Numbers Whose Powers Are Arbitrarily Close to Integers

Abstract

In this paper, it is proved that, for any sequence of positive numbers ${\xi }_n$, $n=1,2,\dots$, which does not converge to zero faster than the exponential function, and any sequence of positive numbers ${\delta }_n$, $n=1,2,3,\dots$, there is an uncountable set of positive numbers S such that, for each $\alpha >1$ in S, there are infinitely many $n\in \mathbb{N}$ for which the fractional parts $\left\{{\xi }_n{\alpha }^n\right\}$ are smaller than ${\delta }_n$, regardless of how fast the sequence ${\delta }_n$ tends to zero. In particular, for any sequence bounded away from zero, namely, ${\xi }_n\ge \xi >0$ for $n\ge 1$, it is shown that infinitely many integers n for which the inequality $\left\{{\xi }_n{\alpha }^n\right\}<{\delta }_n$ is true can be extracted from an arbitrary subsequence $\mathcal{N}$ of positive integers.

1. Introduction

Starting from Weyl’s 1916 paper [1], various problems related to the distribution of the sequence of fractional parts

$$\left\{\xi {\alpha }^n\right\},n=1,2,3,\dots ,$$

(1)

where $\xi >0$ and $\alpha >1$ are two real numbers, were studied. Weyl’s result implies that, for each $\alpha >1$, the sequence (1) is uniformly distributed for almost all $\xi >0$. See also [2] for a more precise version of this result. In the opposite direction, Koksma [3] proved that, if $\xi >0$ is fixed, then the sequence (1) is uniformly distributed for almost all $\alpha >1$. In this respect, the exceptional $\alpha$ are Pisot and Salem numbers. Recall that an algebraic integer $\alpha >1$ is called a Pisot number if its conjugates over $\mathbb{Q}$ other than $\alpha$ itself (if any) all lie in the open unit disc $\left|z\right|<1$. An algebraic integer $\alpha >1$ is called a Salem number if its degree over $\mathbb{Q}$ is an even number $d\ge 4$ and $d-2$ of its conjugates lie on the unit circle $\left|z\right|=1$. (Since such $\alpha$ is reciprocal its other conjugates are $\alpha$ and ${\alpha }^{-1}$). See, for instance, the paper of Pisot and Salem themselves [4], where they proved that, if $\xi =1$ and $\alpha >1$ is a Salem number, then the sequence (1) is everywhere dense in $[0,1]$, but not uniformly distributed in $[0,1]$. The monographs [5,6] contain some basic information about Pisot and Salem numbers, while in Smyth’s review paper [7] there are more recent references. In some literature, Pisot numbers are also called Pisot–Vijayaraghavan numbers or PV numbers; see, for instance, some early papers of Vijayaraghavan on this subject [8,9,10,11].

For algebraic numbers $\alpha$, at least something can be said about the distribution of (1). Extending an earlier result of Flatto, Lagarias, and Pollington for rational $\alpha >1$ [12], in [13], it was proved that, for each $\xi >0$ and each algebraic number $\alpha >1$, there should be a gap between the largest and the smallest limit points of the sequence (1) which depends only on $\alpha$ except if $\alpha$ is a Pisot number or a Salem number when we need an extra condition $\xi \notin \mathbb{Q}\left(\alpha \right)$. This result is not only of interest itself but also has several applications. In particular, it seems to be useful in a so-called Erdos similarity conjecture [14]. The cases when $\xi \in \mathbb{Q}\left(\alpha \right)$ and $\alpha$ is a Pisot number or a Salem number were treated in [15,16], respectively; see also [17]. Nevertheless, for example, Mahler’s $3/2$-problem [18], where he asks whether, for $\alpha =3/2$, there is a so-called Z-number, namely, $\xi >0$, such that all elements of (1) lie in $(0,1/2)$, is unsolved (see [19,20,21]). However, the situation with any specific transcendental number $\alpha$ is more complicated and less known. For example, it is not known if (1) with $(\xi ,\alpha )=(1,e)$ has one or more than one limit point. Determining whether there is a transcendental number $\alpha >1$ for which the sequence $\left\{{\alpha }^n\right\}$, $n=1,2,3,\dots$, has only finitely many limit points is still a completely open problem as well.

We remark that the behavior of the sequence (1) for $\xi =1$ and a transcendental number $\alpha >1$ can be very different depending on $\alpha$. In [22], it was shown that, for any sequence of real numbers $r_n$, $n=1,2,3,\dots$, and any $\epsilon >0$, there is a transcendental number $\alpha >1$ such that

$$\parallel {\alpha }^n-r_n\parallel <\epsilon$$

(2)

for all $n\in \mathbb{N}$. (Here, $\parallel y\parallel$ is the distance from $y\in \mathbb{R}$ to the nearest integer.) See also two subsequent papers [23,24]. In fact, if we want (2) to hold not for all $n\in \mathbb{N}$, but only for infinitely many n, then this follows from another paper of Koksma [25] with $\epsilon$ replaced by a sequence of positive numbers ${\epsilon }_n$ such that the series ${\sum }_{n=1}^{\infty }{\epsilon }_n$ are divergent. In [22], it was also shown that, for any sequence of positive numbers ${\delta }_n$, $n=1,2,3,\dots$, there is a transcendental number $\alpha >1$ for which the inequality

$$\left\{{\alpha }^n\right\}<{\delta }_n$$

holds for infinitely many $n\in \mathbb{N}$. This time, there are no conditions or restrictions whatsoever on the rate of convergence of ${\delta }_n$ to zero.

In this paper, it will be shown that, even if we replace in (1) a fixed number $\xi >0$ by any sequence of positive numbers ${\xi }_n$, $n=1,2,3,\dots$, which is not converging to zero faster than the exponential function, then there are “many” numbers $\alpha >1$ such that $\left\{{\xi }_n{\alpha }^n\right\}$ is smaller than an arbitrary positive number ${\delta }_n$ for infinitely many $n\in \mathbb{N}$, regardless of how fast the sequence ${\delta }_n$, $n=1,2,3,\dots$, converges to zero. (This type of sequence, specifically, with ${\xi }_n=1/n$ and an integer $\alpha \ge 2$, was considered before; see [26], where their density in $[0,1]$ was established, and [27,28,29].) Of course, the theorem stated below holds in the special case when ${\xi }_n=\xi >0$ for each $n\in \mathbb{N}$ and, more generally, when ${\xi }_n$ is bounded away from zero, namely, ${\xi }_n\ge \xi >0$ for $n\in \mathbb{N}$.

Theorem 1. 

Let $\delta =\{{\delta }_1,{\delta }_2,{\delta }_3,\dots \}$ and $\xi =\{{\xi }_1,{\xi }_2,{\xi }_3,\dots \}$ be two sequences of positive numbers such that

$$\underset{n\to \infty }{lim\; sup}{\displaystyle \frac{log{\xi }_n}{n}}\ge 0.$$

(3)

Then, for any interval $I=[a,b]$, where $1\le a<b$, there is an uncountable set $S(\delta ,\xi ,I)\subset I$ such that, for each $\alpha \in S(\delta ,\xi ,I)$, the inequalities

$$0<\left\{{\xi }_n{\alpha }^n\right\}<{\delta }_n$$

(4)

hold for infinitely many $n\in \mathbb{N}$.

Note that the condition (3) of Theorem 1 cannot be omitted. For example, if $\tau >0$, ${\xi }_n=e^{-\tau n}$ and ${\delta }_n=1/n!$, then, for each $\alpha \in [1,e^{\tau })$ and each sufficiently large $n\in \mathbb{N}$, we have

$$\left\{{\xi }_n{\alpha }^n\right\}=\left\{e^{(log\alpha -\tau )n}\right\}=e^{(log\alpha -\tau )n}>{\displaystyle \frac{1}{n!}}={\delta }_n,$$

so there is no $\alpha$ in the interval $I=[1,e^{\tau })$ for which the inequality (4) is true for infinitely many $n\in \mathbb{N}$. Similarly, if ${\xi }_n={\delta }_n=1/n!$, then there is no $\alpha >1$ at all for which (4) is true for infinitely many $n\in \mathbb{N}$.

Of course, since the set $S(\delta ,\xi ,I)$ is uncountable and $\mathbb{Q}$ is countable, $S(\delta ,\xi ,I)$ contains an uncountable subset of transcendental numbers $\alpha$ with the property (4). Therefore, Theorem 1 is already more general than Theorem 3 of [22] for ${\xi }_n=1$.

Replace each ${\delta }_j$ by

$${\delta }_j^{\prime }:=min(1/j,{\delta }_1,\dots ,{\delta }_j)$$

and set

$${\mathsf{\Phi }}_j=\lceil 1/{\delta }_j^{\prime }\rceil$$

for every $j\in \mathbb{N}$. (Here and below, $\lceil y\rceil$ is the ceiling function, namely, the smallest integer greater than or equal to $y\in \mathbb{R}$.) It is clear then that ${\mathsf{\Phi }}_1,{\mathsf{\Phi }}_2,{\mathsf{\Phi }}_3,\dots$ is an unbounded nondecreasing sequence of positive integers such that each element of the sequence

$$U=\{1/{\mathsf{\Phi }}_1,1/{\mathsf{\Phi }}_2,1/{\mathsf{\Phi }}_3,\dots \}$$

does not exceed the corresponding element of the sequence $\delta$. Therefore, in order to prove Theorem 1, it suffices to show that for each interval $I\subset {\mathbb{R}}_{>1}$, there is an uncountable set of real numbers $S\subset I$ such that, for every $\alpha \in S$, the inequalities

$$0<\left\{{\xi }_n{\alpha }^n\right\}<{\displaystyle \frac{1}{{\mathsf{\Phi }}_n}}$$

(5)

hold for infinitely many $n\in \mathbb{N}$.

We will prove the following more general statement:

Theorem 2. 

Let $\xi =\{{\xi }_1,{\xi }_2,{\xi }_3,\dots \}$ be a sequence of positive numbers satisfying (3), and let ${\mathsf{\Phi }}_1\le {\mathsf{\Phi }}_2\le {\mathsf{\Phi }}_3\le \dots$ be an unbounded sequence of positive integers. Then, for any interval $I=[a,b]$, $1\le a<b$, and any real number $\eta >1$, there is an uncountable set $S(\xi ,\mathsf{\Phi },I,\eta )\subset I$ such that, for each $\alpha \in S(\xi ,\mathsf{\Phi },I,\eta )$, the inequalities

$${\eta }^m<{\xi }_n{\alpha }^n<{\eta }^m+{\displaystyle \frac{1}{{\mathsf{\Phi }}_n}}$$

(6)

hold for infinitely many pairs $(n,m)\in {\mathbb{N}}^2$.

It is clear that (6) implies (5), with, for example, $\eta =2$, so Theorem 2 immediately implies Theorem 1.

We will derive Theorem 2 from the following proposition of independent interest.

Proposition 1. 

Let ${\gamma }_1,{\gamma }_2,{\gamma }_3,\dots$ be a sequence of real numbers, and let $1={\epsilon }_1\ge {\epsilon }_2\ge {\epsilon }_3\ge \dots$ be a sequence of positive numbers. Assume that $I=[a,b]$, where $0\le a<b$, and let $\mathcal{N}$ be an infinite subset of $\mathbb{N}$. Then there is an uncountable set $B\subset I$ such that, for each $\beta \in B$, the inequalities

$$0<\{n\beta -{\gamma }_n\}<{\epsilon }_n$$

(7)

hold for infinitely many $n\in \mathcal{N}$.

Note that, if ${\gamma }_1={\gamma }_2={\gamma }_3=$ … $=0$ and if the sequence ${\epsilon }_n$ tends to zero faster than any constant power of $1/n$, then the numbers $\beta$ satisfying (7) are Liouville numbers. Recall that a Liouville number is a real number whose irrationality exponent is infinite, see p. 248 in [30]. This means that, for any $C>1$, there is a pair of integers $k,n$, where $n>1$, such that

$$0<|\beta -{\displaystyle \frac{k}{n}}|<{\displaystyle \frac{1}{n^C}}.$$

Therefore, Proposition 1 is the construction of uncountably many Liouville type numbers with good approximation not just by rational fractions $k/n$, $k,n\in \mathbb{N}$, but by fractions with “moving numerator” $(k+{\gamma }_n)/n$. The author thanks Prof. Nikolay Moshchevitin for a useful advice towards this construction. Note that the approximation $(k+{\gamma }_n)/n$ to those special Liouville type numbers is with $k\in \mathbb{Z}$ and with n being not just in $\mathbb{N}$ but in any infinite sequence of positive integers $\mathcal{N}$. For instance, $\mathcal{N}$ can be the set of squares or the set of primes.

Next, we will prove Proposition 1 (Section 2) and then derive Theorem 2 from this proposition (Section 3). In Section 4, we will give a stronger version of Theorem 1 under a condition slightly stronger than that in (3). Then, in Section 5, we provide another application of Proposition 1. Section 6 contains some final remarks.

2. Proof of Proposition 1

We begin with the following simple observation:

Lemma 1. 

Let $I=[a,b]$ be a closed real interval with $a<b$, and let $u<v$ be two real numbers. Then, for each sufficiently large positive integer n, there is an integer $k=k\left(n\right)$ such that $(k+u)/n,(k+v)/n\in I$.

Proof. 

Take any integer n satisfying

$$n\ge {\displaystyle \frac{v-u+1}{b-a}}.$$

(8)

Note that $n>0$. Select

$$k=k\left(n\right)=\lceil na-u\rceil .$$

(9)

Then $k\ge na-u$, and hence $a\le (k+u)/n$. Next, from (8) and (9), it follows that

$${\displaystyle \frac{k+v}{n}}<{\displaystyle \frac{na-u+1+v}{n}}\le {\displaystyle \frac{na+n(b-a)}{n}}={\displaystyle \frac{nb}{n}}=b.$$

Therefore, the numbers $(k+u)/n$ and $(k+v)/n$ both belong to the interval I, which completes the proof of the lemma. □

Next, for any sequence of real numbers ${\gamma }_n$, $n=1,2,3,\dots$, and any infinite sequence of positive integers $\mathcal{N}$, we will prove the existence of a real number that is very close to the fraction $(k+{\gamma }_n)/n$ for infinitely many pairs $(n,k)$, where $n\in \mathcal{N}$ and $k\in \mathbb{Z}$.

Lemma 2. 

Let ${\gamma }_1,{\gamma }_2,{\gamma }_3,\dots$ be a sequence of real numbers, and let $1={\epsilon }_1\ge {\epsilon }_2\ge {\epsilon }_3\ge \dots$ be a sequence of positive numbers. Let $I=[a,b]$, where $0<a<b$, and let $\mathcal{N}\subseteq \mathbb{N}$ be infinite. Then, for any sequence $\mathbf{u}=\{u_1,u_2,u_3,\dots \}$, where $u_j\in \{2,4\}$ for each $j\in \mathbb{N}$, there is a positive real number $\beta \left(\mathbf{u}\right)\in I$ and a sequence $n_1<n_2<n_3<\dots$ in $\mathcal{N}$ such that, for every $j\in \mathbb{N}$, we have

$${\displaystyle \frac{{\epsilon }_{n_j}}{u_j+1}}\le \{n_j\beta \left(\mathbf{u}\right)-{\gamma }_{n_j}\}\le {\displaystyle \frac{{\epsilon }_{n_j}}{u_j}}.$$

(10)

Proof. 

We will construct the number $\beta \left(\mathbf{u}\right)$ using the method of nested intervals. Set $I_0=I=[a,b]$. Take the least integer $n_1$ in $\mathcal{N}$ satisfying

$$n_1\ge {\displaystyle \frac{13}{10(b-a)}},$$

(11)

and set

$$u={\gamma }_{n_1}+{\displaystyle \frac{{\epsilon }_{n_1}}{5}}\mathrm{and}v={\gamma }_{n_1}+{\displaystyle \frac{{\epsilon }_{n_1}}{2}}.$$

Note that

$$0<v-u={\displaystyle \frac{3{\epsilon }_{n_1}}{10}}\le {\displaystyle \frac{3}{10}},$$

so $n_1$ chosen in (11) satisfies the inequality (8). Choosing $k_1$ as in (9), namely,

$$k_1=\lceil n_{1}a-{\gamma }_{n_1}-{\epsilon }_{n_1}/5\rceil$$

and applying Lemma 1, we find that both endpoints of the interval

$$J_1=[n_1^{-1}(k_1+{\gamma }_{n_1}+{\epsilon }_{n_1}/5),n_1^{-1}(k_1+{\gamma }_{n_1}+{\epsilon }_{n_1}/2)]$$

belong to the interval $I_0$. Consequently, as $u_1\in \{2,4\}$,

$$I_1=[n_1^{-1}(k_1+{\gamma }_{n_1}+{\epsilon }_{n_1}/(u_1+1)),n_1^{-1}(k_1+{\gamma }_{n_1}+{\epsilon }_{n_1}/u_1)]$$

is its subinterval, so it satisfies $I_1\subset I_0$. Furthermore, for any number $\zeta \in I_1$, we have

$${\displaystyle \frac{{\epsilon }_{n_1}}{u_1+1}}\le n_1\zeta -{\gamma }_{n_1}-k_1\le {\displaystyle \frac{{\epsilon }_{n_1}}{u_1}}.$$

From ${\epsilon }_{n_1}\le 1$ and $u_1\in \{2,4\}$, it follows that $k_1=\lfloor n_1\zeta -{\gamma }_{n_1}\rfloor$ (where $\lfloor y\rfloor$ is the integral part of $y\in \mathbb{R}$), so (10) is true for $j=1$ and any number $\zeta$ from the interval $I_1$.

We now argue by induction on j. Assume that $l\ge 1$ is an integer such that, for $j=1,2,\dots ,l$, there is a nested collection of intervals

$$I_j=[n_j^{-1}(k_j+{\gamma }_{n_j}+{\epsilon }_{n_j}/(u_j+1),n_j^{-1}(k_j+{\gamma }_{n_j}+{\epsilon }_{n_j}/u_j)]=[a_j,b_j]$$

(12)

with uniquely chosen $n_1<n_2<n_3<$ … $<n_l$ in $\mathcal{N}$ and $k_1,\dots ,k_l\in \mathbb{Z}$ such that

$$I_l\subseteq I_{l-1}\subseteq \cdots \subseteq I_1\subseteq I_0=[a,b].$$

For any $\zeta \in I_j$, we clearly have

$${\displaystyle \frac{{\epsilon }_{n_j}}{u_j+1}}\le n_j\zeta -{\gamma }_{n_j}-k_j=\{n_j\zeta -{\gamma }_{n_j}\}\le {\displaystyle \frac{{\epsilon }_{n_j}}{u_j}},$$

so (10) is true for $j=1,2,\dots ,l$ and any number $\zeta$ from the interval $I_l$.

Next, we will show how to choose the interval $I_{l+1}$ of the form

$$I_{l+1}=[n_{l+1}^{-1}(k_{l+1}+{\gamma }_{n_{l+1}}+{\epsilon }_{n_{l+1}}/(u_{l+1}+1)),n_{l+1}^{-1}(k_{l+1}+{\gamma }_{n_{l+1}}+{\epsilon }_{n_{l+1}}/u_{l+1})]$$

(13)

contained in $I_l$, with $k_{l+1}\in \mathbb{Z}$ and $n_{l+1}>n_l$ in $\mathcal{N}$. To this end, we will apply Lemma 1, with $a=a_l$ being the left endpoint of $I_l$, $b=b_l$ being the right endpoint of $I_l$, and the smallest integer $n=n_{l+1}>n_l$ in $\mathcal{N}$ satisfying

$$n_{l+1}\ge {\displaystyle \frac{13}{10(b_l-a_l)}}.$$

(14)

As above, applying Lemma 1 to

$$u={\gamma }_{n_{l+1}}+{\displaystyle \frac{{\epsilon }_{n_{l+1}}}{5}}\mathrm{and}v={\gamma }_{n_{l+1}}+{\displaystyle \frac{{\epsilon }_{n_{l+1}}}{2}},$$

due to $0<v-u\le 3/10$, we can choose an appropriate integer $k_{l+1}$ by (9), namely,

$$k_{l+1}=\lceil n_{l+1}{a}_l-{\gamma }_{n_{l+1}}-{\epsilon }_{n_{l+1}}/5\rceil .$$

(15)

Then, by Lemma 1, both endpoints of the interval

$$J_{l+1}=[n_{l+1}^{-1}(k_{l+1}+{\gamma }_{n_{l+1}}+{\epsilon }_{n_{l+1}}/5),n_{l+1}^{-1}(k_{l+1}+{\gamma }_{n_{l+1}}+{\epsilon }_{n_{l+1}}/2)]$$

belong to $I_l$. Consequently, the subinterval $I_{l+1}$ of $J_{l+1}$, which we defined in (13), satisfies $I_{l+1}\subset J_{l+1}\subseteq I_l$.

By this construction, since the length of $I_j$, namely, ${\epsilon }_{n_j}/\left(u_j(u_j+1)n_j\right)$, tends to zero as $j\to \infty$, the unique point of the intersection ${\cap }_{j=1}^{\infty }{I}_j$ is the required positive real number $\beta \left(\mathbf{u}\right)$. (It is clear that $\beta \left(\mathbf{u}\right)\in I_1\subseteq I_0=[a,b]$.) □

We now show that the numbers $\beta \left(\mathbf{u}\right)$ and $\beta \left({\mathbf{u}}^{\prime }\right)$ are distinct for distinct vectors

$$(u_1,u_2,u_3,\dots )\mathrm{and}(u_1^{\prime },u_2^{\prime },u_3^{\prime },\dots ).$$

Indeed, let ℓ be the smallest positive integer for which $u_{\ell }\ne u_{\ell }^{\prime }$. Without restriction of generality, we may assume that $u_{\ell }=2$ and $u_{\ell }^{\prime }=4$. Since $(u_1,\dots ,u_{\ell -1})=(u_1^{\prime },\dots ,u_{\ell -1}^{\prime })$, the intervals $I_j$ and $I_j^{\prime }$ constructed in (12) are the same for $j=1,2,\dots ,\ell -1$. Furthermore, the integers $n_{\ell }$ and $k_{\ell }$ are also the same. (In view of (14) and (15), they do not depend on $u_{\ell }$.) Therefore, by (12) and $(u_{\ell },u_{\ell }^{\prime })=(2,4)$, we find that

$$I_{\ell }=[n_{\ell }^{-1}(k_{\ell }+{\gamma }_{n_{\ell }}+{\epsilon }_{n_{\ell }}/3),n_{\ell }^{-1}(k_{\ell }+{\gamma }_{n_{\ell }}+{\epsilon }_{n_{\ell }}/2)]$$

and

$$I_{\ell }^{\prime }=[n_{\ell }^{-1}(k_{\ell }+{\gamma }_{n_{\ell }}+{\epsilon }_{n_{\ell }}/5),n_{\ell }^{-1}(k_{\ell }+{\gamma }_{n_{\ell }}+{\epsilon }_{n_{\ell }}/4)].$$

Note that the intervals $I_{\ell }$ and $I_{\ell }^{\prime }$ are disjoint. Since $\beta \left(\mathbf{u}\right)\in I_{\ell }$ and $\beta \left({\mathbf{u}}^{\prime }\right)\in I_{\ell }^{\prime }$, the numbers $\beta \left(\mathbf{u}\right)$ and $\beta \left({\mathbf{u}}^{\prime }\right)$ are distinct. In fact, we always have the inequality $\beta \left({\mathbf{u}}^{\prime }\right)<\beta \left(\mathbf{u}\right)$ if the vector $\mathbf{u}$ is lexicographically smaller than the vector ${\mathbf{u}}^{\prime }$.

Clearly, there is a continuum of such distinct sequences $\mathbf{u}$ when $\mathbf{u}$ runs over all possible infinite sequences consisting of 2 and 4. As we have shown above, the numbers $\beta \left(\mathbf{u}\right)$ are all distinct, so there are continuum of numbers $\beta \left(\mathbf{u}\right)$. This completes the proof of Proposition 1, because in (10) we have $0<{\epsilon }_{n_j}/(u_j+1)$ and ${\epsilon }_{n_j}/u_j<{\epsilon }_{n_j}$, so (7) is true for each $n=n_j$.

3. Proof of Theorem 2

Fix any $\eta >1$ and an interval $I=[a,b]$, where $1\le a<b$. Note that, without restriction of generality, we may assume that $a>1$, because in case $a=1$, one can consider the subinterval $\left[\right(a+b)/2,b]$ of I instead of I itself.

In order to apply Proposition 1, we will consider the sequence of positive numbers ${\epsilon }_1=1$,

$${\epsilon }_n=min\left(1,{\displaystyle \frac{1}{{\mathsf{\Phi }}_{2}2!{\xi }_2}},{\displaystyle \frac{1}{{\mathsf{\Phi }}_{3}3!{\xi }_3}},\dots ,{\displaystyle \frac{1}{{\mathsf{\Phi }}_{n}n!{\xi }_n}}\right)$$

for each $n\ge 2$. Then, $1={\epsilon }_1\ge {\epsilon }_2\ge {\epsilon }_3\ge \dots$ is a sequence of positive numbers. Let also

$${\gamma }_n=-{\displaystyle \frac{log{\xi }_n}{log\eta }}$$

(16)

for each $n\in \mathbb{N}$.

Fix any $ϵ$ in the interval $(0,loga)$. Then, by (3), there is an infinite sequence $\mathcal{N}\subset \mathbb{N}$ such that, for each $n\in \mathcal{N}$, we have

$${\xi }_n\ge e^{-ϵn}.$$

(17)

Furthermore, by (17), for $\alpha \ge a$, we have

$${\xi }_n{\alpha }^n\to \infty \mathrm{as}n\in \mathcal{N}\mathrm{tends}\mathrm{to}\mathrm{infinity}.$$

(18)

Now, by Proposition 1 applied to the interval

$$J=[loga/log\eta ,logb/log\eta ]$$

(19)

and (16), there is an uncountable in J set of positive numbers B such that, for each $\beta \in B$, the inequalities

$$0<n\beta +{\displaystyle \frac{log{\xi }_n}{log\eta }}-m<{\epsilon }_n\le {\displaystyle \frac{1}{{\mathsf{\Phi }}_{n}n!{\xi }_n}}$$

hold for infinitely many pairs $(n,m)$, where $n\in \mathcal{N}$ and $m\in \mathbb{Z}$. Multiplying all this by $log\eta >0$, we derive that the inequalities

$$0<n\beta log\eta +log{\xi }_n-mlog\eta <{\displaystyle \frac{1}{{\mathsf{\Phi }}_n(n-1)!{\xi }_n}}$$

(20)

hold for infinitely many pairs $(n,m)$, where $n\in \mathcal{N}$ and $m\in \mathbb{Z}$. Note that, by (17), we have

$${\mathsf{\Phi }}_n(n-1)!{\xi }_n\to \infty \mathrm{as}n\in \mathcal{N}\mathrm{tends}\mathrm{to}\mathrm{infinity}.$$

(21)

Let S be the set of numbers of the form $\alpha ={\eta }^{\beta }$, where $\beta$ runs over every element of B. Note that the map $x\mapsto {\eta }^x$ maps the interval J defined in (19) into the interval $[a,b]$. Therefore, the set S is a subset of $[a,b]$. Moreover, the set S is uncountable because so is the set B.

Consider the difference

$${\xi }_n{\alpha }^n-{\eta }^m=e^{n\beta log\eta +log{\xi }_n}-e^{mlog\eta }={\eta }^m(e^{n\beta log\eta +log{\xi }_n-mlog\eta }-1).$$

By (20), the exponent here is in the interval $(0,1/({\mathsf{\Phi }}_n(n-1)!{\xi }_n))$. Additionally, ${\eta }^m<{\xi }_n{\alpha }^n$ by (20) as well. Consequently, from (21), it follows that

$$0<{\xi }_n{\alpha }^n-{\eta }^m={\eta }^m(e^{n\beta log\eta +log{\xi }_n-mlog\eta }-1)<{\xi }_n{\alpha }^n·{\displaystyle \frac{2}{{\mathsf{\Phi }}_n(n-1)!{\xi }_n}}={\displaystyle \frac{2{\alpha }^n}{{\mathsf{\Phi }}_n(n-1)!}}<{\displaystyle \frac{1}{{\mathsf{\Phi }}_n}}$$

for a sufficient large $n\in \mathcal{N}$. Here, in view of (18), for each sufficiently large $n\in \mathcal{N}$, the corresponding integer m must be positive. This completes the proof of (6).

4. A Different Version of the Main Result

Note that, in the proof of Theorem 2, we did not use (3) but rather the condition (18), with a subsequence $\mathcal{N}$. Therefore, we can change the initial condition (3) for the sequence ${\xi }_n$, $n=1,2,3,\dots$, by the condition

$${\xi }_n{a}^n\to \infty \mathrm{as}n\in \mathcal{N}\mathrm{tends}\mathrm{to}\mathrm{infinity},$$

(22)

where $\mathcal{N}$ is an arbitrary infinite sequence of positive integers and $a>1$ is a fixed number. Observe that (22) is true for any infinite sequence $\mathcal{N}\subseteq \mathbb{N}$ and any $a>1$ if, say, ${\xi }_n\ge \xi >0$ for every $n\in \mathbb{N}$.

Then, by the argument given in Section 3, we obtain the following version of Theorem 1:

Theorem 3. 

Let $I=[a,b]$ be an interval with $1<a<b$. Assume that $\mathcal{N}$ is an infinite sequence of positive integers, and $\xi =\{{\xi }_1,{\xi }_2,{\xi }_3,\dots \}$ is a sequence of positive numbers satisfying (22) with this $\mathcal{N}$. Then, for any sequence of positive numbers $\delta =\{{\delta }_1,{\delta }_2,{\delta }_3,\dots \}$, there is an uncountable set $S(I,\mathcal{N},\xi ,\delta )\subset I$ such that, for each $\alpha \in S(I,\mathcal{N},\xi ,\delta )$, the inequalities

$$0<\left\{{\xi }_n{\alpha }^n\right\}<{\delta }_n$$

(23)

hold for infinitely many $n\in \mathcal{N}$.

We omit the proof, since it is exactly the same as that above.

5. An Application of Proposition 1

Recently, in [31], we studied the following problem. Given $\theta \in {\mathbb{R}}_{>0}\setminus \mathbb{N}$, let $R_{\theta }\left(N\right)$ be the least nonzero value of $\parallel a^{\theta }\parallel$ as $a=1,2,\dots ,N$. Define

$$E_{\theta }=\underset{N\to \infty }{lim\; sup}{\displaystyle \frac{log(1/R_{\theta }\left(N\right))}{logN}}.$$

In Theorem 5 of [31], we provided several estimates for the quantity $E_{\theta }$ for some $\theta$. For example, it was shown that $E_{2/3}\ge 1$, with the equality holding under assumption of the $abc$-conjecture.

Then Iyer [32] showed that $E\left(\theta \right)$ can be infinite for some $\theta \in {\mathbb{R}}_{>0}\setminus \mathbb{N}$. This follows from Theorem 1.9 of [32], where it was shown that, for any sequence of positive numbers ${\delta }_n$, $n=1,2,3,\dots$, there are many $\tau \in {\mathbb{R}}_{>0}\setminus \mathbb{N}$ for which the inequalities

$$0<\parallel n^{\tau }\parallel <{\delta }_n$$

hold for infinitely many $n\in \mathbb{N}$. Indeed, selecting ${\delta }_n=1/n!$ and all the corresponding numbers $\tau$, we see that $E_{\tau }=\infty$ for each of those $\tau$, because $log(n!)/log\left(n\right)\to \infty$ as $n\to \infty$.

We will derive the following more general result:

Theorem 4. 

Let $I=[a,b]$ be an interval with $0<a<b$. Assume that $\xi =\{{\xi }_1,{\xi }_2,{\xi }_3,\dots \}$ is a sequence of positive numbers satisfying

$${\xi }_n{n}^a\to \infty \mathrm{as}n\to \infty .$$

(24)

Then, for any sequence of positive numbers $\delta =\{{\delta }_1,{\delta }_2,{\delta }_3,\dots \}$, there is an uncountable set $W(I,\xi ,\delta )\subset I$ such that, for each $\tau \in W(I,\xi ,\delta )$, the inequalities

$$0<\left\{{\xi }_n{n}^{\tau }\right\}<{\delta }_n$$

(25)

hold for infinitely many $n\in \mathbb{N}$.

Proof. 

Without restriction of generality, we may assume that ${\delta }_n<1$ for each $n\in \mathbb{N}$. In all that follows, it will be shown that (25) holds for infinitely many powers of 2, namely, the inequalities

$$0<\left\{{\xi }_{2^n}{2}^{n\tau }\right\}<{\delta }_{2^n}$$

(26)

are true for infinitely many $n\in \mathbb{N}$.

To this end, we will apply Proposition 1 to $I=[a,b]$,

$${\gamma }_n=-{\displaystyle \frac{log{\zeta }_{2^n}}{log2}},$$

and the sequence of positive numbers ${\epsilon }_n$, $n=1,2,3,\dots$, where ${\epsilon }_1=1$ and

$${\epsilon }_n=min\left({\epsilon }_1,\dots ,{\epsilon }_{n-1},{\displaystyle \frac{{\delta }_{2^n}}{{\xi }_{2^n}{2}^{nb}}}\right)$$

(27)

for $n=2,3,4,\dots$. It is clear that $1={\epsilon }_1\ge {\epsilon }_2\ge {\epsilon }_3\ge \dots$.

By Proposition 1, it follows that there is an uncountable in I set of positive numbers W such that, for each $\tau \in W$, the inequalities

$$0<n\tau +{\displaystyle \frac{log{\xi }_{2^n}}{log2}}-m<{\epsilon }_n$$

hold for infinitely many pairs $(n,m)$, where $n\in \mathbb{N}$ and $m\in \mathbb{Z}$. Multiplying by $log2$, we obtain that the inequalities

$$0<n\tau log2+log{\xi }_{2^n}-mlog2<{\epsilon }_{n}log2$$

(28)

hold for infinitely many pairs $(n,m)$, where $n\in \mathbb{N}$ and $m\in \mathbb{Z}$.

Now, we consider the difference

$${\xi }_{2^n}{2}^{n\tau }-2^m=e^{n\tau log2+log{\xi }_{2^n}}-e^{mlog2}=2^m(e^{n\tau log2+log{\xi }_{2^n}-mlog2}-1).$$

By (28), the exponent here is in the interval $(0,{\epsilon }_{n}log2)$. Additionally, $2^m<{\xi }_{2^n}{2}^{n\tau }$ by (28), and ${\xi }^{2^n}{2}^{nb}{\epsilon }_n\le {\delta }_{2^n}$ by (27). Therefore,

$$0<{\xi }_{2^n}{2}^{n\tau }-2^m=2^m(e^{n\tau log2+log{\xi }_{2^n}-mlog2}-1)<{\xi }_{2^n}{2}^{n\tau }·(2{\epsilon }_{n}log2)<{\xi }_{2^n}{2}^{nb}{\epsilon }_n\le {\delta }_{2^n}$$

for each of those $n\in \mathbb{N}$.

Here, we have

$${\zeta }_{2^n}{2}^{n\tau }\to \infty \mathrm{as}n\to \infty ,$$

because

$${\zeta }_{2^n}{2}^{na}\to \infty \mathrm{as}n\to \infty .$$

Consequently, as

$$2^m>{\xi }_{2^n}{2}^{n\tau }-{\delta }_{2^n}>{\xi }_{2^n}{2}^{n\tau }-1,$$

the corresponding integer m must be positive for each sufficiently large $n\in \mathbb{N}$, and hence $2^m\in \mathbb{Z}$.

Combined with $0<{\delta }_{2^n}<1$, this implies that $2^m$ is the integer part of the number ${\xi }_{2^n}{2}^{n\tau }$. Thus, for infinitely many $n\in \mathbb{N}$, we have

$$0<\left\{{\xi }_{2^n}{2}^{n\tau }\right\}={\xi }_{2^n}{2}^{n\tau }-2^m<{\delta }_{2^n}$$

which is (26). This completes the proof of the theorem. □

Note that we cannot omit the condition (24). Indeed, select, for instance, ${\xi }_n={\delta }_n=1/n!$. Then, for each $\tau >0$, we have $0<{\xi }_n{n}^{\tau }<1$ for each sufficiently large $n\in \mathbb{N}$. Therefore, for each $\tau >0$ and all sufficiently large $n\in \mathbb{N}$, we have

$$\left\{{\xi }_n{n}^{\tau }\right\}={\xi }_n{n}^{\tau }>{\xi }_n={\displaystyle \frac{1}{n!}}={\delta }_n,$$

so (25) does not hold for $\tau >0$.

6. Concluding Remarks

In particular, Theorem 3 implies that, for any $\xi >0$ and any sequence of positive numbers ${\delta }_n$, $n=1,2,3,\dots$, there are uncountably many $\alpha >1$ for which the inequalities

$$0<\left\{\xi {\alpha }^p\right\}<{\delta }_p$$

hold for infinitely many primes p, and uncountably many $\gamma >1$ for which the inequalities

$$0<\left\{\xi {\gamma }^{n^2}\right\}<{\delta }_{n^2}$$

hold for infinitely many $n\in \mathbb{N}$.

On the other hand, we do not know whether our method can be extended to conclude the same as stated in Theorem 3 with inequality (23) replaced by

$$0<\{{\xi }_n{\alpha }^n-{\eta }_n\}<{\delta }_n,$$

where ${\eta }_n$, $n=1,2,3,\dots$, is an arbitrary sequence of real numbers. This problem is open even if ${\xi }_n=1$ for $n\in \mathbb{N}$. More precisely, we do not know whether for any sequence of real numbers ${\eta }_n$, $n=1,2,3,\dots$, and any sequence of positive numbers ${\delta }_n$, $n=1,2,3,\dots$, there is a real number $\alpha >1$ for which we have

$$0<\{{\alpha }^n-{\eta }_n\}<{\delta }_n$$

for infinitely many $n\in \mathbb{N}$.

Similarly, with respect to Theorem 4, we may ask whether for any sequence of real numbers ${\eta }_n$, $n=1,2,3,\dots$, and any sequence of positive numbers ${\delta }_n$, $n=1,2,3,\dots$, there is a real number $\tau >0$ for which the inequalities

$$0<\{n^{\tau }-{\eta }_n\}<{\delta }_n$$

hold for infinitely many $n\in \mathbb{N}$.