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	\begin{center}
		\vskip 1cm{\LARGE\bf Approximating Sums of Consecutive\\
			\vskip .1in
		Integral Roots}
		\vskip 1cm
		\large
		Pongpol Ruankong and Kantaphon Kuhapatanakul\\
     	Department of Mathematics\\
     	Faculty of Science\\
     	Kasetsart University\\
		Bangkok 10900\\
		Thailand\\ 
		\href{mailto:fscippru@ku.ac.th}{fscippru@ku.ac.th}\\
		\href{mailto:fscikpkk@ku.ac.th}{fscikpkk@ku.ac.th}	
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	\begin{abstract}
		 We present an alternative proof of Saltzman and Yuan's result on the sums of consecutive integral roots. We use the AM-GM-HM inequality to prove the main result. Moreover, the lower bound for which the result holds is greatly improved.
	\end{abstract}
	
	\section{Introduction}

     For each real number $x$, let $\left\lfloor x\right\rfloor$ denote the greatest integer not exceeding $x$. The sums of consecutive integral roots have been studied by many mathematicians (see \cite{ha}-\cite{zh}). For instance, the following identities hold for every positive integer $n$:
     \begin{itemize}
     	\item [$(1)$] $\left\lfloor\sqrt n +\sqrt {n+1}\right\rfloor=\left\lfloor\sqrt{4n+1}\right\rfloor$
     	\item [$(2)$] $\left\lfloor\sqrt n +\sqrt {n+1}+\sqrt {n+2}\right\rfloor=\left\lfloor\sqrt{9n+8}\right\rfloor$
     	\item [$(3)$] $\left\lfloor\sqrt n +\sqrt {n+1}+\sqrt {n+2}+\sqrt {n+3}\right\rfloor=\left\lfloor\sqrt{16n+20}\right\rfloor$
     	\item [$(4)$] $\left\lfloor\sqrt n +\sqrt {n+1}+\sqrt {n+2}+\sqrt {n+3}+\sqrt {n+4}\right\rfloor=\left\lfloor\sqrt{25n+49}\right\rfloor$
     	\item [$(5)$] $\left\lfloor\sqrt[3]n +\sqrt[3]{n+1}\right\rfloor=\left\lfloor\sqrt[3]{8n+3}\right\rfloor$
     	\item [$(6)$] $\left\lfloor\sqrt[3]n +\sqrt[3]{n+1}+\sqrt[3]{n+2}\right\rfloor=\left\lfloor\sqrt[3]{27n+26}\right\rfloor$.	
     \end{itemize} 
     
     On the contrary, Zhan \cite{zh} showed that for any real number $c$, there is a positive integer $n$ such that
     $$\left\lfloor\sqrt n +\sqrt {n+1}+\sqrt {n+2}+\sqrt {n+3}+\sqrt {n+4}+\sqrt{n+5}\right\rfloor\neq\left\lfloor\sqrt{36n+c}\right\rfloor.$$ 
     
     Saltzman and Yuan \cite{sy} presented the following similar formula for $n\ge4$.
     $$\left\lfloor\sqrt n +\sqrt {n+1}+\sqrt {n+2}+\sqrt {n+3}+\sqrt {n+4}+\sqrt{n+5}\right\rfloor=\left\lfloor\sqrt{36n+89}\right\rfloor,$$
     and gave the general formula: for all integers $p,m\ge2$, 
     \begin{equation}  \label{root}
     \left\lfloor\sqrt[p]{n}+\sqrt[p]{n+1}+\sqrt[p]{n+2}+\cdots+\sqrt[p]{n+m-1}\right\rfloor
     =\left\lfloor\sqrt[p]{m^pn+\frac{m^p(m-1)}{2}-1}\right\rfloor
     \end{equation} 
     holds for all positive integers
     \begin{equation}  \label{bound}
     n&gt;\frac{m^p(m-1)(2m-1)(p-1)}{12p}.
     \end{equation}
     They utilized properties of concave functions to approximate the sums of consecutive integral roots.
     
     In this paper, we give an alternative proof of \eqref{root} using the AM-GM-HM inequality and obtain a new lower bound on $n$, which is approximately half of \eqref{bound}.

\section{Results}

In the sequel, let $m$ and $p$ be two positive integers greater than $1$, let $M=\frac{(p-1)m^p(m^2-1)}{12p}$, and let $i_1,i_2,\dots,i_p\in\{0,1,2,\dots,m-1\}$. Denoted by $A_p,G_p$ and $H_p$ the arithmetic mean, the geometric mean, and the harmonic mean of the positive integers $n+i_1, n+i_2,\dots,n+i_p$, respectively, i.e.,
\begin{align*}
A_p &amp;= \frac{1}{p}\sum_{j=1}^p (n+i_j),\\
G_p &amp;= \sqrt[p]{(n+i_1)\dots(n+i_p)},\\
H_p &amp;= \frac{p}{\sum_{j=1}^p\frac{1}{n+i_j}}.
\end{align*}
By the well-known AM-GM-HM inequality,
$$H_p\le G_p \le A_p$$
where equality holds if and only if $n+i_1=n+i_2=\cdots=n+i_p$. To prove the main theorem, we first present two technical lemmas.
\begin{lemma}   \label{lem1}
	For every positive integer $n$,
	$$	\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}\left(A_p-H_p\right) &lt; \dfrac{M}{n^{p-1}}\left(n+\frac{m-1}{2}\right)^{p-2}.$$
\end{lemma}
\begin{proof}
	Observe that
	\begin{align*}
	A_p-H_p&amp;=\frac{1}{p}\sum_{j=1}^p (n+i_j)-\frac{p}{\sum_{j=1}^p\frac{1}{n+i_j}}\\
	&amp;=\frac{\left(\left(n+\frac{1}{p}\sum_{j=1}^pi_j\right)\left(\frac{1}{p}\sum_{k=1}^p\frac{\prod_{j=1}^p(n+i_j)}{n+i_k}\right)-\prod_{j=1}^p(n+i_j)\right)}{\frac{1}{p}\sum_{k=1}^p\frac{\prod_{j=1}^p(n+i_j)}{n+i_k}}.
	\end{align*}
	Hence
	$$\frac{A_p-H_p}{p}\sum_{k=1}^p\frac{\prod_{j=1}^p(n+i_j)}{n+i_k}=\left(n+\frac{1}{p}\sum_{j=1}^pi_j\right)\left(\frac{1}{p}\sum_{k=1}^p\frac{\prod_{j=1}^p(n+i_j)}{n+i_k}\right)-\prod_{j=1}^p(n+i_j).$$
	Next, we determine the coefficients of  
	\begin{equation}   \label{coef}
	\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}\left(\left(n+\frac{1}{p}\sum_{j=1}^pi_j\right)\left(\frac{1}{p}\sum_{k=1}^p\frac{\prod_{j=1}^p(n+i_j)}{n+i_k}\right)-\prod_{j=1}^p(n+i_j)\right)
	\end{equation}
	as a polynomial in variable $n$.
	
	It is easy to see that the coefficients of $n^p$ and $n^{p-1}$ in \eqref{coef} are zero. 
	Moreover, we claim that, for each $t=2,3,\dots,p$, the coefficient of $n^{p-t}$ in \eqref{coef} is equal to $M{p-2\choose t-2}\left(\frac{m-1}{2}\right)^{t-2}$ implying that the polynomial is $M\left(n+\frac{m-1}{2}\right)^{p-2}$.
	
	The coefficient of $n^{p-t}$ in \eqref{coef}, for $t=2,3,\dots,p$, is equal to the sums of the following:
	\begin{itemize}
		\item [(i)] $\displaystyle\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}\left(\frac{1}{p^2}\sum_{j=1}^pi_j\left(\sum_S\sum_{\substack{j_1&lt;\cdots&lt;j_{t-1}\\ j_1,\dots,j_{t-1}\in S}}i_{j_1}\cdots i_{j_{t-1}}\right)\right)$
		\item [(ii)] $\displaystyle\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}\left(\frac{1}{p}\sum_S\sum_{\substack{j_1&lt;\cdots&lt;j_t\\ j_1,\dots,j_t\in S}}i_{j_1}\cdots i_{j_t}\right)$
		\item [(iii)] $\displaystyle\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}\left(-\sum_{\substack{j_1&lt;\cdots&lt;j_t\\ j_1,\dots,j_t\in \mathbb N_p}}i_{j_1}\cdots i_{j_t}\right)$
	\end{itemize}
	where $\sum_S$ represents the sum over all $(p-1)$-subsets $S$ of $\mathbb N_p=\{1,2,\dots,p\}$. (Note that for the case $t=p$, (ii) is an empty sum which is conventionally zero. This coincides with the fact that (ii) does not occur in the case $t=p$.)
	
	Distributing $\sum_{j=1}^p i_j$ inside the double sum, (i) becomes
	$$\frac{1}{p^2}\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}\left(\sum_S\sum_{\substack{j_1&lt;\cdots&lt;j_{t-1}\\ j_1,\dots,j_{t-1}\in S}}\left(\sum_{r=1}^{t-1}i_{j_r}(i_{j_1}\cdots i_{j_{t-1}})+\sum_{r\notin\{1,\dots,t-1\}}i_{j_r}(i_{j_1}\cdots i_{j_{t-1}})\right)\right).$$
	Since the double sum consists of exactly $p{p-1 \choose t-1}$ terms, using the symmetry of the $i_j$ and combining the like terms, (i) amounts to
	\begin{equation}   \label{sum1}
	\frac{1}{p}{p-1\choose t-1}\left((t-1)\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}i_1^2\cdots i_{t-1}+(p-t+1)\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}i_1\cdots i_t\right).
	\end{equation}
	Similarly, (ii) amounts to 
	\begin{equation}   \label{sum2}
	{p-1\choose t}\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}i_1\cdots i_t,
	\end{equation}
	and (iii) amounts to 
	\begin{equation}   \label{sum3}
	-{p\choose t}\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}i_1\cdots i_t.
	\end{equation}
	Note that  
	$$\frac{p-t+1}{p}{p-1\choose t-1}+{p-1\choose t}-{p\choose t}=-\frac{t-1}{p}{p-1\choose t-1}$$ for $t =2,\dots, p.$ Hence adding \eqref{sum1}, \eqref{sum2} and \eqref{sum3} yields
	\begin{align*}
	\frac{t-1}{p}{p-1\choose t-1}&amp;\left(\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}i_1^2\cdots i_{t-1}-\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}i_1\cdots i_t\right)\\
	&amp;=\frac{p-1}{p}{p-2\choose t-2}\left(m^{p-t+1}\sum_{i=0}^{m-1}i^2\left(\sum_{i=0}^{m-1}i\right)^{t-2}-m^{p-t}\left(\sum_{i=0}^{m-1}i\right)^t\right)\\
	&amp;=\frac{p-1}{p}{p-2\choose t-2}m^{p-t}\left(\sum_{i=0}^{m-1}i\right)^{t-2}\left(m\sum_{i=0}^{m-1}i^2-\left(\sum_{i=0}^{m-1}i\right)^2\right)\\
	&amp;=M{p-2\choose t-2}\left(\frac{m-1}{2}\right)^{t-2},
	\end{align*} which is the coefficient of $n^{p-t}$ in $M\left(n+\frac{m-1}{2}\right)^{p-2}$. Therefore,
	\begin{align*}
	\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}\left(A_p-H_p\right)&amp;&lt; \frac{1}{n^{p-1}}\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}\left(\frac{A_p-H_p}{p}\sum_{k=1}^p\frac{\prod_{j=1}^p(n+i_j)}{n+i_k}\right)\\
	&amp;&lt;\dfrac{M}{n^{p-1}}\left(n+\frac{m-1}{2}\right)^{p-2}.
	\end{align*}
\end{proof}

\begin{lemma}   \label{lem2}
	For every positive integer $n\ge M+\frac{(p-2)(m-1)}{2}$,
	$$n^{p-1}\ge M\left(n+\frac{m-1}{2}\right)^{p-2}.$$	
\end{lemma}
\begin{proof}
	For $n\ge M+\frac{(p-2)(m-1)}{2}$,
	\begin{align*}
	n^{p-1}&amp;\ge M n^{p-2}+\frac{(p-2)(m-1)}{2}n^{p-2}\\
	&amp;\ge M n^{p-2}+\frac{(p-2)(m-1)}{2}Mn^{p-3}+\left(\frac{(p-2)(m-1)}{2}\right)^2n^{p-3}\\ 
	&amp;\,\,\,\vdots\\
	&amp;\ge M \sum_{j=0}^{p-3}\left(\frac{(p-2)(m-1)}{2}\right)^j n^{p-2-j} +                                         \left(\frac{(p-2)(m-1)}{2}\right)^{p-2}n\\
	&amp;\ge M \sum_{j=0}^{p-2}\left(\frac{(p-2)(m-1)}{2}\right)^j n^{p-2-j}.  
	\end{align*} 
	Since $(p-2)^j\ge {p-2 \choose j}$ for all $0\le j\le p-2$,
	$$n^{p-1}\ge M \sum_{j=0}^{p-2}{p-2 \choose j}\left(\frac{m-1}{2}\right)^j n^{p-2-j} = M \left(n+\frac{m-1}{2}\right)^{p-2}$$ as desired.
\end{proof}

Now we are ready to state and prove the main result.

\begin{theorem}   
	For every positive integer $n\ge M+\frac{(p-2)(m-1)}{2},$
    $$\lfloor\sqrt[p]{n}+\sqrt[p]{n+1}+\dots+\sqrt[p]{n+m-1}\rfloor = \left\lfloor\sqrt[p]{m^pn+\frac{m^p(m-1)}{2}-1}\right\rfloor.$$
\end{theorem}
\begin{proof} Let 
	$$S=\sqrt[p]{n}+\sqrt[p]{n+1}+\sqrt[p]{n+2}+\cdots+\sqrt[p]{n+m-1}.$$
	One obtains
	$$S^p=\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}\sqrt[p]{(n+i_1)\cdots(n+i_p)}=\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}G_p.$$
	By the AM-GM-HM inequality, 
	$$\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}A_p-\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1} (A_p-H_p) &lt; S^p &lt; \sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1} A_p.$$
	Observe that
	\begin{align*}
	\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}A_p&amp;=\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}n+\sum_{i_1=0}^{m-1}\dots\sum_{i_p=0}^{m-1}\left(\frac{1}{p}\sum_{j=1}^pi_j\right)\\ 
	&amp;= m^pn + \frac{m^p(m-1)}{2}.
	\end{align*}
	Moreover, by Lemma \ref{lem1} and Lemma \ref{lem2}, we have
	$$\sum_{i_1=0}^{m-1}\cdots\sum_{i_p=0}^{m-1}\left(A_p-H_p\right)&lt;\frac{M}{n^{p-1}}\left(n+\frac{m-1}{2}\right)^{p-2}\le 1.$$ 
	Thus
	$$m^pn+\frac{m^p(m-1)}{2}-1&lt;S^p&lt;m^pn+\frac{m^p(m-1)}{2}$$
	holds for all positive integers $n\ge M+\frac{(p-2)(m-1)}{2}$. Equivalently, $$\lfloor\sqrt[p]{n}+\sqrt[p]{n+1}+\dots+\sqrt[p]{n+m-1}\rfloor = \left\lfloor\sqrt[p]{m^pn+\frac{m^p(m-1)}{2}-1}\right\rfloor$$ 
	holds for all positive integers $n\ge M+\frac{(p-2)(m-1)}{2}$. 	
\end{proof}

The table below compares the lower bounds of $n$ between Saltzman and Yuan's work, $N_{SY}=\frac{m^p(m-1)(2m-1)(p-1)}{12p}+1$, and our work, $N=\frac{(p-1)m^p(m^2-1)}{12p}+\frac{(p-2)(m-1)}{2}$, for small values of $p$ and $m$.\\

\begin{tabular}{ c|c|c||c|c||c|c||c|c||}
	\cline{2-9}
	&amp;\multicolumn{2}{c||}{$p=3$} &amp;\multicolumn{2}{c||}{$p=4$}  &amp;\multicolumn{2}{c||}{$p=5$}  &amp;\multicolumn{2}{c||}{$p=6$}\\
	\hline
	\multicolumn{1}{|c|}{$m$}   &amp;$N_{SY}$ &amp;$N$  &amp;$N_{SY}$ &amp;$N$  &amp;$N_{SY}$ &amp;$N$  &amp;$N_{SY}$ &amp;$N$\\
	\hline
	\multicolumn{1}{|c|}{$2$}   &amp;$2$     &amp;$2$     &amp;$4$      &amp;$4$      &amp;$7$        &amp;$8$       &amp;$14$        &amp;$16$\\
	\multicolumn{1}{|c|}{$3$}   &amp;$16$    &amp;$13$    &amp;$51$     &amp;$43$     &amp;$163$      &amp;$133$     &amp;$507$       &amp;$409$\\
	\multicolumn{1}{|c|}{$4$}   &amp;$75$    &amp;$55$    &amp;$337$    &amp;$243$    &amp;$1434$     &amp;$1029$    &amp;$5974$      &amp;$4273$\\
	\multicolumn{1}{|c|}{$5$}   &amp;$251$   &amp;$169$   &amp;$1407$   &amp;$942$    &amp;$7501$     &amp;$5006$    &amp;$39063$     &amp;$29050$\\
	\multicolumn{1}{|c|}{$6$}   &amp;$661$   &amp;$423$   &amp;$4456$   &amp;$2840$   &amp;$28513$    &amp;$18152$   &amp;$178201$    &amp;$113410$\\   
	\multicolumn{1}{|c|}{$7$}   &amp;$1487$  &amp;$918$   &amp;$11705$  &amp;$7209$   &amp;$87397$    &amp;$53792$   &amp;$637266$    &amp;$392176$\\ 
	\multicolumn{1}{|c|}{$8$}   &amp;$2587$  &amp;$1796$  &amp;$26881$  &amp;$16135$  &amp;$229377$   &amp;$137637$  &amp;$1911467$   &amp;$1146894$\\
	\multicolumn{1}{|c|}{$9$}   &amp;$5509$  &amp;$3244$  &amp;$55769$  &amp;$32813$  &amp;$535378$   &amp;$314940$  &amp;$5019166$   &amp;$2952466$\\
	\multicolumn{1}{|c|}{$10$}  &amp;$9501$  &amp;$5505$  &amp;$106876$ &amp;$61884$  &amp;$1140001$  &amp;$660014$  &amp;$11875001$  &amp;$6875018$\\
	\hline           
\end{tabular}\\

\section{Acknowledgments}
The authors greatly appreciate the anonymous referee for his/her careful reading and helpful comments. 

\begin{thebibliography}{9}
\bibitem{ha}
F. D. Hammer, Problem E3010, {\it Amer. Math. Monthly} {\bf 95} (1988), 133--134.
\bibitem{kr}
K. Kuhapatanakul and P. Ruankong, Sums of square roots, {\it Math. Gazette}
{\bf 101} (2017), 276--280.
\bibitem{sy}
P. W. Saltzman and P. Yuan, On sums of consecutive integral roots,
{\it Amer. Math. Monthly} {\bf 115} (2008), 254--261.
\bibitem{wa}
Z. Wang, {\it A City of Nice Mathematics} (in Chinese), The Democracy
and Construction Press, Beijing, 2000.

\bibitem{zh} X. Zhan, Formulae
for sums of consecutive square roots, {\it Math. Intelligencer} {\bf
27} (2005), 4--5.

\end{thebibliography}

\bigskip
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\noindent 2010 {\it Mathematics Subject Classification} :  Primary 11A99; Secondary 11Y55.

\noindent \emph{Keywords: } sum of roots.

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\vspace*{+.1in}
\noindent
Received April 30 2019;
revised version received September 2 2019.
Published in {\it Journal of Integer Sequences}, September 25 2019.

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\noindent
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