label{1.1}
\end{equation}
\end{theorem}


\begin{remark}
When $k=1$, (\ref{1.1}) reduces to a particularly simple form:
\begin{equation}
\sum_{\nu=2}^n H_{\nu} =(n+1)(H_n -1). \label{1.2}
\end{equation}
\end{remark}
\begin{remark}
(ii) The referee pointed out the following noteworthy representation of the difference of two harmonic numbers, 
see \cite{B}:
$$
\frac{1}{{n\choose k}} \sum_{\nu=1}^k {n-\nu\choose n-k} \frac{1}{\nu} =H_n - H_{n-k}, \quad 1\leq k\leq n .
$$
\end{remark}




\noindent
Applications of (\ref{1.1}) lead to the following closely related formulas.

\begin{corollary}  \label{C1}
{Let $k$ and $n$ be natural numbers with $2\leq k\leq n$. Then}
\begin{align}
& \sum_{\nu=1}^{n+1} \left\{ {n+k-\nu -1\choose n}  - {n+k-\nu -1\choose k-2} \right\}H_{\nu} \nonumber\\
& \hspace*{15mm} ={n+k \choose k-1} \bigl( H_k -H_{n+1} \bigr)-\frac{1}{n+1}{n+k\choose k}.
\label{1.3}
\end{align}
\end{corollary}

\begin{corollary}  \label{C2}
{Let $k$ and $n$ be natural numbers with $ k\leq n-1$. Then}
\begin{align}
& \sum_{\nu=1}^{n} \left\{ {n+k-\nu -1\choose n-2}  - {n+k-\nu -1\choose k} \right\}H_{\nu} \nonumber\\
& \hspace*{15mm} ={n+k \choose k+1} \bigl( H_k -H_{n} \bigr)+\frac{1}{k+1}{n+k+1\choose k+1}.
\label{1.4}
\end{align}
\end{corollary}







Obaid \cite{O} presented an elegant counterpart of (\ref{1.1}) which involves the famous Fibonacci numbers $F_n$:
\begin{equation}
\sum_{\nu=1}^n (-1)^{\nu} \left\{ {n+k-\nu -1\choose n-1}  - {n+k-\nu -1\choose k-1} \right\}F_{\nu}
= (-1)^{n+1} F_{n-k}, \quad 1\leq k\leq n. \label{1.5}
\end{equation}

With regard to (\ref{1.1}) and (\ref{1.5}) it is natural to ask: do there exist similar formulas if we replace $H_{\nu}$ or $F_{\nu}$ by other special combinatorial or number theoretical sequences?



Our proofs are given in the next section. In Section \ref{S3}, we offer two additional results. Among others, we show that if $n\geq 2$, then $\sum_{\nu=2}^n H_{\nu}$ is not an integer.



\section{Proofs}



To prove  Theorem \ref{T1} we need the following four lemmas. The first two formulas were given by Ramanujan; see
 Berndt \cite[pp.\ 246,  290]{B1}.



 

\begin{lemma}  \label{L1}
{Let $a$, $N$, and $\theta$ be real numbers with $N\geq 0$ and $|\theta|\leq \pi/2$. Then}
\begin{equation}
\sum_{\nu=0}^\infty {N\choose \nu} \sin((a+2\nu)\theta) =2^N \cos^N(\theta) \sin((a+N)\theta). \label{2.1}
\end{equation}
\end{lemma}





\begin{lemma}  \label{L2}
{Let $n$ be a positive integer. Then}
\begin{equation}
\int_0^{\pi/2} x \cos^n(x) \sin(nx) dx=\frac{\pi}{2^{n+2}} H_n. \label{2.2}
\end{equation}
\end{lemma}


The next result is due to Rung and Obaid \cite{RO}. It has applications in the theory of boundary value problems. See also Obaid and Rung \cite{OR}.


\begin{lemma}  \label{L3}
{Let  $k$ and $n$ be integers with $1\leq k\leq n$ and let $\theta$ be a real number. Then}
\begin{align}
& \sum_{\nu=1}^n 2^{\nu}  \left\{   {n+k-\nu -1\choose k-1} - {n+k-\nu-1 \choose n-1} \right\} \cos^{\nu} (\theta) \sin(\nu\theta) \nonumber \\
& \hspace*{15mm} = 2^{n+k}  \cos^{n+k} (\theta) \sin((n-k)\theta). \label{2.3}
\end{align}
\end{lemma}









Moreover, we need the following combinatorial identity which is given in Gould \cite[p.\ 6]{G}.





\begin{lemma}  \label{L4}
{Let $N$ be a nonnegative integer and let $x$ be a real number with $x\notin \{0,1,\ldots,N\}$. Then}
\begin{equation}
\sum_{\nu=0}^N \frac{(-1)^{\nu}}{x-\nu}{N\choose \nu}
=\frac{(-1)^N}{(x-N) {x\choose N}}. \label{2.4}
\end{equation}
\end{lemma}












\begin{proof}[Proof of Theorem \ref{T1}]
Let $1\leq k\leq n$.
We apply (\ref{2.1}) with $a=-2k$, $N=n+k$. Then
\begin{equation}
\sum_{\nu=0}^{n+k} {n+k\choose \nu} \sin(2(\nu-k)\theta)= 2^{n+k} \cos^{n+k}(\theta) \sin((n-k)\theta).  \label{2.5}
\end{equation}

\noindent
Using (\ref{2.3}) and (\ref{2.5}) leads to
\begin{align}
& \sum_{\nu=0}^{n+k} {n+k\choose \nu} \theta \sin(2(\nu-k)\theta) \nonumber\\
 & \hspace*{10mm}  = 
\sum_{\nu=1}^n  \left\{   {n+k-\nu -1\choose k-1} - {n+k-\nu-1 \choose n-1} \right\}2^{\nu} \theta  \cos^{\nu} (\theta) \sin(\nu\theta).  \label{2.6}
\end{align}


\noindent
Next, we 
 integrate and apply the formula
$$
\int_0^{\pi/2} \theta \sin(2p\theta) d\theta= (-1)^{p+1}\frac{\pi}{4p},  \quad p\in \mathbb{Z}\setminus\{0\}.
$$
This gives
\begin{align}
\int_0^{\pi/2} 
\sum_{\nu=0}^{n+k} {n+k\choose \nu} \theta \sin(2(\nu-k)\theta) d \theta
& =   \int_0^{\pi/2}  \sum_{\nu=0  \atop \nu\neq k }^{n+k} {n+k\choose \nu} \theta \sin(2(\nu-k)\theta) d\theta \nonumber \\ \nonumber
& =  \sum_{\nu=0 \atop \nu\neq k }^{n+k} {n+k\choose \nu}\int_0^{\pi/2}  \theta \sin(2(\nu-k)\theta)  d\theta \\ 
& =  - \frac{\pi}{4} \sum_{\nu=0  \atop \nu\neq k}^{n+k} \frac{(-1)^{\nu - k}}{\nu -k}  {n+k\choose \nu}. \label{2.7} 
\end{align}

\noindent
Using (\ref{2.2}) yields
\begin{align}
\int_0^{\pi/2}
& \sum_{\nu=1}^n  \left\{   {n+k-\nu -1\choose k-1} - {n+k-\nu-1 \choose n-1} \right\}2^{\nu} \theta  \cos^{\nu} (\theta) \sin(\nu\theta)d\theta 
\nonumber\\
& \hspace*{10mm} = \frac{\pi}{4}
\sum_{\nu=1}^n  \left\{   {n+k-\nu -1\choose k-1} - {n+k-\nu-1 \choose n-1} \right\} H_{\nu}. \label{2.8}
\end{align}

\noindent
From (\ref{2.6}), (\ref{2.7}) and (\ref{2.8}) we conclude that
\begin{equation}
\sum_{\nu=1}^n  \left\{   {n+k-\nu -1\choose k-1} - {n+k-\nu-1 \choose n-1} \right\} H_{\nu}
=   \sum_{\nu=0  \atop \nu\neq k}^{n+k} \frac{(-1)^{\nu - k}}{k-\nu}  {n+k\choose \nu}.  \label{2.9}
\end{equation}



\noindent
We define
$$
\phi_N(x)={x\choose N}, \quad N\in \mathbb{N}, \, x\in \mathbb{R}.
$$
Let  $k<N$. Then

$$
\phi'_N(k)= \frac{(-1)^{N-k-1}}{(k+1){N\choose k+1}}, \quad 
\phi''_N(k)= 2  \phi'_N(k) \sum_{\nu=0 \atop \nu\neq k}^{N-1} \frac{1}{k-\nu}.
$$

From (\ref{2.4}) we obtain
$$
\sum_{\nu=0 \atop \nu\neq k}^N\frac{(-1)^{\nu}}{x-\nu}{N\choose \nu}
=\frac{(-1)^N}{(x-N) {x\choose N }}-\frac{(-1)^k }{x-k}{N\choose k}.
$$
Using l'Hospital's rule leads to


\begin{eqnarray}\nonumber
\sum_{\nu=0 \atop \nu\neq k}^N
 \frac{(-1)^{\nu}}{k-\nu}{N\choose \nu}
& = & \lim_{x\to k} \left(   \frac{(-1)^N}{(x-N) {x\choose N}}-\frac{    (-1)^k }{x-k}{N\choose k}   \right) \\  \nonumber
& = & \frac{(-1)^{k+1}}{k-N} {N\choose k} \left(1+\frac{ (k-N)\phi''_N(k)}{2 \phi'_N(k)}  \right) \\ \nonumber
& = & 
{(-1)^{k+1}} {N\choose k} \left(  \frac{1}{k-N}+\sum_{\nu=0 \atop \nu\neq k}^{N-1}\frac{ 1}{k-\nu}\right). \nonumber
\end{eqnarray}

We set $N=n+k$ and obtain
\begin{equation}
\sum_{\nu=0 \atop \nu\neq k}^{n+k} \frac{(-1)^{\nu}}{k-\nu}{n+k\choose \nu}
=(-1)^k {n+k\choose k}\bigl( H_n - H_k\bigr).   \label{2.10}
\end{equation}
From (\ref{2.9}) and (\ref{2.10}) we conclude that (\ref{1.1}) holds.
\end{proof}







\begin{proof}[Proof of  Corollary \ref{C1}.]
Let $2\leq k\leq n$. We define
\begin{equation}
T(k,n)=
\sum_{\nu=1}^n \left\{ {n+k-\nu -1\choose n-1}  - {n+k-\nu -1\choose k-1} \right\}H_{\nu}.   \label{2.11}
\end{equation}
Applying Pascal's formula 
$$
{N+1\choose \nu} ={N\choose \nu}+{N\choose \nu-1} 
$$
gives
\begin{align}
T(k,n+1) & = \sum_{\nu=1}^{n+1}  \left\{ {n+k-\nu -1 \choose n} +{n+k-\nu-1\choose n-1} \right\} H_{\nu} \label{2.12} \nonumber\\ \nonumber
 & \hspace*{5mm} - \sum_{\nu=1}^{n+1}  \left\{  {n+k-\nu-1\choose k-1} + {n+k-\nu -1 \choose k-2} \right\}H_{\nu}   \\ 
& = 
T(k,n)+\sum_{\nu=1}^{n+1} \left\{ {n+k-\nu -1 \choose n}  - {n+k-\nu -1 \choose k-2} \right\}H_{\nu}. 
\end{align}

From (\ref{1.1}) we obtain
\begin{equation}
T(k,n+1)-T(k,n)   =  {n+k  \choose k-1} (H_k - H_{n+1} ) -\frac{1}{n+1} {n+k  \choose k}.    \label{2.13}
\end{equation}


Combining (\ref{2.12}) and (\ref{2.13}) leads to (\ref{1.3}).
\end{proof}



\begin{proof}[Proof of  Corollary \ref{C2}.]
Let $1\leq k\leq n-1$ and let $T(k,n)$ be the sum given in (\ref{2.11}). Then
\begin{align}
T(k+1,n) & =  \sum_{\nu=1}^n \left\{   {n+k-\nu \choose n-1}-{n+k-\nu\choose k}\right\} H_{\nu} \label{2.14} \nonumber\\ \nonumber
& =   \sum_{\nu=1}^{n}  \left\{ {n+k-\nu -1 \choose n-1} +{n+k-\nu-1\choose n-2} \right\} H_{\nu} \\ \nonumber
& \hspace*{5mm} - \sum_{\nu=1}^{n}  \left\{  {n+k-\nu-1\choose k} + {n+k-\nu -1 \choose k-1} \right\}H_{\nu}   \\ 
& = 
T(k,n)+\sum_{\nu=1}^{n} \left\{ {n+k-\nu -1 \choose n-2}  - {n+k-\nu -1 \choose k} \right\}H_{\nu}. 
\end{align}
Using (\ref{1.1}) gives
\begin{equation}
T(k+1,n)-T(k,n)= {n+k\choose k+1} (H_k - H_n) +\frac{1}{k+1}{n+k+1\choose k+1}.   \label{2.15}
\end{equation}
From (\ref{2.14}) and (\ref{2.15}) we conclude that (\ref{1.4}) holds.
\end{proof}





\section{Additional results} \label{S3}

There exists a kind of converse of Theorem \ref{T1}. We use (\ref{1.1}) with $k=n-1$ and assume that $(K_{\nu})_{\nu\geq 1}$ is a real or complex sequence. Then we obtain the following uniqueness theorem.


\begin{theorem}   \label{T2}
{If
\begin{equation}
\sum_{\nu=2}^n \left\{ {2n-\nu -2\choose n-1}  - {2n-\nu -2\choose n-2} \right\}K_{\nu}
={2n-1 \choose n-1} \bigl( K_{n-1} -K_n \bigr) \label{3.1}
\end{equation}
holds for integers $n\geq 2$, then there exists a constant $\lambda$ such that}
\begin{equation}
K_r = \lambda H_r, \quad r=1,2, \ldots . \label{3.2}
\end{equation}
\end{theorem}


\begin{proof}
We consider two cases.

\smallskip\noindent {\underline{Case 1.}} \, $K_1=0$. We use induction to prove that $K_r=0$ for $r\geq 1$. Let $K_1=K_2 = \cdots = K_{r-1}=0$. From (\ref{3.1}) with $n=r\geq 2$ we obtain
$$
-K_r= {2r-1\choose r-1} (-K_r).
$$
Since ${2r-1\choose r-1}>1$, we get $K_r=0$.


\smallskip\noindent {\underline{Case 2.}} \, $K_1\neq 0$. Since (\ref{3.1}) remains valid, if we replace $K_j$ by $K_j/K_1$, we may assume that $K_1=1$. We show that (\ref{3.2}) holds with $\lambda=1$. Let $K_j=H_j$ $(j=1,\ldots,r-1)$. Then we obtain from (\ref{3.1}) with $n=r$:
\begin{align}
& \sum_{\nu=2}^{r-1} \left\{ {2r-\nu -2\choose r-1}  - {2r-\nu -2\choose r-2} \right\}H_{\nu}
+  \left\{ {r -2\choose r-1}  - {r -2\choose r-2} \right\}K_{r} \nonumber\\
& \hspace*{10mm} ={2r-1 \choose r-1} \bigl( H_{r-1} -K_r \bigr). \label{3.3}
\end{align}
Since the solution of a linear equation is unique, we conclude from (\ref{1.1}) (with $n=r$, $k=r-1$) that (\ref{3.3}) implies that $K_r=H_r$.
\end{proof}


In 1915, Theisinger \cite{T} proved that if $n>1$, then $H_n$ is not an integer. K\"ursch\'ak offered an interesting extension. He showed that if $1\leq n<m$, then the difference $H_m-H_{n-1}$ is never an integer; see P\'olya and Szeg\"o \cite[pp.\ 159, 381]{PS}. We apply (\ref{1.2}) to prove the following related result.



\begin{theorem} \label{T3}
{Let $n\geq 2$ be a natural number. Then the sum $\sum_{\nu=2}^n H_{\nu}$ is not an integer.}
\end{theorem}

\begin{proof} 
We define $J_n=(n+1)H_n$. Then
$$
J_2=\frac{9}{2} \quad\mbox{and} \quad J_3 = \frac{22}{3}.
$$
Let $n\geq 4$.
We assume that $J_n$ is an integer. First, we show that there exists a prime number $p$ such that $n+1=2p$. We have
$$
J_n = \frac{1}{n! }\sum_{\nu=1}^n \frac{(n+1)! }{\nu}.
$$
It follows that
$$
n! \,  \Big{|} \,  \sum_{\nu=1}^n \frac{(n+1)! }{\nu}.
$$
An application of Bertrand's postulate gives that there exists a prime number $p$ such that
$$
[n/2] < p < n.
$$
Since $p \, | \, n! $, we obtain
\begin{equation}
p \, \Big{|} \, \sum_{\nu=1}^n \frac{(n+1)! }{\nu}. \label{3.4}
\end{equation}
We have
\begin{equation}
p \, \Big{|} \, \frac{(n+1)! }{\nu}, \quad \nu=1,2,\ldots,p-1, p+1, \ldots, n. \label{3.5}
\end{equation}
From (\ref{3.4}) and (\ref{3.5}) we conclude that
$$
p \, \Big{|} \, \frac{(n+1)! }{p}.
$$
This means that
$$
p \, \Big{|} 1 \cdot 2 \cdots (p-1) \cdot (p+1) \cdots n \cdot (n+1).
$$
It follows that there exists a number $k\in \{ p+1,\ldots, n+1\}$ such that $p|k$. This implies that $2p \leq n+1$. Otherwise, if $n+1<2p$, then there exists a positive integer $m$ with $pm=k$ and
$$
p\leq pm =k \leq n+1 <2p.
$$
Thus, $m=1$ and $k=p$. A contradiction. Hence $2p\leq n+1$.


We  assume that $n$ is even. Let $n=2N$. Then
$$
N+1=[n/2] +1 \leq p \leq \frac{n+1}{2} =N+1/2.
$$
A contradiction.  It follows that $n$ is odd. Let $n=2N+1$. Then
$$
N+1= [n/2] +1 \leq p \leq \frac{n+1}{2}=N+1.
$$
Thus, $p=N+1$. This gives $n+1=2N+2=2p$.


We suppose that there exists a natural number $n\geq 4$ such that $J_n$ is an integer. Then $n=2p-1$. Here, $p$ is an odd prime number.
We have the representation
$$
H_n= \frac{1}{2^{ [\log_2(n)]}} \frac{a_n}{b_n},
$$
where $a_n$ and $b_n$ are odd integers.
It follows that
$$
J_{2p-1} = 2p H_{2p-1}  =  \frac{2p}{2^{ [\log_2(2p-1)]}} \frac{a_{2p-1}}{b_{2p-1}}=
\frac{p}{2^{ [\log_2(2p-1)]-1}} \frac{a_{2p-1}}{b_{2p-1}}.
$$
From
$$
\log_2(2p-1) =\frac{\log(2p-1)}{\log(2)} \geq \frac{\log(5)}{\log(2)}>2
$$
we conclude that
$$
[\log_2(2p-1) ]\geq 2.
$$
Thus
$$
J_{2p-1} = \frac{c_{2p-1}}{2^r b_{2p-1}},
$$
where $r$ is a natural number and  $c_{2p-1} = p a_{2p-1}, b_{2p-1}$ are odd integers. A contradiction. This implies that
 $\sum_{\nu=2}^n H_{\nu} =J_n-(n+1)$ is not an integer for all $n\geq 2$.
\end{proof}


\section{Acknowledgment} We thank the referee for helpful comments.



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A. Sofo, Some more identities involving rational sums, \emph{Appl. Anal. Discr. Math.} {\bf 2} (2008), 56--66.

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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 05A19; Secondary 11B83, 33B10.

\noindent \emph{Keywords: } combinatorial identity, harmonic number, trigonometric function.

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\noindent (Concerned with sequence
\seqnum{A001008} and
\seqnum{A002805}.)

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\noindent
Received  May 7 2026; 
revised versions received  August 4 2026; August 5 2026.
Published in {\it Journal of Integer Sequences}, August 14 2026.

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