\documentclass[10pt,a4paper]{article} \usepackage[a4paper]{geometry} \usepackage{amssymb,latexsym,amsmath,amsfonts,amsthm} \usepackage{graphicx,comment} \usepackage{epsfig} %\usepackage{showkeys} \usepackage{tikz} \usepackage{subcaption} %\usepackage{comment} \usepackage[T1]{fontenc} %\usepgflibrary{decorations.markings} %\usetikzlibrary{decorations.markings} \title{Analysis of the Asymptotics of Potentials of Greedy Energy Sequences} \author{Ryan McCleary} \newcommand{\Tr}{\mathrm{Tr}\,} \newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\supp}{\mathrm{supp}} \newcommand{\ir}{\text{\rm{i}}} \newcommand{\diag}{\mathrm{diag}\,} \newcommand{\Imag}{\mathrm{Im}\,} \newcommand{\Real}{\mathrm{Re}\,} \newcommand{\area}{\text{\rm{area}}} \newcommand{\wh}{\widehat} \newcommand{\wtil}{\widetilde} \newcommand{\ud}{\,\mathrm{d}} \newcommand{\mo}{\text{\rm{mod}}} \newcommand{\crit}{\text{\rm{crit}}} \newcommand{\wt}{\widetilde} \newcommand{\Id}{\text{\rm{Id}}} \newcommand{\Res}{\mathrm{Res}} \newcommand{\lf}{\left\lfloor} \newcommand{\rf}{\right\rfloor} \newcommand{\lc}{\left\lceil} \newcommand{\rc}{\right\rceil} \newcommand{\sign}{\mathrm{sign}} \newcommand{\kt}{\tilde{k}} \newcommand{\Fkr}{\widetilde{F}_{k}^{(\rho)}} \newcommand{\Fk}{F_{k}^{(\rho)}} \newcommand{\Fkmr}{\widetilde{F}_{k-1}^{(\rho)}} \newcommand{\Fkm}{F_{k-1}^{(\rho)}} \newcommand{\Fkpr}{\widetilde{F}_{k+1}^{(\rho)}} \newcommand{\Fkp}{F_{k+1}^{(\rho)}} \newcommand{\Skr}{S_{k}^{(\rho)}} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{example}{Example} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{rhp}[theorem]{RH problem} \theoremstyle{remark} \newtheorem{note}[theorem]{Note} \newtheorem{remark}[theorem]{Remark} \newtheorem{problem}[theorem]{Problem} \newtheorem{convention}[theorem]{Convention} \numberwithin{equation}{section} \hyphenation{pa-ra-me-tri-za-tion} \begin{document} \begin{center} Honors Thesis\\ \vspace{3mm} On \\ \vspace{3mm} Analysis of the Asymptotics of Potentials\\ \vspace{3mm} of Greedy Energy Sequences\\ \vspace{3mm} Ryan McCleary \end{center} \vspace{3cm} \begin{tabular}{p{0.6\textwidth}} \hrulefill \\ Abey Lopez-Garcia, Ph.D. \\ Thesis Committee Chair \\ College of Sciences \\ \end{tabular}\\ \vspace{1cm} \begin{tabular}{p{0.6\textwidth}} \hrulefill \\ Robert Jenkins, Ph.D. \\ Committee Member From Major \\ College of Sciences \\ \end{tabular} \vspace{1cm} \begin{tabular}{p{0.6\textwidth}} \hrulefill \\ Xin Li, Ph.D. \\ Department Chair, Department of Mathematics \\ College of Sciences \\ \end{tabular} \maketitle \tableofcontents \section{Introduction} The study of greedy energy sequences can be motivated by a problem from physics known as the Thompson Problem. Given a $N$ point charges, we would like to know which configuration of these charges on the sphere will give the smallest possible electrostatic energy. Given two charges $Q_0$ and $Q_1$ separated by a distance $r$, recall that the electrostatic energy of this system is given by Coulomb's lab as $E = K\frac{Q_0Q_1}{r},$ where $K$ is Coulomb's constant. If we place a charge $Q$ on the unit circle at the point $a_0=1$, where on the circle can we place a second point $a_1$ with charge $Q$ so that the energy of the system is minimized? It is clear that this minimum will be attained by choosing $a_1=-1$. After placing $a_1$, we can again ask where we can place a third charge of $Q$ so that the energy of the system is minimized. It turns out that this minimum can be attained in two different ways: either by choosing $a_2=i$ by choosing $a_2=-i$. If we want to place a fourth charge of $Q$ on the sphere with minimum energy, however, there is now only one choice. We must choose $a_3=-a_2$. After placing the fourth point, there are four places in which we could put a charge of $Q$ that would minimize the energy of the collection of 5 points. These four points are the midpoints of the four points we have already placed on the circle. This iterative process we have just outlined of placing point charges on a circle (or more generally, a sphere) is the defining property of a greedy energy sequence. \section{Definitions and notation} In this section, we introduce the necessary definitions to discuss greedy energy sequences on the unit circle. Let us denote the unit circle by $S^1 := \{ z \in \mathbb{C} : |z| = 1\}$. \begin{definition} Let $s$ be a positive real number. A greedy $s$-energy sequence $(a_{n})_{N=0}^{\infty} \subset S^1$ is a sequence possessing the following properties: (1) $a_{0}$ is selected arbitrarily (we will adopt the convention $a_0=1$ unless indicated otherwise); and (2) assuming that $(a_{0},a_{1},\ldots,a_{n})$ have been selected, $a_{n+1}$ is chosen to satisfy $\sum_{k=0}^{n} \frac{1}{|a_{n+1} - a_{k}|^s} = \inf_{x \in S^1} \sum_{k=0}^{n} \frac{1}{|x- a_{k}|^s},$ for all $n \geq 0$. \end{definition} Our next definition formalized the notion of potential that was discussed in the introduction. \begin{definition} If $(a_{n})_{n=0}^{\infty}$ is a greedy $s$-energy sequence on $S^1$ and $n \geq 1$, define $U_{n,s}(x) := \sum_{k=0}^{n-1} \frac{1}{|x - a_{k}|^s},$ We call $U_{n,s}(x)$ the (Riesz) potential of $x$ with respect to $a_0,a_1,\ldots,a_{n-1}$. If $n \geq 1$, $a_{n}$ satisfies $U_{n,s}(a_{n}) = \inf_{x \in S^1} U_{n,s}(x).$ \end{definition} \begin{definition} Let $\omega = (x_1,x_2,\ldots,x_N)$ be a configuration of $N \geq 2$ points of $S^1$, and let $E_s(\omega) := \sum_{1\leq i \neq j \leq N} \frac{1}{|x_i - x_j|^s} = 2 \sum_{i=1}^{N-1}\sum_{j=i+1}^{N}\frac{1}{|x_i-x_j|^s}.$ (This quantity is understood to be infinite if there exist indices $1 \leq i \neq j \leq N$ such that $x_i = x_j$.) We say that an $N$-point configuration $\omega \in (S^1)^N$ is an $N$-point minimal $s$-energy configuration on $S^1$ if $E_{s}(\omega) = \inf\{ E_{s}(\omega^*) : \omega^* \in (S^1)^N \}.$ \end{definition} \begin{remark} For $s > 0$, let $(a_n)_{n=0}^{\infty}$ be a greedy $s$-energy sequence on $S^1$. For $N\geq 2$, let $\alpha_{N,s}=(a_0,a_1,\ldots,a_{N-1})$ consist of the first $N$ points of $(a_n)$. Then $E_{s}(\alpha_{N,s}) = \sum_{0 \leq i \neq j \leq N - 1} \frac{1}{|a_{i} - a_{j}|^{s}} = 2\sum_{i=1}^{N-1}\sum_{j=0}^{i-1} \frac{1}{|a_i - a_j|^{s}} = 2\sum_{i=1}^{N-1} U_{i,s}(a_i).$ \end{remark} It is easy to see that $N$-point minimal $s$-energy configurations exist on $S^1$ for all $N\geq 2$. Indeed, for $N\geq 2$, let us define the function $\Phi_N : (S^1)^N \to \mathbb{R}$ by $\Phi_N(\omega) = E_s(\omega)$ for all $\omega \in (S^1)^N$. Since $S^1$ is compact, $(S^1)^N$ is also compact. Moreover, $\Phi_N$ is a lower semicontinuous function on $(S^1)^N$, so $\Phi_N$ attains a minimum value on $(S^1)^N$. Hence, $N$-point minimal $s$-energy configurations exist on $S^1$. Let $\mathcal{M}_{1}$ be the set of all probability measures defined on $S^1$. Define the $s$-energy of a probability measure $\sigma$ on $S^1$ as follows: $I_{s}(\sigma) = \iint_{S^d \times S^d} \frac{1}{|x - y|^s}\,d\sigma(x)\,d\sigma(y).$ We call $\sigma$ a minimal $s$-energy distribution on $S^1$ if the following is true: $I_{s}(\sigma) = \inf_{\nu \in \mathcal{M}_{d}} I_{s}(\nu).$ It is shown in \cite[Proposition 4.6.4]{BorodachovHardinSaff}, that the normalized surface area (Lebesgue) measure on $S^1$ is the unique minimal $s$-energy distribution on $S^1$ for every $0 < s < 1$. We denote this measure by $\sigma_1$. If $s \geq 1$, then $I_{s}(\nu) = \infty$ for all $\nu \in \mathcal{M}_{1}$. The $s$-energy of $\sigma_1$ is given explicitly by $I_{s}(\sigma_1) = \frac{\Gamma(1-s)}{\Gamma(-s/2)\Gamma(1 - s/2)} = \frac{2^{-s}}{\sqrt{\pi}} \frac{\Gamma((1-s)/2)}{\Gamma(1 - s/2)}.$ \section{Geometric structure} The asymptotic behavior of the energy $E_{s}(\alpha_{N,s})$, where $\alpha_{N,s}$ is the configuration of the first $N$ points of a greedy $s$-energy sequence $(a_n)$, were investigated in \cite{LopWag}. However, the asymptotic behavior of the potentials $U_{N,s}(a_N)$ have not been analyzed. In this work, we will examine the asymptotic behavior of $U_{N,s}(a_N)$, where $s > 0$ and $(a_n) \subset S^1$ is a greedy $s$-energy sequence. Our reason for restricting this analysis to $S^1$ is that there exists a convenient geometric description of greedy $s$-energy sequences on $S^1$ which does not have an analog in higher dimensions. For an $M$-tuple $A = (a_1,a_2,\ldots,a_M)$ and an $N$-tuple $B = (b_1,b_2,\ldots,b_N)$, define $(A,B) := (a_1,a_2,\ldots,a_M,b_1,b_2,\ldots,b_N)$. In \cite{BialasCalvi}, the following result was given for Leja sequences (see \cite{BialasCalvi} for the definition of a Leja sequence). It was remarked in \cite{LopWag} that for any $s > 0$, this result holds for greedy $s$-energy sequences, since they coincide with Leja sequences on $S^1$. We now state this result in terms of greedy $s$-energy sequences on $S^1$. Recall that we always take the first point of a greedy $s$-energy sequence $(a_n)_{n=0}^{\infty}$ on $S^1$ to be the point $a_0 = 1$. \begin{proposition} \label{prop:geodesc} For an arbitrary $s > 0$, let $(a_n)_{n=0}^{\infty}$ be a greedy $s$-energy sequence on $S^1$. Then \begin{enumerate} \item Any $2^n$th section of $(a_n)$ is formed by the $2^n$th roots of unity. \item For any $n \geq 0$, let $\alpha_{2^{n+1}}$ be the $2^{n+1}$th section of $(a_n)$. Then $\alpha_{2^{n+1}}$ contains $\alpha_{2^n}$ as its first $2^n$ points, and there exists a $2^n$th root $\rho$ of $-1$ and a greedy $s$-energy sequence $({a'}_{n})_{n=0}^{\infty}$ on $S^1$ such that $\alpha_{2^{n+1}} = (\alpha_{2^n}, \rho {\alpha'}_{2^n})$. (Here, ${\alpha'}_{N}$ denotes the $N$th section of $({a'}_{n})$.) \item Iterating (2), we see that for any $N = 2^{n_1} + 2^{n_2} + \cdots + 2^{n_p}$, $n_1 > n_2 > \cdots > n_p \geq 0$, there exists for each integer $k$ with $1 \leq k \leq p$ a greedy $s$-energy sequence $(a_n^k)_{n=0}^{\infty} \subset S^1$ (with $a_0^k = 1$) such that $\alpha_{N} = (\alpha_{2^{n_1}}^{1}, \rho_1\alpha_{2^{n_2}}^{2}, \rho_1\rho_2\alpha_{2^{n_3}}^{3},\ldots,\left(\prod_{i=1}^{p-1} \rho_i\right)\alpha_{2^{n_p}}^{p})$ for some numbers $\rho_i$ that are $2^{n_i}$th roots of $-1$. Following our usual convention, we have denoted the $N$th section of $(a_n^k)$ by $\alpha_N^k$. \end{enumerate} \end{proposition} \section{Expressing $U_{N,s}(a_N)$ in terms of the binary expansion of $N$} Since we have now established that greedy $s$-energy sequences on $S^1$ are composed of rotated configurations of equally spaced points, it will be important to examine the potential of configurations of equally spaced points. To this end, we introduce the next definition: \begin{definition} For $N\geq 1$, define $\mathcal{U}_{s}(N) := \sum_{k=0}^{N-1} \left|\exp(\pi i/N) - \exp(2k\pi i/N)\right|^{-s}.$ \end{definition} It is easy to check that $\mathcal{U}_{s}(N)$ is the potential of $N$ equally spaced points on $S^1$ evaluated at a midpoint of an arc between two adjacent points. The next proposition gives a formula for $U_{N,s}(a_N)$ in terms of the binary expansion of $N$. it will be used extensively in our asymptotic analysis of $U_{N,s}(a_N)$. \begin{proposition} \label{prop:descUnan} Let $s>0$, and let $(a_{n})_{n=0}^{\infty}$ be a greedy $s$-energy sequence on $S^{1}$. If $N = 2^{n_1} + \cdots + 2^{n_p}$, $n_1 > \cdots > n_p \geq 0$, then $$\label{eq:descUnan} U_{N,s}(a_{N}) = \sum_{k=1}^{p}\mathcal{U}_{s}(2^{n_{k}}).$$ \end{proposition} \begin{proof} The proof is by induction on $p$. If $p=1$ and $N=2^{n}$, then by \ref{prop:geodesc}, the points in $\alpha_{N} = (a_{0},\ldots,a_{N-1})$ are the $N$-th roots of unity. The point $a_{N}$ is the midpoint of one of the arcs between adjacent points in $\alpha_{N}$, hence $U_{N,s}(a_{N}) = \mathcal{U}_{s}(N)$. For a fixed $p\geq 1$, assume as induction hypothesis that \eqref{eq:descUnan} is valid for every greedy $s$-energy sequence on $S^1$ and every $N\geq 1$ with binary representation of length $p$. Let $\widetilde{N}=2^{n_{1}}+\cdots+2^{n_{p}}+2^{n_{p+1}}$, $n_{1}>\cdots>n_{p}>n_{p+1}\geq 0$. Then, by \ref{prop:geodesc}, we can write $\alpha_{\widetilde{N}}=(\alpha_{2^{n_{1}}},\rho {\alpha'}_{\widetilde{N}-2^{n_1}}),$ where $\rho$ satisfies $z^{2^{n_{1}}}=-1$, and ${\alpha'}_{\widetilde{N}-2^{n_1}}$ is the section of order $\widetilde{N}-2^{n_{1}}$ of a greedy $s$-energy sequence $({a'}_{k})_{k=0}^{\infty}$. So $a_{k}=\rho\,{a'}_{k-2^{n_1}}$, $2^{n_{1}}\leq k\leq \widetilde{N}$, where $({a'}_{0},\ldots,{a'}_{\widetilde{N}-2^{n_1}})={\alpha'}_{\widetilde{N}-2^{n_1}+1}$. We have $U_{\widetilde{N},s}(a_{\widetilde{N}})=U_{2^{n_1},s}(a_{\widetilde{N}})+\sum_{k=2^{n_1}}^{\widetilde{N}-1}|a_{k}-a_{\widetilde{N}}|^{-s}.$ The point $a_{\widetilde{N}}$ is the midpoint of one of the arcs between two adjacent points in $\alpha_{2^{n_1}}$, hence $U_{2^{n_1},s}(a_{\widetilde{N}})=\mathcal{U}_{s}(2^{n_1}).$ We also have $\sum_{k=2^{n_1}}^{\widetilde{N}-1}|a_{k}-a_{\widetilde{N}}|^{-s}= \sum_{k=2^{n_1}}^{\widetilde{N}-1}|\rho\, {a'}_{k-2^{n_1}}-\rho\, {a'}_{\widetilde{N}-2^{n_1}}|^{-s}=\sum_{k=0}^{\widetilde{N}-2^{n_1}-1}|{a'}_{k}-{a'}_{\widetilde{N}-2^{n_1}}|^{-s}=U_{M,s}({a'}_{M}),$ where $M:=\widetilde{N}-2^{n_{1}}$. The number $M=2^{n_{2}}+\cdots+2^{n_{p+1}}$ has a binary representation of length $p$, so by induction hypothesis we obtain $U_{M,s}(b_{M})=\sum_{k=2}^{p+1}\mathcal{U}_{s}(2^{n_{k}}).$ In conclusion, $U_{\widetilde{N},s}(a_{\widetilde{N}})=\sum_{k=1}^{p+1}\mathcal{U}_{s}(2^{n_k})$, which finishes the proof of \eqref{eq:descUnan}. \end{proof} \section{Formula for $\mathcal{U}_{s}(N)$ in terms of the energy of equally spaced points} \begin{definition} For $N \geq 2$, let $\mathcal{L}_{s}(N) := E_s((e_1,e_2,\ldots,e_N)),$ where $e_k = \exp(2\pi k i/N)$ ($1 \leq k \leq N$). We will also agree that $\mathcal{L}_{s}(1)=0$. \end{definition} It can be readily verified that $\mathcal{L}_{s}(N) = 2^{-s} N \sum_{k=1}^{N-1} \left(\sin\frac{k\pi}{N}\right)^{-s}.$ Much is already known about the asymptotic properties of $\mathcal{L}_{s}(N)$. Once $\mathcal{U}_{S}(N)$ is expressed in terms of $\mathcal{L}_{S}(N)$, the asymptotic behavior of $\mathcal{U}_{s}(N)$ will be easily understood. \begin{lemma} \label{prop:lformula} If $N \geq 2$ and $\{e_1,e_2,\ldots,e_N\}$ are equally spaced points on $S^1$, then $\sum_{k=2}^{N} \frac{1}{|e_k - e_1|^s} = \frac{\mathcal{L}_{s}(N)}{N}.$ \end{lemma} \begin{proof} Fix $N \geq 2$. Let $S = (e_1,e_2,\ldots,e_N)$ be a set of $N$ equally spaced points on $S^1$. We have \begin{align*} \mathcal{L}_{s}(N) = \sum_{k=1}^{N}\sum_{z \in S \setminus e_k} \frac{1}{|z - e_k|^{s}}. \end{align*} By symmetry, this implies $\mathcal{L}_{s}(N) = N \sum_{k=2}^{N} \frac{1}{|e_k - e_1|^{s}},$ and dividing by $N$ yields the desired equality. \end{proof} \begin{proposition} \label{prop:uInTermsOfL} If $N \geq 1$, then for any $s > 0$, $\mathcal{U}_{s}(N) = \frac{\mathcal{L}_{s}(2N)}{2N} - \frac{\mathcal{L}_{s}(N)}{N}.$ \end{proposition} \begin{proof} Let $e_k = \exp(2\pi i k / 2N)$ for $1 \leq k \leq 2N$. By the definition of $\mathcal{U}_s(N)$, we evidently have \begin{align*} \mathcal{U}_{s}(N) = \sum_{k=1}^{N} \frac{1}{|e_{2k} - e_1|^{s}} = \sum_{k=2}^{2N} \frac{1}{|e_k - e_1|^{s}} - \sum_{k=1}^{N-1}\frac{1}{|e_{2k+1} - e_1|^{s}}. \end{align*} By Proposition~\ref{prop:lformula}, this implies $\mathcal{U}_{s}(N) = \frac{\mathcal{L}_{s}(2N)}{2N} - \frac{\mathcal{L}_{s}(N)}{N}.$ \end{proof} \section{The case $0 \cdots > n_p \geq 0$ be the binary representation of $N$. By Proposition~\ref{prop:descUnan}, we have $\left|\frac{U_{N,s}(a_N) - N I_{s}(\sigma_1)}{N^s}\right| = \left|\sum_{k=1}^{p} \frac{\mathcal{U}_{s}(2^{n_k}) - 2^{n_k} I_{s}(\sigma_1)}{N^s}\right|.$ Applying the triangle inequality, we now obtain $\left|\frac{U_{N,s}(a_N) - N I_{s}(\sigma_1)}{N^s}\right| \leq \sum_{k=1}^{p} \left|\frac{\mathcal{U}_{s}(2^{n_k}) - 2^{n_k} I_{s}(\sigma_1)}{M^s}\right|.$ Now observe that, for every $1 \leq k \leq p$, $(2^{n_1})^s = (2^{n_k})^s (2^s)^{n_1-n_k}$. Moreover, $N \geq 2^{n_1}$. Hence, \begin{align*} \sum_{k=1}^{p} \left|\frac{\mathcal{U}_{s}(2^{n_k}) - 2^{n_k} I_{s}(\sigma_1)}{N^s}\right| &\leq \sum_{k=1}^{p} \left|\frac{\mathcal{U}_{s}(2^{n_k}) - 2^{n_k} I_{s}(\sigma_1)}{(2^{n_1})^s}\right| \\ &= \sum_{k=1}^{p} \left|\frac{\mathcal{U}_{s}(2^{n_k}) - 2^{n_k} I_{s}(\sigma_1)}{(2^{n_k})^s (2^s)^{n_1-n_k}}\right| \\ &= \sum_{k=1}^{p} |\mathcal{W}_{s}(2^{n_k})| (2^s)^{n_k - n_1} \leq C \sum_{k=1}^{p} (2^s)^{n_k-n_1}\\ &< C \sum_{k=0}^{\infty} (2^s)^{-k} = C\frac{2^s}{2^s - 1}. \end{align*} We have therefore shown that the sequence \eqref{eq:secordseqs01} is bounded, since for all $N \geq 1$, $\left|\frac{U_{N,s}(a_N) - N I_{s}(\sigma_1)}{N^s}\right| < C\frac{2^s}{2^s - 1}.$ \end{proof} Now that the matter of boundedness is settled, we will show that \ref{eq:secordseqs01} is divergent by considering two subsequences which converge to different limits. \begin{proposition} \label{prop:secorddivs01} Let $0 n_2 > \cdots > n_p \geq 0$, such that $$\lim_{N\in\mathcal{N}} \frac{2^{n_k}}{N} = \theta_k \label{eq:thetadef}$$ for all $1 \leq k \leq p$. \end{definition} \begin{definition} Define the function $G : \Theta_p \times (0,\infty) \to \mathbb{R}$ by $G((\theta_1,\ldots,\theta_p); s) = \sum_{k=1}^{p} \theta_k^s.$ \end{definition} We now establish bounds on $G(\vec{\theta}; s)$, which will be needed for the next definition. We start by introducing a definition: \begin{lemma} If $\vec{\theta}=(\theta_1,\ldots,\theta_p)\in\Theta_p$, then $$1 \leq G(\vec{\theta}; s) < \frac{2^s}{2^s-1}.\label{eq:gbounds}$$ \end{lemma} \begin{proof} Since $\sum_{k=1}^{p} \theta_k = 1$ and the function $x\mapsto x^s$ is concave, we obtain \begin{align*} 1 = (\theta_1+\cdots+\theta_p)^s \leq \theta_1^s+\cdots+\theta_p^s = G(\vec{\theta}; s). \end{align*} Fix a vector $\vec{\theta} = (\theta_1,\ldots,\theta_p) \in \Theta_p$. We now use the following property of $\Theta_p$: for every $\vec{\theta}\in \Theta_p$, there is an odd integer $N = 2^{n_1}+\cdots+2^{n_t}$, where $t \leq p$ and $n_1 > n_2 > \cdots > n_t = 0$, such that $\theta_k = \frac{2^{n_k}}{N}$ for all $1\leq k\leq t$ and $\theta_k=0$ for $t+1\leq k\leq p$. Using this property, we now obtain \begin{align*} G(\vec{\theta};s) &= \sum_{k=1}^{p} \theta_k^s \leq \sum_{k=1}^{t} \left(\frac{2^{n_k}}{2^{n_1}}\right)^s = \sum_{k=1}^{t} 2^{s(n_k-n_1)} \\ &< \sum_{j=0}^{\infty} (2^s)^{-j} = \frac{2^s}{2^s-1}. \end{align*} \end{proof} Since we now know that $G(\vec{\theta}; s)$ is bounded above by a function of $s$, we can make the following definition: \begin{definition} For 0 1 - \sum_{k=M+1}^{\infty} 2^{-(k-1)} > 1 - \epsilon, \end{align*} which implies $\left(\frac{\widetilde{N}}{N}\right)^{s} > (1 - \epsilon)^{s}.$ For each1\leq k\leq M$, the sequence$2^{n_k}/\widetilde{N}$is bounded, so we may choose a subsequence$\widehat{\mathcal{N}}\subset \mathcal{N}$such that $\lim_{N\in\widehat{\mathcal{N}}}\frac{2^{n_k}}{\widetilde{N}} = \theta_k$ for some$\vec{\theta}=(\theta_1,\ldots,\theta_M)\in\Theta_M$and all$1\leq k\leq M. Since $S_{N,1} = \sum_{k=1}^{M}\mathcal{W}_{s}(2^{n_k})\left(\frac{2^{n_k}}{\widetilde{N}}\right)^{s}\left(\frac{\widetilde{N}}{N}\right)^{s} < (1-\epsilon)^{s}\sum_{k=1}^{M}\mathcal{W}_{s}(2^{n_k})\left(\frac{2^{n_k}}{\widetilde{N}}\right)^{s},$ we have \begin{align*} \limsup_{N\in\widehat{\mathcal{N}}} S_{N,1} &\leq \limsup_{N\in\widehat{\mathcal{N}}} \left((1-\epsilon)^{s}\sum_{k=1}^{M}\mathcal{W}_{s}(2^{n_k})\left(\frac{2^{n_k}}{\widetilde{N}}\right)^{s}\right) = (1-\epsilon)^{s}G(\vec{\theta};s)(2^{s}-1)\frac{2\zeta(s)}{(2\pi)^s}\\ &\leq (1-\epsilon)^{s}(2^{s}-1)\frac{2\zeta(s)}{(2\pi)^s}. \end{align*} It now follows that \begin{align*} \limsup_{N\in\widehat{\mathcal{N}}} \frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s} &= \limsup_{N\in\widehat{\mathcal{N}}}\left(S_{N,1}+S_{N,2}\right) \leq \limsup_{N\in\widehat{\mathcal{N}}} S_{N,1} + \limsup_{N\in\widehat{\mathcal{N}}} S_{N,2}\\ &\leq (1-\epsilon)^{s}(2^{s}-1)\frac{2\zeta(s)}{(2\pi)^s} + C\epsilon. \end{align*} Since\epsilonwas arbitrary, we obtain \begin{align*} \limsup_{N\in\widehat{\mathcal{N}}}\frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s} \leq (2^{s}-1)\frac{2\zeta(s)}{(2\pi)^s}. \end{align*} By our choice of\mathcal{N}, the sequence \eqref{eq:ndefseq} converges. Therefore, \begin{align*} \lim_{N\in\mathcal{N}}\frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s} = \limsup_{N\in\widehat{\mathcal{N}}} \frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s} \leq (2^s-1)\frac{2\zeta(s)}{(2\pi)^s}, \end{align*} so we have proved \eqref{eq:limsupunan}. %SupposeN_0$is an integer such that, whenever$N\geq %N_0, %$%\left|\sum_{k=1}^{M}\mathcal{W}_{s}(2^{n_k})\left(\frac{2^{n_k}}{\widetilde{N}}\right)^{s}-%G(\vec{\theta};s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^{s}}\right| %< \epsilon. %$ %Applying the triangle inequality, we now obtain %\begin{align*} %&\left|\sum_{k=1}^{M}\mathcal{W}_{s}(2^{n_k})\left(\frac{2^{n_k}}{\widetilde{N}}\right)^{s}\left(\frac{\widetilde{N}}{N}\right)^{s}-G(\vec{\theta};s)%(2^s-1)\frac{2\zeta(s)}{(2\pi)^{s}}\right|\\ %&\leq \left(\frac{\widetilde{N}}{N}\right)^{s}\left|\sum_{k=1}^{M}\mathcal{W}_{s}(2^{n_k})\left(\frac{2^{n_k}}{\widetilde{N}}\right)^{s}-G(\vec{\theta};s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^{s}}\right|\\ %&+ G(\vec{\theta};s)(2^s-1)\left|\frac{2\zeta(s)}{(2\pi)^{s}}\right|\left|\left(\frac{\widetilde{N}}{N}\right)^{s}-1\right|\\ %&< \epsilon + \overline{g}(s)(2^s-1)\left|\frac{2\zeta(s)}{(2\pi)^s}\right|\left(1-(1-\epsilon)^s\right). %\end{align*} We now prove \eqref{eq:liminfunan}. As before, ifp\in\mathbb{N}$and$\vec{\theta}=(\theta_1,\ldots,\theta_p)\in\Theta_p$for some$p\in\mathbb{N}$, then there is a sequence$\mathcal{N}$of natural numbers$N=2^{n_1}+\cdots+2^{n_p}$such that $\lim_{N\in\mathcal{N}} \frac{2^{n_k}}{N}=\theta_k$ for$1\leq k\leq p. We then have \begin{align*} \lim_{N\in\mathcal{N}}\frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s} = G(\vec{\theta}; s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}. \end{align*} This means that $G(\vec{\theta};s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}$ is a limit point of the sequence \eqref{eq:secondorderunan}. Sincep$and$\vec{\theta}$were arbitrary, this implies $\liminf_{N\to\infty}\frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s}\leq \overline{g}(s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}.$ Suppose that$\mathcal{N}$is a sequence of natural numbers such that $\left(\frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s}\right)_{N\in\mathcal{N}}\quad\text{converges.}$ First, assume that there is$p\in\mathbb{N}$such that infinitely many elements of$N$have binary expansion of length$p$. As in our argument to prove \eqref{eq:limsupunan}, we may choose a subsequence$\widehat{\mathcal{N}}\subset\mathcal{N}$such that there is a vector$\vec{\theta}=(\theta_1,\ldots,\theta_p)\in\Theta_p$for which $\lim_{N\in\widehat{\mathcal{N}}}\frac{2^{n_k}}{N}=\theta_k$ for$1\leq k\leq p. We then have \begin{align*} \lim_{N\in\mathcal{N}}\frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s} &= \lim_{N\in\widehat{\mathcal{N}}}\frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s}\\ &= G(\vec{\theta}; s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}\\ &\geq \overline{g}(s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}. \end{align*} Now suppose that\tau(N) \to \infty$as$N\to\infty$along the subsequence$\mathcal{N}$. Let$0 < \epsilon < 1$and let$M$be an integer such that $\sum_{k=M}^{\infty} 2^{-sk} < \epsilon.$ Now, if$N=2^{n_1}+2^{n_2}+\cdots+2^{\tau(N)}\in\mathcal{N}$, we write $\frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s} = S_{N,1} + S_{N,2},$ where $S_{M,1} = \sum_{k=1}^{M} \mathcal{W}_{s}(2^{n_k})\left(\frac{2^{n_k}}{N}\right)^s$ and $S_{N,2} = \sum_{k=M+1}^{\tau(N)}\mathcal{W}_{s}(2^{n_k})\left(\frac{2^{n_k}}{N}\right)^s.$ Put$\widetilde{N} = 2^{n_1}+\ldots+2^{M}$. As in our argument to prove \eqref{eq:limsupunan}, we may choose a subsequence$\widehat{\mathcal{N}}\subset\mathcal{N}$so that there exists a vector$\vec{\theta}=(\theta_1,\ldots,\theta_M)\in\Theta_M$such that $\lim_{N\in\widehat{\mathcal{N}}} \frac{2^{n_k}}{\widetilde{N}} = \theta_k$ for all$1\leq k\leq M. Since \begin{align*} S_{N,1} = \sum_{k=1}^{M}\mathcal{W}_{s}(2^{n_k})\left(\frac{2^{n_k}}{\widetilde{N}}\right)^{s}\left(\frac{\widetilde{N}}{N}\right)^{s} \geq \sum_{k=1}^{M}\mathcal{W}_{s}(2^{n_k})\left(\frac{2^{n_k}}{\widetilde{N}}\right)^{s}, \end{align*} We have \begin{align*} \liminf_{N\in\widehat{\mathcal{N}}} S_{N,1} \geq G(\vec{\theta};s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s} \geq \overline{g}(s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}. \end{align*} As we have previously argued, there is a constantC$such that $|S_{N,2}|\leq C\epsilon$ for all$N. Hence, \begin{align*} \liminf_{N\in\widehat{\mathcal{N}}}\frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s} &= \liminf_{N\in\widehat{\mathcal{N}}}\left(S_{N,1}+S_{N,2}\right)\geq \liminf_{N\in\widehat{\mathcal{N}}}S_{N,1} + \liminf_{N\in\widehat{\mathcal{N}}} S_{N,2}\\ &\geq \overline{g}(s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s} - C\epsilon. \end{align*} Since\epsilon$was arbitrary, $\liminf_{N\in\widehat{\mathcal{N}}}\frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s} \geq \overline{g}(s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}.$ As before, it follows that $\lim_{N\in\mathcal{N}} \frac{U_{N,s}(a_N)-NI_{s}(\sigma_1)}{N^s}\geq \overline{g}(s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}.$ This concludes the proof of \eqref{eq:liminfunan}. \end{proof} \section{The case$s=1$} We now turn our attention to the case$s=1$. As w will show, the Riesz potential behaves very differently in this case than the case$0n_2>\cdots>n_p\geq 0. Then \begin{align*} \frac{U_{N,1}(a_N)}{N\log N} = \sum_{k=1}^{p}\frac{\mathcal{U}_{1}(2^{n_k})}{N\log N} = \sum_{k=1}^{p}\mathcal{T}(2^{n_k})\frac{2^{n_k}\log 2^{n_k}}{N\log N} \end{align*} Note that \begin{align*} \left|\frac{U_{N,1}(a_N)}{N\log N}\right| \leq C\sum_{k=1}^{p}2^{-(k-1)}, \end{align*} whereC$is a constant such that$|\mathcal{T}(N)|\leq C$for all$N$. Thus, the sequence \ref{eq:firstorders1} is bounded. Fix$p\in\mathbb{N}$and$\vec{\theta}=(\theta_1,\theta_2\ldots,\theta_p)\in\Theta_p$. Let$\mathcal{N}$be a sequence of natural numbers such that if$N = 2^{n_1} + 2^{n_2} + \cdots + 2^{n_p}$and$n_1>n_2>\cdots >n_p\geq 0$, then $\lim_{N\in\mathcal{N}}\frac{2^{n_k}}{N} = \theta_k$ for$1\leq k\leq p$. Fix$1\leq k\leq p$. First, suppose that$\theta_k>0$. Since $\frac{\log 2^{n_k}}{\log N} = \frac{\log 2^{n_k}-\log N+\log N}{\log N} = \frac{\log(2^{n_k}/N)}{\log N} + 1$ and$2^{n_k}/N\to\theta_k, it follows that $\lim_{N\in\mathcal{N}}\frac{\log 2^{n_k}}{\log N} = 1.$ Then \begin{align*} \lim_{N\in\mathcal{N}}\mathcal{T}(2^{n_k})\frac{2^{n_k}\log 2^{n_k}}{N\log N} = \frac{\theta_k}{\pi}. \end{align*} On the other hand, suppose that\theta_k=0$. We then have$\theta_k = 0$. Since$2^{n_k}\leq N$, $\frac{\log 2^{n_k}}{\log N}$ is bounded. Thus, $\lim_{N\in\mathcal{N}}\mathcal{T}(2^{n_k})\frac{2^{n_k}\log 2^{n_k}}{N\log N} = 0.$ But since$\theta_k=0$, this limit is also equal to$\theta_k/\piin this case. It then follows that \begin{align*} \lim_{N\in\mathcal{N}} \frac{U_{N,1}(a_n)}{N\log N} = \lim_{N\in\mathcal{N}}\sum_{k=1}^{p} \mathcal{T}(2^{n_k})\frac{2^{n_k}\log 2^{n_k}}{N\log N} = \sum_{k=1}^{p}\frac{\theta_k}{\pi} = \frac{1}{\pi}. \end{align*} The final equality follows from the fact that\sum_{k=1}^{p}\theta_k = 1$. Let$\epsilon > 0$. Let$\mathcal{N}$be an infinite sequence of natural numbers. The strategy now is to show that the subsequence of the sequence in \ref{eq:firstorders1} has limit$\frac{1}{\pi}$. Since this sequence is bounded, it will follow that \ref{eq:firstorders1} holds. Let$N = 2^{n_1} + 2^{n_2} + \cdots + 2^{\tau(N)}$, and suppose that$M$is an integer such that $\sum_{k=M}^{\infty} 2^{-k} < \epsilon.$ We can then write $\frac{U_{N,1}(a_N)}{N\log N} = S_{N,1} + S_{N,2},$ where $S_{N,1} = \sum_{k=1}^{M}\mathcal{T}(2^{n_k})\frac{2^{n_k}\log 2^{n_k}}{N\log N}$ and $S_{N,2} = \sum_{k=M+1}^{\tau(N)} \mathcal{T}(2^{n_k})\frac{2^{n_k}\log 2^{n_k}}{N\log N}.$ The second sum is understood to be zero if$M\geq \tau(N)$. Since$(\mathcal{T}(N))$converges, there is a constant$C$such that$|\mathcal{T}(N)|\leq C$for all$N. Since $\left|\frac{2^{n_k}\log 2^{n_k}}{N\log N}\right| \leq 1,$ we have \begin{align*} \left|S_{N,2}\right| \leq C\sum_{k=M+1}^{\tau(N)} \frac{2^{n_k}}{N} < C\sum_{k=M}^{\infty} 2^{-k} < C\epsilon. \end{align*} We now turn our attention toS_{N,1}$. Let$\widetilde{N} = 2^{n_1} + 2^{n_2} + \cdots + 2^{n_M}$. If necessary, we may choose a subsequence$\widehat{\mathcal{N}}\subset\mathcal{N}$such that $\lim_{N\in\widehat{\mathcal{N}}}\frac{2^{n_k}}{\widetilde{N}} = \theta_k$ for$1\leq k\leq M$for some$\vec{\theta}=(\theta_1,\theta_2,\ldots,\theta_M)\in\Theta_M. Note that $\frac{\widetilde{N}\log\widetilde{N}}{N\log N} = \frac{\widetilde{N}}{N}\frac{\log\left(\widetilde{N}/N\right)}{\log N}+\frac{\widetilde{N}}{N}.$ Since $\frac{\widetilde{N}}{N} = 1-\frac{2^{M+1}+\cdots+2^{\tau(N)}}{N} > 1 - \epsilon,$ We have $\left|\frac{\widetilde{N}}{N}-1\right|<\epsilon.$ Hence, \begin{align*} \left|\frac{\widetilde{N}\log\widetilde{N}}{N\log N}-1\right| &= \left|\frac{\widetilde{N}}{N}\frac{\log\left(\widetilde{N}/N\right)}{\log N}+\frac{\widetilde{N}}{N}-1\right| \\ &\leq \left|\frac{\widetilde{N}}{N}\frac{\log\left(\widetilde{N}/N\right)}{\log N}\right| + \left|\frac{\widetilde{N}}{N}-1\right|\\ &< \log(1-\epsilon)+\epsilon. \end{align*} SupposeNis large enough so that \begin{align*} \left|\sum_{k=1}^{M}\mathcal{T}(2^{n_k})\frac{2^{n_k}\log 2^{n_k}}{\widehat{\mathcal{N}}\log\widehat{\mathcal{N}}} - \frac{1}{\pi}\right| < \epsilon. \end{align*} We then have \begin{align*} &\left|\sum_{k=1}^{M}\mathcal{T}(2^{n_k})\frac{2^{n_k}\log 2^{n_k}}{\widetilde{N}\log\widetilde{N}}\frac{\widetilde{N}\log\widetilde{N}}{N\log N} - \frac{1}{\pi}\right|\\ &= \left|\sum_{k=1}^{M}\mathcal{T}(2^{n_k})\frac{2^{n_k}\log 2^{n_k}}{\widetilde{N}\log\widetilde{N}}\frac{\widetilde{N}\log\widetilde{N}}{N\log N} - \frac{1}{\pi}\frac{\widetilde{N}\log\widetilde{N}}{N\log N} + \frac{1}{\pi}\frac{\widetilde{N}\log\widetilde{N}}{N\log N}- \frac{1}{\pi}\right|\\ &\leq \left|\sum_{k=1}^{M}\mathcal{T}(2^{n_k})\frac{2^{n_k}\log 2^{n_k}}{\widetilde{N}\log\widetilde{N}}-\frac{1}{\pi}\right| + \frac{1}{\pi}\left|\frac{\widetilde{N}\log\widetilde{N}}{N\log N}-1\right|\\ &< \epsilon + \frac{1}{\pi}\left(|\log(1-\epsilon)|+\epsilon\right). \end{align*} It then follows that \begin{align*} \left|\frac{U_{N,1}(a_n)}{N\log N}-\frac{1}{\pi}\right| &= \left|\sum_{k=1}^{\tau(N)}\mathcal{T}(2^{n_k})\frac{2^{n_k}\log 2^{n_k}}{N\log N} - \frac{1}{\pi}\right| = \left|S_{N,1} + S_{N,2} - \frac{1}{\pi}\right|\\ &\leq \left|S_{M,1}-\frac{1}{\pi}\right| + \left|S_{M,2}\right| < \epsilon + \frac{1}{\pi}\left(|\log(1-\epsilon)|+\epsilon\right). \end{align*} Since\epsilon$was arbitrary, we have shown that $\lim_{N\in\widehat{\mathcal{N}}}\frac{U_{N,1}(a_N)}{N\log N} = \frac{1}{\pi}.$ Since$\mathcal{N} \subset \mathbb{N}$was arbitrary and we have found a subsequence$\widehat{\mathcal{N}}\subset\mathcal{N}$giving the above limit, it follows that $\lim_{N\in\mathbb{N}}\frac{U_{N,1}(a_N)}{N\log N} = \frac{1}{\pi}.$ \end{proof} \subsection{Second-order asymptotics} \begin{remark} In this section, we will often deal with expressions containing the value$\theta_k\log\theta_k$, for some$1\leq k\leq p$and$(\theta_1,\ldots,\theta_p)\in\Theta_p$. If$\theta_k=0$, such expressions will be understood be be zero. This convention is natural, since$\lim_{x\to 0} x\log x = 0$. \end{remark} \begin{definition} For$p\in\mathbb{N}$and$\vec{\theta}=(\theta_1,\theta_2,\ldots,\theta_p)\in\Theta_p$, define $K(\vec{\theta}) := \sum_{k=1}^{p} \theta_k\log\theta_k,$ where, as mentioned in the above remark,$\theta_k\log\theta_k$is understood to be zero of$\theta_k=0$. \end{definition} We now establish bounds on$K(\vec{\theta})$, which are independent of the vector$\vec{\theta}$. \begin{lemma}\label{lem:boundedk} There exists an absolute constant$C$such that for all$p\in\mathbb{N}$and all$\vec{\theta}\in\Theta_p$, $C < K(\vec{\theta}) \leq 0.$ \end{lemma} \begin{proof} Fix$p\in\mathbb{N}$and a vector$\vec{\theta}\in\Theta_p$. Since$0<\theta_k\leq 1$for all$k$, we have $K(\vec{\theta}) \leq 0.$ We now establish a lower bound for$K(\vec{\theta})$, which does not depend on$\vec{\theta}$. First, note that the function $x \mapsto x\log x$ is decreasing for$0 < x < e^{-1}$. Note also that this function has a minimum value of$-e^{-1}$. Let$M$be large enough so that$2^{-(M-1)} < e^{-1}$. If$p < M, then \begin{align*} K(\vec{\theta}) = \sum_{k=1}^{p} \theta_k\log\theta_k \geq -\sum_{k=1}^{p}e^{-1} > -Me^{-1}. \end{align*} Now, suppose thatp \geq M$. We will use the inequality$\theta_k \leq 2^{-(k-1)}$and our choice of$Mto obtain the following: \begin{align*} K(\vec{\theta}) &= \sum_{k=1}^{M-1}\theta_k\log\theta_k + \sum_{k=M}^{p} \theta_k\log\theta_k \geq -Me^{-1} + \sum_{k=M}^{p}2^{-(k-1)}\log 2^{-(k-1)}\\ &= -Me^{-1} - (\log 2)\sum_{k=M}^{p} (k-1)2^{-(k-1)}\\ &> -Me^{-1} - (\log2) \sum_{k=1}^{\infty} (k-1)2^{-(k-1)} = -Me^{-1} - 2\log 2 \end{align*} Thus, we may takeC = -Me^{-1} - 2\log 2$, and we are done. \end{proof} \begin{definition} In light of the previous result on the boundedness of$K(\vec{\theta})$, let us define $\kappa = \inf_{p\in\mathbb{N}}\inf_{\vec{\theta}\in\Theta_p} K(\vec{\theta}).$ \end{definition} The following result is due to Saff: \begin{proposition}\label{prop:limls1} $\lim_{N\to\infty}\frac{\mathcal{L}_{1}(N)-\frac{1}{\pi}N^2\log N}{N^2} = \frac{1}{\pi}(\gamma-\log(\pi/2)),$ where$\gamma=\lim_{N\to\infty}\left(\sum_{k=1}^{N}\frac{1}{k}-\log N\right)$is the Euler-Mascheroni constant. \end{proposition} \begin{definition} For$N\geq 2$, let $\mathcal{W}_{1}(N) := \frac{\mathcal{U}_{1}(N)-\frac{1}{\pi}N\log N}{N\log N},$ and let$\mathcal{W}_{1}(1)=0$. \end{definition} \begin{proposition} The following limit holds: $\lim_{N\to\infty}\mathcal{W}_{1}(N) = \frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)$ \end{proposition} \begin{proof} If$N\geq 1, then \begin{align*} \frac{\mathcal{U}_{1}(N)-\frac{1}{\pi}N\log N}{N} &= \frac{\mathcal{L}_{1}(2N)}{\frac{1}{2}(2N)^2} - \frac{\mathcal{L}_{1}(N)}{N^2} - \frac{1}{\pi}\log N\\ &= 2\frac{\mathcal{L}_{1}(2N)-\frac{1}{\pi}(2N)^2\log 2N+\frac{1}{\pi}(2N)^2\log 2N}{(2N)^2}\\ &- \frac{\mathcal{L}_{1}(N)-\frac{1}{\pi}N^2\log N+\frac{1}{\pi}N^2\log N}{N^2} - \frac{1}{\pi}\log N\\ &= 2\frac{\mathcal{L}_{1}(2N)-\frac{1}{\pi}(2N)^2\log 2N}{(2N)^2} - \frac{\mathcal{L}_{1}(N)-\frac{1}{\pi}N^2\log N}{N^2}\\ &+ \frac{1}{\pi}\log 4. \end{align*} Hence, we have \begin{align*} \lim_{N\to\infty}\mathcal{W}_{1}(N) = \frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right). \end{align*} \end{proof} \begin{theorem} Let(a_N)_{N=0}^{\infty}$be a greedy$s$-energy sequence on$S^1$, where$s=1$. Then the sequence $\left(\frac{U_{N}(a_N)-\frac{1}{\pi}N\log N}{N}\right)$ is divergent. Moreover, we have $$\limsup_{N\to\infty} \frac{U_{N}(a_N)-\frac{1}{\pi}N\log N}{N} = \frac{1}{\pi}\left(\gamma + \log\left(\frac{8}{\pi}\right)\right).\label{eq:secordlimsups1}$$ and $$\liminf_{N\to\infty} \frac{U_{N}(a_N)-\frac{1}{\pi}N\log N}{N} = \frac{1}{\pi}\left(\gamma + \log\left(\frac{8}{\pi}\right)+\kappa\right).\label{eq:secordliminfs1}$$ \end{theorem} \begin{proof} We start by proving \ref{eq:secordlimsups1}. Fix$\vec{\theta}=(\theta_1,\ldots,\theta_p)\in\Theta_p$. Let$\mathcal{N}$be a sequence of natural numbers such that, for$N=2^{n_1}+2^{n_k}+\cdots+2^{n_p}\in\mathcal{N}$,$n_1>n_2>\cdots>n_p\geq 0$, $\lim_{N\in\mathcal{N}} \frac{\theta_k}{N} = \theta_k$ for$1\leq k\leq p. We then have \begin{align*} \frac{U_{N,1}(a_N)-\frac{1}{\pi}N\log N}{N} &= \sum_{k=1}^{p}\left(\frac{\mathcal{U}_{1}(2^{n_k})-\frac{1}{\pi}2^{n_k}\log 2^{n_k} - \frac{1}{\pi}2^{n_k}\log\left(2^{-n_k}N\right)}{N}\right)\\ &= \sum_{k=1}^{p}\frac{2^{n_k}}{N}\left(\mathcal{W}_{1}(2^{n_k}) + \frac{1}{\pi}\log\left(\frac{2^{n_k}}{N}\right)\right), \end{align*} Fix1\leq k\leq p$. First, suppose$\theta_k > 0. It then follows that \begin{align*} \lim_{N\in\mathcal{N}} \frac{2^{n_k}}{N}\left(\mathcal{W}_{1}(2^{n_k}) + \frac{1}{\pi}\log\left(\frac{2^{n_k}}{N}\right)\right) = \frac{\theta_k}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)+\log\theta_k\right). \end{align*} Now, assume\theta_k=0. \begin{align*} \lim_{N\in\mathcal{N}} \frac{2^{n_k}}{N}\left(\mathcal{W}_{1}(2^{n_k}) + \frac{1}{\pi}\log\left(\frac{2^{n_k}}{N}\right)\right)=0. \end{align*} Hence, \begin{align*} \lim_{N\in\mathcal{N}}\frac{U_{N,1}(a_N)-\frac{1}{\pi}N\log N}{N}&=\lim_{N\in\mathcal{N}}\sum_{k=1}^{p}\frac{2^{n_k}}{N}\left(\mathcal{W}_{1}(2^{n_k}) + \frac{1}{\pi}\log\left(\frac{2^{n_k}}{N}\right)\right)\\ &=\sum_{k=1}^{p}\frac{\theta_k}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)+\log\theta_k\right)\\ &= \frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)+\frac{1}{\pi}K(\vec{\theta}). \end{align*} Suppose now that\mathcal{N}$is any sequence of natural numbers such that $\left(\frac{U_{N}(a_N)-\frac{1}{\pi}N\log N}{N}\right)_{N\in\mathcal{N}}\quad\text{converges.}$ First, suppose that there is$p\in\mathbb{N}$such that$\tau(N)=p$for infinitely many$N\in\mathcal{N}$. If necessary, we may choose a subsequence$\widehat{\mathcal{N}}\subset\mathcal{N}$such that for$N=2^{n_1}+2^{n_2}+\cdots+2^{n_p}\in\widehat{\mathcal{N}}, n_1>n_2>\cdots>n_p\geq 0$, we have $\lim_{N\in\widehat{\mathcal{N}}}\frac{2^{n_k}}{N}=\theta_k$ for some$\vec{\theta}=(\theta_1,\theta_2,\ldots,\theta_p)\in\Theta_p$and all$1\leq k\leq p. Then \begin{align*} \lim_{N\in\mathcal{N}}\frac{U_{N}(a_N)-\frac{1}{\pi}N\log N}{N} &= \lim_{N\in\widehat{\mathcal{N}}} \frac{U_{N}(a_N)-\frac{1}{\pi}N\log N}{N}\\ &= \frac{1}{\pi}\left(\gamma + \log\left(\frac{8}{\pi}\right)+K(\vec{\theta};s)\right). \end{align*} This, we have \begin{align*} \frac{1}{\pi}\left(\gamma + \log\left(\frac{8}{\pi}\right)+\kappa\right)&\leq\lim_{N\in\mathcal{N}}\frac{U_{N}(a_N)-\frac{1}{\pi}N\log N}{N} \leq\frac{1}{\pi}\left(\gamma + \log\left(\frac{8}{\pi}\right)\right). \end{align*} Now, suppose that\tau(N)\to\infty$as$N\to\infty$along the subsequence$\mathcal{N}$. Fix$\epsilon \in \left(0,\frac{1}{2}\right)$. Let$M$be an integer large enough so that$2^{-M} < e^{-1}$and $\sum_{k=M}^{\infty} k2^{-k} < \epsilon.$ Put $S_{N,1} := \sum_{k=1}^{M}\frac{2^{n_k}}{N}\left(\mathcal{W}_{1}(2^{n_k}) + \frac{1}{\pi}\log\left(\frac{2^{n_k}}{N}\right)\right)$ and $S_{N,2} := \sum_{k=M+1}^{\tau(N)}\frac{2^{n_k}}{N}\left(\mathcal{W}_{1}(2^{n_k}) + \frac{1}{\pi}\log\left(\frac{2^{n_k}}{N}\right)\right),$ where$S_{N,2}$is understood to be zero if$M>\tau(N)$. Since$(\mathcal{W}_{1}(N))$converges, there is a constant$C$such that$|\mathcal{W}_{1}(N)|\leq C$for all$N$. Hence, by our choice of$M, \begin{align*} |S_{N,2}| &= \left|\sum_{k=M+1}^{\tau(N)}\frac{2^{n_k}}{N}\left(\mathcal{W}_{1}(2^{n_k}) + \frac{1}{\pi}\log\left(\frac{2^{n_k}}{N}\right)\right)\right|\\ &\leq C\sum_{k=M+1}^{\tau(N)}\frac{2^{n_k}}{N} + \frac{1}{\pi}\left|\sum_{k=M+1}^{\tau(N)}\frac{2^{n_k}}{N}\log\left(\frac{2^{n_k}}{N}\right)\right|\\ &< C\epsilon + \frac{\log 2}{\pi}\sum_{k=M+1}^{\tau(N)} (k-1)2^{-(k-1)}\\ &< \left(C+\frac{\log 2}{\pi}\right)\epsilon. \end{align*} We now turn our attention toS_{N,1}$. We may choose a subsequence$\widehat{\mathcal{N}}\subset\mathcal{N}$such that for$N=2^{n_1}+\cdots+2^{n_{\tau(N)}}\in\widehat{\mathcal{N}}$and all$1\leq k\leq M$, $\lim_{N\in\widehat{\mathcal{N}}}\frac{2^{n_k}}{N}=\theta_k$ for some$\vec{\theta}=(\theta_1,\ldots,\theta_M)\in\Theta_M$. Fix$1\leq k\leq M$and suppose that$\theta_k>0$. Let$N_0$be large enough so that $\left|\frac{2^{n_k}}{\widetilde{N}}-\theta_k\right| < 2^{-k}\epsilon$ whenever$N \geq N_0$. Furthermore, there is an integer$N_1$such that $\left|\frac{2^{n_k}}{\widetilde{N}}\log\frac{2^{n_k}}{\widetilde{N}}-\theta_k\log\theta_k\right|<2^{-k}\epsilon$ whenever$N\geq N_1$. Note that since$\widehat{N}/N>1-\epsilon$, $\left|\log\frac{\widehat{N}}{N}\right|<|\log(1-\epsilon)|.$ The triangle inequality then gives, for$N\geq\max\{N_0,N_1\}, \begin{align*} \left|\frac{2^{n_k}}{\widetilde{N}}\log\frac{2^{n_k}}{N}-\theta_k\log\theta_k\right| &= \left|\frac{2^{n_k}}{\widetilde{N}}\log\frac{2^{n_k}}{\widetilde{N}}+\frac{2^{n_k}}{\widetilde{N}}\log\frac{\widetilde{N}}{N}-\theta_k\log\theta_k\right|\\ &\leq \left|\frac{2^{n_k}}{\widetilde{N}}\log\frac{2^{n_k}}{\widetilde{N}}-\theta_k\log\theta_k\right|+\frac{2^{n_k}}{N}\left|\log\frac{\widetilde{N}}{N}\right|\\ &< 2^{-k}\epsilon + 2^{-(k-1)}|\log(1-\epsilon)| \end{align*} LetN_2$be large enough so that $\left|\mathcal{W}_{1}(N)-\frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)\right| < 2^{-k}\epsilon$ whenever$N \geq N_1$. Since$\theta_k>0$, we have$n_k \to \infty$as$N\to\infty$. So there is an integer$N_3$such that$2^{n_k} \geq N_1$whenever$N \geq N_2$. Let$C$be a constant such that$|\mathcal{W}_{1}(N)|\leq C$for all$N$. It then follows that, if$N\geq\max\{N_1,N_3\}, \begin{align*} &\left|\frac{2^{n_k}}{\widetilde{N}}\mathcal{W}_{1}(2^{n_k})-\frac{\theta_k}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)\right|\\ &= \left|\frac{2^{n_k}}{\widetilde{N}}\mathcal{W}_{1}(2^{n_k})-\theta_k\mathcal{W}_{1}(2^{n_k})+\theta_k\mathcal{W}_{1}(2^{n_k})-\frac{\theta_k}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)\right|\\ &\leq C\left|\frac{2^{n_k}}{\widehat{N}}-\theta_k\right| + \theta_k\left|\mathcal{W}_{1}(2^{n_k})-\frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)\right|\\ &< \left(C+\theta_k\right)2^{-k}\epsilon. \end{align*} Finally, it follows that \begin{align} &\left|\frac{2^{n_k}}{\widehat{N}}\left(\mathcal{W}_{1}(2^{n_k})+\frac{1}{\pi}\log\frac{2^{n_k}}{N}\right)-\frac{\theta_k}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)+\log\theta_k\right)\right|\\ &\leq \left|\frac{2^{n_k}}{\widehat{N}}\mathcal{W}_{1}(2^{n_k})-\frac{\theta_k}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)\right| + \frac{1}{\pi}\left|\frac{2^{n_k}}{\widehat{N}}\log\frac{2^{n_k}}{N}-\theta_k\log\theta_k\right|\\ &< (C+1+\theta_k)2^{-k}\epsilon+2^{-(k-1)}|\log(1-\epsilon)|.\label{eq:thetanonz} \end{align} On the other hand, suppose\theta_k=0$. Since $\frac{2^{n_k}}{N} \leq \frac{2^{n_k}}{\widehat{N}},$$2^{n_k}/N \to 0$as$N\to \infty$along$\widehat{\mathcal{N}}$, so we obtain $\lim_{N\in\widehat{\mathcal{N}}}\frac{2^{n_k}}{\widehat{N}}\left(\mathcal{W}_{1}(2^{n_k})+\frac{1}{\pi}\log\left(\frac{2^{n_k}}{N}\right)\right)=0.$ Hence, there is an integer$N_4$such that, whenever$N\geq N_4, \begin{align*} \left|\frac{2^{n_k}}{\widehat{N}}\left(\mathcal{W}_{1}(2^{n_k})+\frac{1}{\pi}\log\left(\frac{2^{n_k}}{N}\right)\right)\right|<(C+1+\theta_k)2^{-k}\epsilon+2^{-(k-1)}|\log(1-\epsilon)|. \end{align*} Since\theta_k=0, this is equivalent to \ref{eq:thetanonz}. Thus, we have \begin{align*} &\left|S_{N,1}-\frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)+K(\vec{\theta})\right)\right|\\ &= \left|\frac{\widehat{N}}{N}\sum_{k=1}^{M}\frac{2^{n_k}}{\widehat{N}}\left(\mathcal{W}_{1}(2^{n_k})+\frac{1}{\pi}\log\left(\frac{2^{n_k}}{N}\right)\right)-\sum_{k=1}^{M}\left(\frac{\theta_k}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)+\theta_k\log\theta_k\right)\right|\\ &\leq \left|\sum_{k=1}^{M}\frac{2^{n_k}}{\widehat{N}}\left(\mathcal{W}_{1}(2^{n_k})+\frac{1}{\pi}\log\left(\frac{2^{n_k}}{N}\right)\right)-\sum_{k=1}^{M}\left(\frac{\theta_k}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)+\theta_k\log\theta_k\right)\right|\\ &+ \frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)\left|\frac{\widetilde{N}}{N}-1\right|\\ &< \sum_{k=1}^{M}\left((C+1+\theta_k)2^{-k}\epsilon+2^{-(k-1)}|\log(1-\epsilon)|\right)+\frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)\epsilon\\ &< (C+1+\theta_k)\epsilon + 2|\log(1-\epsilon)|\frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)\epsilon. \end{align*} Thus, we have shown that \begin{align*} \lim_{N\in\widehat{\mathcal{N}}} \frac{U_{N,1}(a_N)-\frac{1}{\pi}N\log N}{N} &= \limsup_{N\in\widehat{\mathcal{N}}}\left(S_{N,1}+S_{N,2}\right) \leq \limsup_{N\in\widehat{\mathcal{N}}}S_{N,1} + \limsup_{N\in\widehat{\mathcal{N}}}S_{N,2}\\ &< \frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right) - \epsilon\left(C+\theta_k+\theta_k\log\theta_k+2\log 2 + 1\right)\\ &+ \epsilon\left(C + \frac{1}{\pi}\log 2\right). \end{align*} Since\mathcal{N}$was arbitrary, letting$\epsilon \to 0gives $\limsup_{N\in\widehat{\mathcal{N}}} \frac{U_{N,1}(a_N)-\frac{1}{\pi}N\log N}{N} \leq \frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right).$ But since $\lim_{p\to\infty} \frac{U_{2^p,1}(a_{2^p})-\frac{1}{\pi}2^p\log 2^p}{2^p} = \lim_{p\to\infty} \mathcal{W}_{1}(2^p) = \frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right),$ we obtain $\limsup_{N\in\widehat{\mathcal{N}}} \frac{U_{N,1}(a_N)-\frac{1}{\pi}N\log N}{N} \geq \frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right).$ And \ref{eq:secordlimsups1} is proved. We also have \begin{align*} \lim_{N\in\widehat{\mathcal{N}}}\frac{U_{N,1}(a_N)-\frac{1}{\pi}N\log N}{N} &= \liminf_{N\in\widehat{\mathcal{N}}} \left(S_{N,1}+S_{N,2}\right) \geq \liminf_{N\in\widehat{\mathcal{N}}}S_{N,1} + \liminf_{N\in\widehat{\mathcal{N}}}S_{N,2}\\ &> \frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)+K(\vec{\theta})\right)\\ &- (C+1+\theta_k)\epsilon + 2|\log(1-\epsilon)|\frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)\right)\epsilon\\ &+\left(C+\frac{\log 2}{\pi}\right)\epsilon. \end{align*} Letting\epsilon \to 0, we obtain \begin{align*} \lim_{Nin\mathcal{N}} \frac{U_{N,1}(a_N)-\frac{1}{\pi}N\log N}{N} = \lim_{Nin\mathcal{\widehat{N}}} \frac{U_{N,1}(a_N)-\frac{1}{\pi}N\log N}{N}\\ \geq \frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)+K(\vec{\theta})\right). \end{align*} Because\mathcal{N}$is arbitrary, it now follows that $\liminf_{N\to\infty} \frac{U_{N,1}(a_N)-\frac{1}{\pi}N\log N}{N}\geq \frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)+\kappa\right)$ And since $\frac{1}{\pi}\left(\gamma+\log\left(\frac{8}{\pi}\right)+K(\vec{\theta})\right)$ is a limit point of \ref{eq:secordseqs1}, \ref{eq:secordliminfs1} follws. \end{proof} \section{The case$s>1$} \begin{proposition}\label{prop:sgeq1ucal} If$s > 1$, then $\lim_{N \to \infty} \frac{\mathcal{U}_{s}(N)}{N^{1+s}} = (2^s - 1)\frac{2\zeta(s)}{(2\pi)^s}.$ \end{proposition} \begin{proof} It follows from [\cite{MMRS}, Theorem 3.2] that if$s > 1, then $\lim_{N \to \infty} \frac{\mathcal{L}_{s}(N)}{N^{1+s}} = \frac{2\zeta(s)}{(2\pi)^s}.$ We now have \begin{align*} \lim_{N \to \infty} \frac{\mathcal{U}_{s}(N)}{N^s} &= \lim_{N \to \infty} \frac{1}{N^s}\left(\frac{\mathcal{L}_{s}(2N)}{2N} - \frac{\mathcal{L}_{s}(N)}{N}\right) \\ &= \lim_{N \to \infty} \left(2^s \frac{\mathcal{L}_{s}(2N)}{(2N)^{1+s}} - \frac{\mathcal{L}_{s}(N)}{N^{1+s}}\right) \\ &= (2^s - 1)\frac{2\zeta(s)}{(2\pi)^s}. \end{align*} \end{proof} \begin{theorem} Let(a_N)_{N=0}^{\infty}$be a greedy$s$-energy sequence for$s>1$. Then the sequence $$\label{eq:seqsgeq1} \left(\frac{U_{N,s}(a_N)}{N^s}\right)_{N=1}^{\infty}$$ is bounded and divergent. Moreover, we have $$\label{eq:limsupsge1} \limsup_{N\to\infty} \frac{U_{N,s}(a_N)}{N^s} = (2^s-1)\frac{2\zeta(s)}{(2\pi)^s}$$ and $$\label{eq:liminfsge1} \liminf_{N\to\infty}\frac{U_{N,s}(a_N)}{N^s} = \frac{2\zeta(s)}{(2\pi)^s}.$$ \end{theorem} \begin{proof} For$N\geq 1$and$s>1$, let $T_{s}(N) := \frac{\mathcal{U}_{s}(N)}{N^{1+s}}.$ By Proposition~\ref{prop:sgeq1ucal}, $T_{s}(N) \to (2^s-1)\frac{2\zeta(s)}{(2\pi)^s}\quad\text{as}\quad N\to\infty.$ Note that if$N=2^{n_1}+2^{n_2}+\cdots+2^{n_p}$,$n_1>n_2>\cdots>n_p\geq 0, then \begin{align*} \frac{U_{s}(N)}{N^s} = \sum_{k=1}^{p}\mathcal{T}_{s}(2^{n_k})\left(\frac{2^{n_k}}{N}\right)^{s}. \end{align*} Because(T_{s}(N))_{N=1}^{\infty}$converges and$2^{n_k}/N \leq 2^{-(k-1)}$, the boundedness of \ref{eq:seqsgeq1} follows immediately. Fix$p\in\mathbb{N}$and$\vec{\theta}\in\Theta_p$. Then there is a sequence$\mathcal{N}\subset\mathbb{N}$such that $\lim_{N\in\mathcal{N}}\frac{2^{n_k}}{N} = \theta_k$ for$1\leq k\leq p$. Since$\sum_{k=1}^{p} \theta_k=1$, it follows that $\lim_{N\in\mathcal{N}}\frac{U_{N,s}(a_N)}{N^{1+s}}=G(\vec{\theta};s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}.$ Thus, $G(\vec{\theta};1+s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}$ is a limit point of the sequence \ref{eq:seqsgeq1}. Now, suppose that$\mathcal{N}$is a sequence of natural numbers such that the sequence \ref{eq:seqsgeq1} indexed by$\mathcal{N}$converges. First, suppose that there is$p\in\mathbb{N}$such that$\tau(N)=p$for infinitely many$N\in\mathcal{N}$. If necessary, we may now choose a subsequence$\widehat{\mathcal{N}}\subset\mathcal{N}$such that $\lim_{N\in\widehat{\mathcal{N}}}\frac{2^{n_k}}{N}=\theta_k$ for some$\theta_k\in\Theta_p$and all$1\leq k\leq p$. From our previous considerations, it follows that $\lim_{N\in\mathcal{N}} \frac{U_{N,s}(a_N)}{N^{1+s}} =\lim_{N\in\widehat{\mathcal{N}}} \frac{U_{N,s}(a_N)}{N^{1+s}} \leq (2^s-1)\frac{2\zeta(s)}{(2\pi)^s}.$ Suppose, on the other hand,$\tau(N) \to \infty$as$N\to\infty$along$\mathcal{N}$. Let$\epsilon\in (0,1)$and let$M$be an integer such that $\sum_{k=M}^{\infty} 2^{-sk} < \epsilon.$ Put $S_{N,1} := \sum_{k=1}^{M}\mathcal{T}_{s}(2^{n_k})\left(\frac{2^{n_k}}{N}\right)^{1+s}$ and $S_{N,1} := \sum_{k=M=1}^{\tau(N)}\mathcal{T}_{s}(2^{n_k})\left(\frac{2^{n_k}}{N}\right)^{1+s}.$ Let$\widehat{\mathcal{N}}\subset\mathcal{N}$be such that, for$N=2^{n_k}+2^{n_2}+\cdots+2^{n_k}\in\widehat{\mathcal{N}}$,$n_1>n_2>\cdots>n_p\geq 0$, $\lim_{N\in\widehat{\mathcal{N}}}\frac{2^{n_k}}{N} = \theta_k$ for some$\vec{\theta}=(\theta_1,\ldots,\theta_M)\in\Theta_M$and all$1\leq k\leq M$. Let$C$be a constant such that $|\mathcal{T}_{s}(N)| \leq C$ for all natural numbers$N$. By our choice of$M, it then follows that \begin{align*} |s_{N,2}| = \left|\sum_{k=M+1}^{\tau(N)}\mathcal{T}_{s}(2^{n_k})\left(\frac{2^{n_k}}{N}\right)^{s}\right| \leq C\sum_{k=M+1}^{\tau(N)} 2^{-sk} < C\epsilon. \end{align*} Since\widetilde{N}/N \leq 1, \begin{align*} S_{N,1} = \sum_{k=1}^{M}\mathcal{T}_{s}(2^{n_k})\left(\frac{2^{n_k}}{\widehat{N}}\right)^{s}\left(\frac{\widehat{N}}{N}\right)^{s} \leq \sum_{k=1}^{M}\mathcal{T}_{s}(2^{n_k})\left(\frac{2^{n_k}}{\widehat{N}}\right)^{s}. \end{align*} So it follows that \begin{align*} \lim_{N\in\mathcal{N}}\frac{U_{N,s}(a_N)}{N^{s}} &= \limsup_{N\in\widehat{\mathcal{N}}}\left(S_{N,1}+S_{N,2}\right) \leq \limsup_{N\in\widehat{\mathcal{N}}} S_{N,1} + \limsup_{N\in\widehat{\mathcal{N}}} S_{N,2}\\ &< \limsup_{N\in\widehat{\mathcal{N}}}\sum_{k=1}^{M}\mathcal{T}_{s}(2^{n_k})\left(\frac{2^{n_k}}{N}\right)^{s} + C\epsilon\\ &\leq \limsup_{N\in\widehat{\mathcal{N}}}\sum_{k=1}^{M}\mathcal{T}_{s}(2^{n_k})\left(\frac{2^{n_k}}{\widetilde{N}}\right)^{s} + C\epsilon\\ &\leq (2^s-1)\frac{2\zeta(s)}{(2\pi)^s} + C\epsilon \end{align*} Letting\epsilon \to 0$gives $\lim_{N\in\mathcal{N}}\frac{U_{N,s}(a_N)}{N^{1+s}} \leq (2^s-1)\frac{2\zeta(s)}{(2\pi)^s}.$ Since$\mathcal{N}was arbitrary, this implies $\limsup_{N\to\infty}\frac{U_{N,s}(a_N)}{N^{1+s}} \leq (2^s-1)\frac{2\zeta(s)}{(2\pi)^s}.$ But since $\lim_{p\to\infty} \frac{U_{2^p,1}(a_{2^p})}{(2^p)^{1+s}} = \lim_{p\to\infty} \mathcal{T}_{s}(2^p) = (2^s-1)\frac{2\zeta(s)}{(2^\pi)^s},$ we must have $\limsup_{N\to\infty}\frac{U_{N,s}(a_N)}{N^{1+s}} \geq (2^s-1)\frac{2\zeta(s)}{(2\pi)^s}.$ Thus, \ref{eq:limsupsge1} is proved. Similarly, since $\frac{\widehat{N}}{N} > 1-\epsilon,$ we obtain \begin{align*} \lim_{N\in\mathcal{N}}\frac{U_{N,s}(a_N)}{N^{s}} &= \liminf_{N\in\widehat{\mathcal{N}}}\left(s_{N,1}+S_{N,2}\right) \geq \liminf_{N\in\widehat{\mathcal{N}}} S_{N,1} + \liminf_{N\in\widehat{\mathcal{N}}} S_{N,2}\\ &\geq \liminf_{N\in\widehat{\mathcal{N}}}\sum_{k=1}^{M}\mathcal{T}_{s}(2^{n_k})\left(\frac{2^{n_k}}{N}\right)^{s} - C\epsilon\\ &\geq (1-\epsilon)^{s}\liminf_{N\in\widehat{\mathcal{N}}}\sum_{k=1}^{M}\mathcal{T}_{s}(2^{n_k})\left(\frac{2^{n_k}}{\widetilde{N}}\right)^{s} - C\epsilon\\ &\geq (1-\epsilon)^{s}\overline{g}(\vec{\theta};s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s} - C\epsilon. \end{align*} Now, letting\epsilon \to 0$gives $\lim_{N\in\mathcal{N}}\frac{U_{N,s}(a_N)}{N^{s}} \geq \overline{g}(\vec{\theta};s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}.$ Because$\mathcal{N}$is arbitrary, we have $\liminf_{N\in\mathcal{N}}\frac{U_{N,s}(a_N)}{N^{s}} \geq \overline{g}(\vec{\theta};s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}$ But since $G(\vec{\theta};s)(2^s-1)\frac{2\zeta(s)}{(2\pi)^s}$ is a limit point of the sequence \ref{eq:seqsgeq1}, we also have $\liminf_{N\in\mathcal{N}}\frac{U_{N,s}(a_N)}{N^{s}} \leq \overline{g}(\vec{\theta};s)(2^s-1)\frac{2\zeta(s)}.{(2\pi)^s}.$ So we have proved \ref{eq:liminfsge1}. \end{proof} \begin{thebibliography}{99} \bibitem{BorodachovHardinSaff} S.V. 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