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\newtheorem{corl}{Corollary}
\newtheorem*{ackn}{Acknowledgement}

\theoremstyle{namedthrm}
\newtheorem{thrm}{Theorem}

\title{Irrational Sums}
\author{Tom C. Brown, D.-Y. Pei and Peter Jau-Shyong Shiue}
\maketitle
\begin{center}{\small {\bf Citation data:} T.C. {Brown} and P.J.-S. {Shiue}, \emph{Irrational sums}, Rocky Mountain J. Math. \textbf{25} (1995),
  1219--1223.}\bigskip\end{center}

\section{Introduction\label{s1}}

In this note we give some sufficient conditions for the irrationality of
the sum of the series $\sum_{n=1}^\infty 1/H(f(n))$, where $H(k))_{k\geq0}$
is a sequence of integers, positive from some point on, satisfying a
homogeneous linear recurrence relation with integer coefficients, and
$f$ is a strictly increasing function from the set of positive integers
to the set of nonnegative integers.

We will refer to such a sequence $(H(k))_{k\geq0}$ simply as a ``recurrent
sequence,'' and the symbol $f$ will always denote a strictly increasing
function from the set of positive integers to the set of nonnegative integers.

Let us agree that the symbol $\sum1/H(f(n))$ denotes the summation of all
those terms $1/H(f(n))$ for which $H(f(n))>0$.

All of our results are based on the following theorem of C. Badea
\cite{badea1987}.

\begin{thrm}[A]\namedthmlabel{tA} (Badea \cite{badea1987}). If $(a_k)_{k\geq0}$
is a sequence of positive integers such that $a_{k+1}>a_k^2 - a_k+1$ for all
sufficiently large $k$, then $\sum1/a_k$ is irrational.\end{thrm}

A simple example to show that the converse of Badea's Theorem \ref{tA} is
false is the series $\sum1/n! = e$. Another easy example to see that the
converse of Badea's result is false is the following. Let $\{c_n\}$, $n\geq1$,
be a nonperiodic sequence of 2's and 5's, and let $a_n = 10^n/c_n$, $n\geq1$.
Then $\sum1/a_n$ is irrational, and $a_{n+1} < a_n^2 - a_n + 1$, $n\geq3$.

Thus our goal is to find simple conditions on $H(k)$ and $f(n)$ which ensure
that $H(f(n+1)) > H(f(n))^2 - H(f(n)) + 1$ for all sufficiently large $n$.

To avoid complications, \emph{from now on we will always assume that the
characteristic polynomial of the recurrent sequence $H(k)$ has a unique
(real) root $\beta>1$ of maximum modulus}.

It then follows from standard properties of recurrence relations (see,
for example, \cite{liu1985}) that there exist numbers $A>0$ and $c\geq0$
such that $\lim_{k\to\infty} H(k)/(k^c\beta^k) = A$. (If $\beta$ is a
root of multiplicity 1, then $c=0$.)

\section{Main results.}

\begin{thrm}[1]\namedthmlabel{t1} If $f(n+1)-2f(n)\to\infty$ as $n\to\infty$
and $f(n+1)\geq f(n)^2$ for all sufficiently large $n$, then $\sum1/H(f(n))$
is irrational for every recurrent sequence $H(k)$.\end{thrm}
\begin{proof} Assume that $H(k)/k^c\beta^k\to A$ as $k\to\infty$ (where
$\beta>1$, $A>0$ and $c\geq0$). To apply Badea's result, we need to show
that $H(f(n+1))/(H(f(n))^2 - H(f(n))+1)>1$ for sufficiently large $n$.
We do this by dividing the numerator and denominator of the lefthand side
of this inequality by $f(n+1)^c\beta^{f(n+1)}$.

Since $H(f(n+1))/f(n+1)^c\beta^{f(n+1)}\to A > 0$ as $n\to\infty$, then
$H(f(n+1))/f(n+1)^c\beta^{f(n+1)} > (2/3)A$ for all sufficiently large
$n$.

Next,
\begin{align*}
\frac{H(f(n))^2 - H(f(n)) + 1}{f(n+1)^c\beta^{f(n+1)}}
&= \frac{f(n)^{2c}}{f(n+1)^c}\frac1{\beta^q}\left(\frac{H(f(n))^2}{f(n)^{2c}\beta^{2f(n)}} - \frac{H(f(n))}{f(n)^{2c}\beta^{2f(n)}}\right) \\
&\qquad + \frac1{f(n+1)^c\beta^{f(n+1)}},
\end{align*}
where $q = f(n+1)-2f(n)$. Since the expression inside the large brackets
converges to $A^2$ and the other term converges to 0, for sufficiently
large $n$ (using also $f(n)^{2c}/f(n+1)^c\leq1$)
\[ \frac{H(f(n))^2 - H(f(n)) + 1}{f(n+1)^c\beta^{f(n+1)}} < \beta^{-q}(A^2 + 1) + (1/3)A. \]

Finally,
\[ \frac{H(f(n+1))}{H(f(n))^2 - H(f(n)) + 1} > \frac{(2/3)A}{\beta^{-q}(A^2 + 1) + (1/3)A} > 1, \]
as required.\end{proof}

\begin{corl} For every recurrent sequence $H(k)$, $\sum1/H(2^{2^n})$ is
irrational.\end{corl}

For the next result, we weaken the condition on $f$ and strengthen the
condition on $H(k)$.

\begin{thrm}[2]\namedthmlabel{t2} If $f(n+1)-2f(n)\to\infty$ as $n\to\infty$,
then $\sum1/H(f(n))$ is irrational for every recurrent sequence $H(k)$ for
which $\beta$ has multiplicity 1. (Recall that $\beta>1$ is the unique root
of maximum modulus of the characteristic polynomial of $H(k)$.)\end{thrm}
\begin{proof} The proof of Theorem \ref{t1}, with $c$ set equal to 0
throughout, gives a proof of Theorem \ref{t2}.\end{proof}

\begin{corl}Let $H(k)$ be a recurrent sequence for which $\beta$ has multiplicity
1. Then for every $\epsilon>0$, $\sum1/H([(2+\epsilon)^n])$ is irrational. For
every $0<\epsilon<1$, $\sum1/H(2^n - [(2-\epsilon)^2])$ is irrational.\end{corl}

\begin{thrm}[3]\namedthmlabel{t3} Let $H(k)$ be a recurrent sequence for
which $\beta$ has multiplicity 1. Then there exists an integer $P$ such that
for every pair of fixed integers $s,p$ with $s>0$, $-\infty<p\leq P$,
$\sum 1/H(s2^n+p)$ is irrational.\end{thrm}
\begin{proof} Assume that $H(k)/\beta^k\to A$ as $k\to\infty$, where $\beta>1$
and $A>0$. Let $s,p$ be given with $s>0$ and $p<-\log A/\log\beta$. Let $f(n)
= s2^n+p$, $n\geq1$. Since $f(n+1) - 2f(n) = -p$,
\[ \frac{H(f(n))^2 - H(f(n)) + 1}{\beta^{f(n+1)}}
= \frac{1}{\beta^{-p}}\left(\frac{H(f(n))^2}{\beta^{2f(n)}} - \frac{H(f(n))}{\beta^{2f(n)}}\right)
+ \frac1{\beta^{f(n+1)}}\to \beta^pA^2. \]
Thus, since $H(f(n+1))/\beta^{f(n+1)}\to A$,
\[ \frac{H(f(n+1))}{H(f(n))^2 - H(f(n)) + 1} \to \frac1{\beta^pA}. \]
Since $\beta^pA < 1$ by the choice of $p$,
\[ \frac{H(f(n+1))}{H(f(n))^2 - H(f(n)) + 1} > 1 \]
for sufficiently large $n$, and therefore $1/\sum H(f(n)) = \sum1/H(s2^n+p)$
is irrational, by Badea's theorem.\end{proof}

\section{Remarks\label{s3}}

For the Fibonacci sequence $F(k)$, where
\[ F(0) = 0,\qquad F(1) = 1,\qquad F(k+2) = F(k+1) + F(k), ~~ k\geq0 \]
\[ F(k) = (1/\sqrt5)(((1 + \sqrt5)/2)^k - ((1 - \sqrt5)/2)^k), \]
\[ \beta = (1 + \sqrt5)/2, \qquad A = 1/\sqrt5, \]
$-\log A/\log\beta = 1.67\ldots$. Thus, according to the proof of Theorem
\ref{t3}, $\sum1/F(s2^n+p)$ is irrational for every fixed pair of integers
$s>0$ and $p\leq1$. This is a generalization of a result of C. Badea
\cite{badea1987}, who showed, answering a question of Erd\H{o}s and Graham
\cite{erdos+graham1980}, that $\sum1/F(2^n+1)$ is irrational.

More generally, let $H(0)=0$, $H(1)=1$, $H(k+2) = aH(k+1) + bH(k)$, $k\geq0$,
where $a\geq1$, $b\geq1$. Then $H(k) = (1/\sqrt{a^2 + 4b})(((a + \sqrt{a^2+4b})
/2)^k - ((a - \sqrt{a^2 + 4b})/2)^k)$, $\beta = (a + \sqrt{a^2 + 4b})/2$,
$A = 1/\sqrt{a^2 + 4b}$, and $\beta^pA < 1$ for $p\leq 1$, so again $\sum
1/H(s2^n+p)$ is irrational for every fixed pair of integers $s>0$ and $p\leq1$.
This extends a result of Kuipers \cite{kuipers1977} (see also
\cite{laohakosol+roenrom1984}), who showed this in the case $b=1$ and $p=0$.
(One can relax the requirement $a\geq1$, $b\geq1$ to $a=1$, $b\geq1$ or
$a\geq2$, $a^2 + 4b >0$. In these cases $A<1<\beta$, so that $\beta^pA < 1$
holds for $p\leq0$ and $\sum1/H(s2^n+p)$ is irrational for $s>0$ and $p\leq0$.)

If $a^2+4b<0$, so that the characteristic polynomial $x^2-ax-b$ of the
sequence $H(k)$ no longer has a unique root of maximum modulus, it is
easy to verify that the sequence $H(k)$ has infinitely many negative
terms, for any nontrivial initial values $H(0)$, $H(1)$. For such a
sequence the present methods give no information about the irrationality
of $\sum1/H(f(n))$ for any function $f$.

Some examples of polynomials for which $\beta>1$ and $b$ has multiplicity
1 ($\beta$ is the unique root of maximum modulus for the given polynomial)
are discussed in Hua and Wang \cite{hua+wang1981}, including the polynomials
$x^d-x^{d-1}-\cdots-x-1$, $d\geq2$, (which come from the generalized
Fibonacci sequences $F(0) = F(1) = \cdots F(d-2) = 0$, $F(d-1) = 1$,
$F(k+d) = F(k+d-1) + F(k+d-2) + \cdots + F(k+1) + F(k)$, $k\geq0$), $x^d
- Lx^{d-1}-1$, $d\geq2$, $L\geq2$, and $x^t - t^2r^{t-1}x^{t-1} + (-1)^{t-2}
A_{t-2}r^{t-2}x^{t-2} + \cdots - A_1rx - 1 = 0$, $t\geq2$, where
\[ A_1 = \binom{2t}1,\qquad A_k = \binom{2t}k - A_1\binom{2t-2}{k-1} - \cdots - A_{k-1}\binom{2t-2k+2}1, \]
$t - 2\geq k>1$, and the positive integer $r$ satisfies $t^2>2/r^{t-1} +
|A_1|/r^{t-2} + \cdots + |A_{t-2}|/r$.

\begin{ackn} We are grateful to the referee for several helpful remarks.
\end{ackn}


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