On the denseness of certain reciprocal power sums

1.   Introduction

Let $\mathbb{Z}$, $\mathbb{Z}^+$ and $\mathbb{Q}$ be the set of integers, the set of positive integers and the set of rational numbers, respectively. Let $n\in \mathbb{Z}^+$. In 1915, Theisinger ^[9] showed that the $n$-th harmonic sum $1+\frac{1}{2}+...+\frac{1}{n}$ is never an integer if $n > 1$. In 1923, Nagell ^[8] extended Theisinger's result by showing that if $a$ and $b$ are positive integers and $n\ge 2$, then the reciprocal sum $\sum_{i = 0}^{n-1}\frac{1}{a+bi}$ is never an integer. Erdős and Niven ^[2] generalized Nagell's result by considering the integrality of the elementary symmetric functions of $\frac{1}{a}, \frac{1}{a+b}, ..., \frac{1}{a+(n-1)b}$. In the recent years, Erdős and Niven's result ^[2] was extended to the general polynomial sequence, see ^[1], ^[4], ^[7], ^[10] and ^[11]. Another interesting and related topic is presented in ^[12].

By $(\mathbb{Z}^+)^{\infty}$ we denote the set of all the infinite sequence $\{s_i\}_{i = 1}^{\infty}$ of positive integers (note that all the $s_i$ are not necessarily distinct and not necessarily monotonic). For any given $\mathcal{S} = \{s_i\}_{i = 1}^{\infty}\in (\mathbb{Z}^+)^{\infty}$, we let $\mathcal{S}_n: = \{s_1, ..., s_n\}.$ Associated to the infinite sequence $\mathcal{S}$ of positive integers and a polynomial $f(x)$ of nonnegative integer coefficients, one can form an infinite sequence $\{H_f(\mathcal{S}_n)\}_{n = 1}^{\infty}$ of positive rational fractions with $H_f(\mathcal{S}_n)$ being defined as follows:

Feng, Hong, Jiang and Yin ^[3] showed that when $f(x)$ is linear, the reciprocal power sum $H_f(\mathcal{S}_n)$ is never an integer if $n\ge 2$. Associated to any given infinite sequence $\mathcal{S}$ of positive integers, we let

and

Put

Note that $\alpha_f$ may be $+\infty$. Then $\alpha_f(\mathcal{S})\le\alpha_f$ and $H_f(\mathcal{S})\subseteq (\inf H_f(\mathcal{S}), \alpha_f(\mathcal{S}))$. It is clear that $H_f(\mathcal{S})$ is not dense (nowhere dense) in the interval $(\inf H_f(\mathcal{S}), \alpha_f(\mathcal{S}))$. However, if we put all the sets $H_f(\mathcal{S})$ together, then one arrives at the following interesting dense result that is the main result of this paper.

Theorem 1.1. Let $f(x)$ be a polynomial of nonnegative integer coefficients and let $U_f$ be the union set defined by

(ⅰ). If $\deg f(x) = 1$, then $U_f$ is dense in the interval $(\delta, +\infty)$ with $\delta: = 1$ if $f(x) = x$, and $\delta: = 0$ otherwise.

(ⅱ). If $\deg f(x)\ge 2$ and

holds for all positive integers $k$, then $U_f$ is dense in the interval $(0, \alpha_f)$ with $\alpha_f$ being given in (1).

It is well known that (see, for instance, ^[6])

Furthermore, $\alpha\approx 1.076674$. Evidently, for any positive integer $n$, we have

One can easily check that (2) is true when $f(x) = x^2+1$. So Theorem 1.1 infers that for any sufficiently small $\varepsilon > 0$, there are positive integers $n_1$ and $n_2$ and infinite sequences $\mathcal{S}^{(1)}$ and $\mathcal{S}^{(2)}$ of positive integers such that $1-\varepsilon < H_{x^2+1}(\mathcal{S}^{(1)}_{n_1}) < 1$ and $1 < H_{x^2+1}(\mathcal{S}^{(2)}_{n_2}) < 1+\varepsilon$. But it is unclear whether $H_{x^2+1}(\mathcal{S}_n)$ can take 1 as its value. We guess that the answer to this question is negative.

This paper is organized as follows. First, in Section 2, we recall the results due to Kakeya ^[5], and then show some preliminary lemmas which are needed in the proof of Theorem 1.1. Then in Section 3, we supply the proof of Theorem 1.1. The final section is devoted to some remarks. Actually, two conjectures are proposed there.

2.   Auxiliary lemmas

In this section, we present several auxiliary lemmas that are needed in the proof of Theorem 1.1. Now let us state a result obtained by Kakeya in 1914.

Lemma 2.1. ^[5] Let $\sum_{k = 1}^{\infty}a_k$ be an absolutely convergent infinite series of real numbers and let the set, denoted by $SPS$, of all the partial sums of the series $\sum_{k = 1}^{\infty}a_k$ be defined by

Let $u: = \inf SPS$ and $v: = \sup SPS$ (note that $u$ may be $-\infty$ and $v$ may be $+\infty$). Then the set $U$ consists of all the values in the interval $(u, v)$ if and only if

holds for all $k\in \mathbb{Z}^+$.

Using Lemma 2.1, we can prove the following two useful results that play key roles in the proof of Theorem 1.1.

Lemma 2.2. Let $\sum_{k = 1}^{\infty}a_k$ be a convergent infinite series of positive real numbers and

If

holds for all $k\in \mathbb{Z}^+$, then the set $V$ is dense in the interval $(0, v)$ with $v: = \sum_{k = 1}^{\infty}a_k$.

Proof. From the condition (4) and Lemma 2.1, we know that the set

consists of all the values in the interval $(0, v)$ since here $\inf SPS = 0$. Let $r$ be any given real number in $(0, v)$ and $\varepsilon$ be any sufficiently small positive number (one may let $\varepsilon < \min(r, v-r)$). Then $r\in SPS$ which implies that there is an integer $m\in\mathbb{Z}^+\cup \{ \infty \}$ and there are $m$ integers $k_1, ..., k_m$ with $1\le k_1 < ... < k_m$ such that $r = \sum_{i = 1}^{m}a_{k_i}$.

If $m\in\mathbb{Z}^+$, then $r\in V$. So Lemma 2.2 is true in this case.

If $m = \infty$, then $r = \sum_{i = 1}^{\infty}a_{k_i}$. That is, ${\rm limit}_{n\rightarrow \infty}\sum_{i = 1}^n a_{k_i} = r$. Thus there is a positive integer $m'$ such that $|r-\sum_{i = 1}^{m'}a_{k_i}| < \varepsilon$. Noticing that all $a_{k_i}$ are positive, we deduce that $r-\varepsilon < \sum_{i = 1}^{m'}a_{k_i} < r$ as desired.

This completes the proof of Lemma 2.2.

Lemma 2.3. Let $\sum_{k = 1}^{\infty}a_k$ be a divergent infinite series of positive real numbers with $a_k$ decreasing as $k$ increasing and $a_k\rightarrow 0$ as $k\rightarrow\infty$. Define

Then the set $V$ is dense in the interval $(0, +\infty)$.

Proof. Let $r$ be any given real number in $(0, +\infty)$ and $\varepsilon$ be any sufficiently small positive number (one may let $\varepsilon < r$). Let $a_0: = 0$ and $m_0 = 0$. Since the series $\sum_{k = 0}^{\infty}a_k$ is divergent, there exists a unique integer $m_1\ge 0$ such that

and

On the one hand, since $a_k$ decreases as $k$ increases and $a_k\rightarrow 0$ as $k\rightarrow\infty$, there is an integer $m_2$ with $m_2 > m_1+1$ and

Moreover, there exists an integer $m_3$ satisfying that $m_3\ge m_2$ and

and

since $\sum_{k = m_2}^{\infty}a_k$ also diverges.

Continuing in this way, we can form an increasing sequence $\{m_k\}_{k = 0}^{\infty}$ such that

but

for any nonnegative integer $t$. Obviously, one has

On the other hand, since $\lim_{k\rightarrow+\infty}a_k = 0$, it follows that there exists a nonnegative integer $t_0$ such that $a_{m_{2t_0+1}+1} < \varepsilon$. That is, we have

Hence $V$ is dense in the interval $(0, +\infty)$.

This concludes the proof of Lemma 2.3.

3.   Proof of Theorem 1.1

In the section, we present the proof of Theorem 1.1.

Proof of Theorem 1.1. Let

and

Pick any given real number $r$ in $(\inf U_f, \sup U_f)$ and let $\varepsilon$ be any sufficiently small positive number (one may let $\varepsilon < \min(r-\inf U_f, \sup U_f-r)$).

(ⅰ). Since $f(x)$ is a polynomial of nonnegative integer coefficients and degree one, it follows that $\sum_{k = 1}^{\infty}\frac{1}{f(k)}$ (resp. $\sum_{k = 2}^{\infty}\frac{1}{f(k)}$) is a divergent infinite series of positive real numbers with $\big\{\frac{1}{f(k)}\big\}_{k = 1}^{\infty}$ (resp. $\big\{\frac{1}{f(k)}\big\}_{k = 2}^{\infty}$) directly decreasing to $0$ as $k$ increases. By Lemma 2.3, we know that $V_f$ (resp. $\bar V_f$) is dense in the interval $(0, +\infty)$. Clearly, we have $\sup U_f = \sup V_f = +\infty$.

If $f(1) = 1$, then $f(x) = x$ which implies that $f(2) > 1$, $\inf U_f = 1$ and $r\in (\inf U_f, \sup U_f) = (1, +\infty)$. Since $\bar V_f$ is dense in the interval $(0, +\infty)$, there is an element

with $2\le k_1 < \dots < k_m$. Now let $s_k = 1$ for $k\in \{k_1, \dots, k_m\}$ and $s_k > \frac{\log\frac{2k_m}{\varepsilon}}{\log f(2)}$ for $k\in \{2, 3, \dots, k_m\}\setminus \{k_1, \dots, k_m\}$. Then

It follows from (5) and (6) that

That is, $U_f$ is dense in the interval $(\inf U_f, \sup U_f) = (1, +\infty)$ in this case.

If $f(1) > 1$, then $\inf U_f = 0$ and $r\in (\inf U_f, \sup U_f) = (0, +\infty)$. Since $V_f$ is dense in the interval $(0, +\infty)$, there is an element

with $1\le k_1 < \dots < k_m$. Now, let $s_k = 1$ for $k\in \{k_1, \dots, k_m\}$ and $s_k > \frac{\log\frac{2k_m}{\varepsilon}}{\log f(1)}$ for $k\in \{1, 2, \dots, k_m\}\setminus \{k_1, \dots, k_m\}$. One has

and so by (7) and (8),

Namely, $U_f$ is dense in the interval $(\inf U_f, \sup U_f) = (0, +\infty)$ in this case.

(ⅱ). First of all, since $f(x)$ is a polynomial of nonnegative integer coefficients and $\deg f(x)\ge 2$, we know that $\sum_{k = 1}^{\infty}\frac{1}{f(k)}$ is a convergent infinite series of positive real numbers. With the hypothesis $\frac{1}{f(k)}\le\sum_{i = 1}^\infty\frac{1}{f(k+i)}$ for any positive integer $k$, Lemma 2.2 yields that $V_f$ is dense in the interval $(0, \sup V_f)$.

We claim that $f(1) > 1$. Otherwise, $f(1) = 1$. Then $f(x) = x^m$ with $m\ge 2$. However,

which contradicts with our hypothesis. So we must have $f(1) > 1$. The claim is proved.

In the following, we let $f(1) > 1$. Then $\inf U_f = 0$, $\sup U_f = \sup V_f = \alpha_f$ and $r\in (\inf U_f, \sup U_f) = (0, \alpha_f)$. Since $V_f$ is dense in the interval $(0, \sup V_f) = (0, \alpha_f)$, there is an element

with $1\le k_1 < \dots < k_m$. Then letting $s_k = 1$ for $k\in \{k_1, \dots, k_m\}$ and $s_k > \frac{\log\frac{2k_m}{\varepsilon}}{\log f(1)}$ for $k\in \{1, 2, \dots, k_m\}\setminus \{k_1, \dots, k_m\}$ gives us that

It infers that

In other words, $U_f$ is dense in the interval $(0, \alpha_f)$. So part (ⅱ) is proved.

The proof of Theorem 1.1 is complete.

4.   Final remarks

We let $f(x)$ be a polynomial of nonnegative integer coefficients and of degree at least two, and let $U_f$ be the union set given in Theorem 1.1. Then part (ii) of Theorem 1.1 says that the condition (2) is a sufficient condition such that the union set $U_f$ is dense in the interval $(0, \alpha_f)$. One may ask the following interesting question: What is the sufficient and necessary condition on $f(x)$ for the union set $U_f$ to be dense in the interval $(0, \alpha_f)$? We propose the following conjecture to answer this problem.

Conjecture 4.1. Let $f(x)$ be not a monomial and be a polynomial of nonnegative integer coefficients and of degree at least two. Then the set $U_f$ is dense in the interval $(0, \alpha_f)$ if and only if the following inequality holds:

By Theorem 1.1, one knows that for any sufficiently small $\varepsilon > 0$, there are positive integers $n_1$ and $n_2$ and infinite sequences $\mathcal{S}^{(1)}$ and $\mathcal{S}^{(2)}$ of positive integers such that $1-\varepsilon < H_{x^2+1}(\mathcal{S}^{(1)}_{n_1}) < 1$ and $1 < H_{x^2+1}(\mathcal{S}^{(2)}_{n_2}) < 1+\varepsilon$. But it is not clear whether $H_{x^2+1}(\mathcal{S}_n)$ can take 1 as its value. We believe that the answer to this question is negative. As the conclusion of this paper, we suggest the following conjecture.

Conjecture 4.2. Let $f(x)$ be a polynomial of integer coefficients satisfying that $f(m)\ne 0$ for any positive integer $m$ and $\mathcal{S} = \{s_i\}_{i = 1}^\infty$ be an infinite sequence of positive integers (not necessarily increasing and not necessarily distinct). Then there is a positive integer $N$ such that for any integer $n$ with $n\ge N$, $H_f(\mathcal{S}_n)$ is not an integer. In particular, for any positive integer $n$, $H_{x^2+1}(\mathcal{S}_n)$ is never equal to 1.

Acknowledgments

This work was supported partially by National Science Foundation of China Grant # 11771304 and by the Fundamental Research Funds for the Central Universities.

Conflict of interest

We declare that we have no conflict of interest.