###### Proof.

Remarks on sums of reciprocals of fractional parts

Bence Borda

Graz University of Technology

Steyrergasse 30, 8010 Graz, Austria

Email<EMAIL_ADDRESS>
Keywords: Diophantine approximation, continued fraction,
small fractional parts, Diophantine sum, metric number theory

Mathematics Subject Classification (2020): 11J54, 11J71, 11J83

###### Abstract

The Diophantine sums $\sum_{n=1}^{N}\|n\alpha\|^{-1}$ and
$\sum_{n=1}^{N}n^{-1}\|n\alpha\|^{-1}$ appear in many different areas
including the ergodic theory of circle rotations, lattice point counting and
random walks, often in connection with Fourier analytic methods. Beresnevich,
Haynes and Velani gave estimates for these and related sums in terms of the
Diophantine approximation properties of $\alpha$ that are sharp up to a
constant factor. In the present paper, we remove the constant factor gap
between the upper and the lower estimates, and thus find the precise
asymptotics for a wide class of irrationals. Our methods apply to sums with
the fractional part instead of the distance from the nearest integer function,
and to sums involving shifts $\|n\alpha+\beta\|$ as well. We also comment on a
higher dimensional generalization of these sums.

##  1 Introduction

The subject of this paper is the asymptotic behavior of the Diophantine sums

$\sum_{n=1}^{N}\frac{1}{\|n\alpha\|}\qquad\textrm{and}\qquad\sum_{n=1}^{N}\frac{1}{n\|n\alpha\|}$
(1)

for various irrational $\alpha$, where $\|\cdot\|$ denotes the distance from
the nearest integer function. These sums appear in many different fields such
as uniform distribution theory [21, 28], multiplicative Diophantine
approximation [4, 13], lattice point counting in polygons [2, 9, 22, 23, 31],
dynamical systems [10, 15, 16, 17] and random walks [6, 7, 8, 33]. We refer to
[3, 4] for a comprehensive survey.

The behavior of the more general sum $\sum_{n=1}^{N}n^{-p}\|n\alpha\|^{-q}$ is
highly sensitive to the value of the exponents $p,q\geq 0$ [27]. Sharp
estimates in the case $p=q=2$ were given in [2], and in the case $p=q>1$ in
[11]. The case $p>q$ leads to convergent series for certain irrationals [12].
For higher dimensional generalizations of the sums in (1), see [1, 19, 20,
29].

Hardy and Littlewood [22, 23, 24], Haber and Osgood [21], Kruse [27] and more
recently Beresnevich, Haynes and Velani [4] gave estimates for the sums in (1)
in terms of the Diophantine approximation properties of $\alpha$ that are
sharp up to a constant factor. The main goal of the present paper is to remove
the constant factor gap between the upper and the lower estimates, thereby
establishing the precise asymptotics.

Recall that an irrational $\alpha$ is called badly approximable if
$\inf_{n\in\mathbb{N}}n\|n\alpha\|>0$. The best known estimates for such an
$\alpha$ are111For the sake of readability, with a slight abuse of notation in
all error terms $\log x:=\log\max\\{e,x\\}$. [4]

$N\log N\leq\sum_{n=1}^{N}\frac{1}{\|n\alpha\|}\ll N\log
N\qquad\textrm{and}\qquad\frac{1}{2}(\log
N)^{2}\leq\sum_{n=1}^{N}\frac{1}{n\|n\alpha\|}\leq 33(\log N)^{2}+O(\log N).$

Our first result improves these.

######  Theorem 1.

Let $\alpha$ be a badly approximable irrational. For any $N\geq 1$,

$\sum_{n=1}^{N}\frac{1}{\|n\alpha\|}=2N\log
N+O(N)\qquad\textrm{and}\qquad\sum_{n=1}^{N}\frac{1}{n\|n\alpha\|}=(\log
N)^{2}+O(\log N)$

with implied constants depending only on $\alpha$.

The only previously known precise asymptotic result (without a constant factor
gap between the upper and the lower estimates) for the sums in (1) appeared in
an obscure paper of Erdős from 1948 [18], in which he showed that

$\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
1/(2N)\end{subarray}}^{N}\frac{1}{\|n\alpha\|}\sim 2N\log
N\qquad\textrm{and}\qquad\sum_{n=1}^{N}\frac{1}{n\|n\alpha\|}\sim(\log
N)^{2}\qquad\textrm{for a.e. }\alpha.$ (2)

This result seems to have escaped the attention of several later authors who
subsequently proved weaker estimates in the metric setting. We learned about
(2) from the recent historical survey [3], and it served as the starting point
of our investigations. Our next result improves (2) by finding the precise
order of the error term.

######  Theorem 2.

Let $c>0$ be an arbitrary constant, and let $\varphi$ be a positive
nondecreasing function on $(0,\infty)$. If
$\sum_{k=1}^{\infty}1/\varphi(k)<\infty$, then for a.e. $\alpha$,

$\begin{split}\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}&=2N\log N+O(N\varphi(\log
N)^{1/2}),\\\ \sum_{n=1}^{N}\frac{1}{n\|n\alpha\|}&=(\log
N)^{2}+O(\varphi(\log N)+\log N\log\log N)\end{split}$

with implied constants depending only on $c$, $\varphi$ and $\alpha$. If
$\sum_{k=1}^{\infty}1/\varphi(k)=\infty$, then for a.e. $\alpha$ the sets

$\begin{split}\Bigg{\\{}N\in\mathbb{N}\,:\,\sum_{\begin{subarray}{c}n=1\\\
\|n\alpha\|\geq c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}&\geq 2N\log
N+N\varphi(\log N)^{1/2}\Bigg{\\}},\\\
\Bigg{\\{}N\in\mathbb{N}\,:\,\sum_{n=1}^{N}\frac{1}{n\|n\alpha\|}&\geq(\log
N)^{2}+\varphi(\log N)\Bigg{\\}}\end{split}$

have upper asymptotic density 1.

Recall that the upper asymptotic density of a set $A\subseteq\mathbb{N}$ is
defined as $\limsup_{N\to\infty}|A\cap[1,N]|/N$. In particular, for a.e.
$\alpha$ we have

$\begin{split}\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}&=2N\log N+O(N(\log
N)^{1/2}(\log\log N)^{1/2+\varepsilon}),\\\
\sum_{n=1}^{N}\frac{1}{n\|n\alpha\|}&=(\log N)^{2}+O(\log N(\log\log
N)^{1+\varepsilon})\end{split}$

with any $\varepsilon>0$, but these fail with $\varepsilon=0$. Note that the
cutoff $\|n\alpha\|\geq c/N$ in the first sum is necessary, as for a.e.
$\alpha$ we have $\|n\alpha\|<(n\log n\log\log n)^{-1}$ for infinitely many
$n\in\mathbb{N}$.

Our main results, Theorems 3 and 4 on the sums in (1) with a general
irrational $\alpha$ are presented in Section 2. We discuss closely related
sums involving fractional parts and shifts $\|n\alpha+\beta\|$, and comment on
a higher dimensional generalization in Section 3.

##  2 Main estimates

For the rest of the paper, $\alpha$ is an irrational number with continued
fraction $\alpha=[a_{0};a_{1},a_{2},\ldots]$ and convergents
$p_{k}/q_{k}=[a_{0};a_{1},a_{2},\ldots,a_{k}]$. Set
$s_{K}=\sum_{k=1}^{K}a_{k}$. We refer to [26] for a general introduction to
continued fractions.

###  2.1 General irrationals

In this section, we prove our main estimates for the sums in (1) with a
general irrational $\alpha$.

######  Theorem 3.

Let $c>0$ be an arbitrary constant. For any $K\geq 0$ and $q_{K}\leq
N<q_{K+1}$,

$\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}=2N\log
N+O\left((a_{K+1}^{1/2}+\log s_{K+1})N\right).$

If in addition $4(ca_{K+1})^{1/2}q_{K}\leq N$, then

$\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}\geq 2N\log N+q_{K+1}-O((\log
s_{K+1})N).$

The implied constants depend only on $c$.

###### Proof.

If $N\leq 4c$, then

$0\leq\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}\leq\frac{N^{2}}{c}\ll 1,$

and the claims trivially follow. We may thus assume that $N>4c$.

Summing the identity
$\|n\alpha\|^{-1}=\int_{0}^{\infty}t^{-2}\mathds{1}_{\\{\|n\alpha\|\leq
t\\}}\,\mathrm{d}t$ over all $1\leq n\leq N$ such that $\|n\alpha\|\geq c/N$
yields

$\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}=\int_{c/N}^{\infty}\frac{1}{t^{2}}\left|\left\\{1\leq
n\leq N\,:\,c/N\leq\|n\alpha\|\leq t\right\\}\right|\,\mathrm{d}t.$ (3)

Let $R=\frac{1}{8}(s_{K}+N/q_{K})+c$, and note that $c/N\leq R/N\leq 1/2$. The
latter inequality follows from the assumption $c<N/4$ and the general estimate
$s_{K}\leq q_{K}$, which can be easily seen e.g. by induction on $K$. We will
estimate the integral in (3) on the intervals $[c/N,R/N]$, $[R/N,1/2]$ and
$[1/2,\infty)$ separately.

The integral on $[1/2,\infty)$ is negligible:

$\int_{1/2}^{\infty}\frac{1}{t^{2}}\left|\left\\{1\leq n\leq
N\,:\,c/N\leq\|n\alpha\|\leq
t\right\\}\right|\,\mathrm{d}t\leq\int_{1/2}^{\infty}\frac{N}{t^{2}}\,\mathrm{d}t=2N.$
(4)

Consider now the integral on $[R/N,1/2]$. Let $D_{N}(\alpha)$ denote the
discrepancy of the point set $\\{n\alpha\\}$, $1\leq n\leq N$. That is,

$D_{N}(\alpha)=\sup_{I\subseteq[0,1]}\left|\sum_{n=1}^{N}\mathds{1}_{I}(\\{n\alpha\\})-\lambda(I)N\right|,$

where the supremum is over all intervals $I\subseteq[0,1]$, and $\lambda$ is
the Lebesgue measure. In particular,

$\left|\left\\{1\leq n\leq N\,:\,c/N\leq\|n\alpha\|\leq
t\right\\}\right|=2\left(t-\frac{c}{N}\right)N+O(D_{N}(\alpha))\quad\textrm{uniformly
in }t\in[R/N,1/2].$

A classical discrepancy estimate [28, p. 126] states that $D_{N}(\alpha)\leq
2(s_{K}+N/q_{K})\ll R$. Therefore

$\begin{split}\int_{R/N}^{1/2}\frac{1}{t^{2}}\left|\left\\{1\leq n\leq
N\,:\,c/N\leq\|n\alpha\|\leq
t\right\\}\right|\,\mathrm{d}t&=\int_{R/N}^{1/2}\frac{2Nt-2c+O(D_{N}(\alpha))}{t^{2}}\,\mathrm{d}t\\\
&=2N\log N+O(N\log R).\end{split}$ (5)

Finally, consider the integral on $[c/N,R/N]$. We will show that

$\int_{c/N}^{R/N}\frac{1}{t^{2}}\left|\left\\{1\leq n\leq
N\,:\,\|n\alpha\|\leq t\right\\}\right|\,\mathrm{d}t\ll(a_{K+1}^{1/2}+\log
R)N.$ (6)

For any $1\leq n<n^{\prime}\leq N$ we have $1\leq n^{\prime}-n<q_{K+1}$, hence
by the best rational approximation property of continued fraction convergents,
$\|n^{\prime}\alpha-n\alpha\|\geq\|q_{K}\alpha\|\geq 1/(2q_{K+1})$. The
pigeonhole principle thus shows that

$\left|\left\\{1\leq n\leq N\,:\,\|n\alpha\|\leq t\right\\}\right|\leq
4q_{K+1}t+1,$

therefore

$\int_{c/N}^{R/N}\frac{1}{t^{2}}\left|\left\\{1\leq n\leq
N\,:\,\|n\alpha\|\leq t\right\\}\right|\,\mathrm{d}t\ll q_{K+1}\log R.$ (7)

The previous formula implies (6) whenever $N\gg q_{K+1}$.

It remains to prove (6) when, say, $N<q_{K+1}/4$. Let $A=\\{1\leq n\leq
N\,:\,q_{K}\mid n\\}$ and $B=\\{1\leq n\leq N\,:\,q_{K}\nmid n\\}$. Consider
an element $n=jq_{K}\in A$ with some integer $1\leq j\leq N/q_{K}$, and
observe that $\|jq_{K}\alpha\|\leq t$ implies that $j\leq 2q_{K+1}t$. In
particular,

$\left|\left\\{n\in A\,:\,\|n\alpha\|\leq
t\right\\}\right|\leq\min\left\\{\frac{N}{q_{K}},2q_{K+1}t\right\\}.$

If $N\leq(q_{K}q_{K+1})^{1/2}$, then

$\int_{c/N}^{R/N}\frac{1}{t^{2}}\left|\left\\{n\in A\,:\,\|n\alpha\|\leq
t\right\\}\right|\,\mathrm{d}t\leq\int_{c/N}^{R/N}\frac{N/q_{K}}{t^{2}}\,\mathrm{d}t\leq\frac{N^{2}}{cq_{K}}\ll
a_{K+1}^{1/2}N.$

If $N>(q_{K}q_{K+1})^{1/2}$, then again

$\begin{split}\int_{c/N}^{R/N}\frac{1}{t^{2}}\left|\left\\{n\in
A\,:\,\|n\alpha\|\leq
t\right\\}\right|\,\mathrm{d}t&\leq\int_{c/N}^{cN/(q_{K}q_{K+1})}\frac{2q_{K+1}t}{t^{2}}\,\mathrm{d}t+\int_{cN/(q_{K}q_{K+1})}^{\infty}\frac{N/q_{K}}{t^{2}}\,\mathrm{d}t\\\
&\ll q_{K+1}\log\frac{N^{2}}{q_{K}q_{K+1}}+q_{K+1}\\\ &\ll
a_{K+1}^{1/2}N.\end{split}$

Consider now $B_{j}=\\{n\in B\,:\,jq_{K}<n<(j+1)q_{K}\\}$ with some integer
$0\leq j<\lceil N/q_{K}\rceil$. Observe that $|n\alpha-np_{K}/q_{K}|\leq
n/(q_{K}q_{K+1})<1/q_{K}$, and that $np_{K}\pmod{q_{K}}$ attains nonzero
residue classes. Therefore the set $\\{n\alpha\\}$, $n\in B_{j}$ coincides
with $\\{\ell/q_{K}\,:\,1\leq\ell\leq q_{K}-1\\}$ (or a subset thereof in case
$j=\lceil N/q_{K}\rceil-1$) up to an error of $1/q_{K}$.

Since $\mathrm{sgn}(\alpha-p_{K}/q_{K})=(-1)^{K}$, for all $n\in B_{j}$ with
$np_{K}\not\equiv(-1)^{K-1}\pmod{q_{K}}$, we have $\|n\alpha\|\geq 1/q_{K}$.
The identity $q_{K}p_{K-1}-q_{K-1}p_{K}=(-1)^{K}$ shows that the unique $n\in
B_{j}$ with $np_{K}\equiv(-1)^{K-1}\pmod{q_{K}}$ is $n=jq_{K}+q_{K-1}$. The
assumption $N<q_{K+1}/4$ ensures that

$\left\|(jq_{K}+q_{K-1})\alpha\right\|=\|q_{K-1}\alpha\|-j\|q_{K}\alpha\|\geq\frac{1}{2q_{K}}-\frac{N}{q_{K}}\cdot\frac{1}{q_{K+1}}>\frac{1}{4q_{K}}.$

In particular, $\|n\alpha\|\geq 1/(4q_{K})$ for all $n\in B$. Since
$\\{n\alpha\\}$, $n\in B_{j}$ is well approximated by (a subset of)
$\\{\ell/q_{K}\,:\,1\leq\ell\leq q_{K}-1\\}$, we have

$\left|\left\\{n\in B_{j}\,:\,\|n\alpha\|\leq t\right\\}\right|\ll
q_{K}t\quad\textrm{uniformly in }1/(4q_{K})\leq t\leq 1/2,$

and by summing over $0\leq j<\lceil N/q_{K}\rceil$,

$\left|\left\\{n\in B\,:\,\|n\alpha\|\leq t\right\\}\right|\ll
Nt\quad\textrm{uniformly in }1/(4q_{K})\leq t\leq 1/2.$

Therefore

$\int_{c/N}^{R/N}\frac{1}{t^{2}}\left|\left\\{n\in B\,:\,\|n\alpha\|\leq
t\right\\}\right|\,\mathrm{d}t\ll\int_{1/(4q_{K})}^{R/N}\frac{Nt}{t^{2}}\,\mathrm{d}t\ll
N\log R.$

Adding our estimates for $n\in A$ and $n\in B$ leads to (6). Formulas (3)–(6)
yield

$\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}=2N\log
N+O\left((a_{K+1}^{1/2}+\log R)N\right)=2N\log N+O\left((a_{K+1}^{1/2}+\log
s_{K+1})N\right),$

as claimed.

Suppose now that $4(ca_{K+1})^{1/2}q_{K}\leq N$, and let us prove the lower
bound. We may assume that $a_{K+1}$ is large in terms of $c$. In particular,
$R/N>8c/N$, and formulas (3)–(5) yield

$\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}\geq 2N\log
N+\int_{c/N}^{8c/N}\frac{1}{t^{2}}\left|\left\\{1\leq n\leq
N\,:\,c/N\leq\|n\alpha\|\leq t\right\\}\right|\,\mathrm{d}t-O(N\log R).$

Let $c/N\leq t\leq 8c/N$. Observe that $(q_{K}+q_{K+1})c/N\leq j\leq tq_{K+1}$
implies both $c/N\leq\|jq_{K}\alpha\|\leq t$ and $j\leq N/q_{K}$, the latter
by the assumption $4(ca_{K+1})^{1/2}q_{K}\leq N$. Therefore

$\left|\left\\{1\leq n\leq N\,:\,c/N\leq\|n\alpha\|\leq t\right\\}\right|\geq
tq_{K+1}-(q_{K}+q_{K+1})c/N-1,$

and so

$\int_{c/N}^{8c/N}\frac{1}{t^{2}}\left|\left\\{1\leq n\leq
N\,:\,c/N\leq\|n\alpha\|\leq t\right\\}\right|\,\mathrm{d}t\geq(\log
8)q_{K+1}-\frac{7}{8}(q_{K}+q_{K+1})-\frac{N}{c}\geq q_{K+1}-O(N).$

Hence

$\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}\geq 2N\log N+q_{K+1}-O(N\log
R)=2N\log N+q_{K+1}-O((\log s_{K+1})N),$

as claimed. ∎

######  Theorem 4.

For any $K\geq 0$ and $q_{K}\leq N<q_{K+1}$,

$\sum_{n=1}^{N}\frac{1}{n\|n\alpha\|}=(\log
N)^{2}+\frac{\pi^{2}}{6}s_{K}+a_{K+1}\sum_{1\leq j\leq
N/q_{K}}\frac{1}{j^{2}}+O\left(\sum_{k=1}^{K+1}a_{k}^{1/2}\log a_{k}+(\log
s_{K+1})\log N\right)$

with a universal implied constant.

###### Proof.

We estimate the contribution of the terms with $\|n\alpha\|<1/(2n)$ and
$\|n\alpha\|\geq 1/(2n)$ separately.

Let $0\leq k\leq K$, and consider the integers $q_{k}\leq n<q_{k+1}$. If
$\|n\alpha\|<1/(2n)$, then by Legendre’s theorem [26, p. 30] we have
$q_{k}\mid n$. A given multiple $n=jq_{k}$ satisfies
$\|jq_{k}\alpha\|<1/(2jq_{k})$ if and only if
$j<(2q_{k}\|q_{k}\alpha\|)^{-1/2}$. Therefore

$\sum_{\begin{subarray}{c}q_{k}\leq n<q_{k+1}\\\
\|n\alpha\|<1/(2n)\end{subarray}}\frac{1}{n\|n\alpha\|}=\sum_{1\leq
j<(2q_{k}\|q_{k}\alpha\|)^{-1/2}}\frac{1}{jq_{k}\|jq_{k}\alpha\|},$

consequently

$\begin{split}\sum_{\begin{subarray}{c}n=1\\\
\|n\alpha\|<1/(2n)\end{subarray}}^{N}\frac{1}{n\|n\alpha\|}&=\sum_{k=0}^{K-1}\sum_{1\leq
j<(2q_{k}\|q_{k}\alpha\|)^{-1/2}}\frac{1}{jq_{k}\|jq_{k}\alpha\|}+\sum_{1\leq
j\leq\min\\{(2q_{K}\|q_{K}\alpha\|)^{-1/2},N/q_{K}\\}}\frac{1}{jq_{K}\|jq_{K}\alpha\|}\\\
&=\sum_{k=0}^{K-1}\frac{1}{q_{k}\|q_{k}\alpha\|}\left(\frac{\pi^{2}}{6}+O(a_{k+1}^{-1/2})\right)+\frac{1}{q_{K}\|q_{K}\alpha\|}\left(\sum_{1\leq
j\leq N/q_{K}}\frac{1}{j^{2}}+O(a_{K+1}^{-1/2})\right)\\\
&=\frac{\pi^{2}}{6}\sum_{k=1}^{K}a_{k}+a_{K+1}\sum_{1\leq j\leq
N/q_{K}}\frac{1}{j^{2}}+O\left(\sum_{k=1}^{K+1}a_{k}^{1/2}\right).\end{split}$
(8)

Letting

$T_{N}=\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
1/(2N)\end{subarray}}^{N}\frac{1}{\|n\alpha\|},$

we have

$\begin{split}\frac{1}{n\|n\alpha\|}\mathds{1}_{\\{\|n\alpha\|\geq
1/(2n)\\}}&=\frac{1}{n}\Bigg{(}\sum_{\begin{subarray}{c}\ell=1\\\
\|\ell\alpha\|\geq
1/(2n)\end{subarray}}^{n}\frac{1}{\|\ell\alpha\|}-\sum_{\begin{subarray}{c}\ell=1\\\
\|\ell\alpha\|\geq
1/(2n)\end{subarray}}^{n-1}\frac{1}{\|\ell\alpha\|}\Bigg{)}\\\
&=\frac{1}{n}\Bigg{(}T_{n}-T_{n-1}-\sum_{\begin{subarray}{c}\ell=1\\\
1/(2n)\leq\|\ell\alpha\|<1/(2(n-1))\end{subarray}}^{n-1}\frac{1}{\|\ell\alpha\|}\Bigg{)}\\\
&=\frac{T_{n}-T_{n-1}}{n}+O\left(\sum_{\ell=1}^{n-1}\mathds{1}_{\\{1/(2n)\leq\|\ell\alpha\|<1/(2(n-1))\\}}\right).\end{split}$

Summation by parts resp. switching the order of summation leads to

$\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
1/(2n)\end{subarray}}^{N}\frac{1}{n\|n\alpha\|}=\sum_{n=1}^{N-1}\frac{T_{n}}{n(n+1)}+\frac{T_{N}}{N}+O\left(\sum_{\ell=1}^{N-1}\mathds{1}_{\\{1/(2N)\leq\|\ell\alpha\|<1/(2\ell)\\}}\right).$

As above, Legendre’s theorem shows that
$\sum_{q_{k}\leq\ell<q_{k+1}}\mathds{1}_{\\{\|\ell\alpha\|<1/(2\ell)\\}}\ll
a_{k+1}^{1/2}$, hence the error term in the previous formula is
$\sum_{\ell=1}^{N-1}\mathds{1}_{\\{1/(2N)\leq\|\ell\alpha\|<1/(2\ell)\\}}\ll\sum_{k=1}^{K+1}a_{k}^{1/2}$.
An application of Theorem 3 with $c=1/2$ gives

$\begin{split}\sum_{n=1}^{N-1}\frac{T_{n}}{n(n+1)}&=\sum_{n=1}^{N-1}\frac{2\log
n}{n+1}+O\left(\sum_{k=0}^{K}\sum_{q_{k}\leq
n<q_{k+1}}\frac{a_{k+1}^{1/2}}{n}+\sum_{n=1}^{N-1}\frac{\log
s_{K+1}}{n}\right)\\\ &=(\log N)^{2}+O\left(\sum_{k=1}^{K+1}a_{k}^{1/2}\log
a_{k}+(\log s_{K+1})\log N\right),\end{split}$

and $T_{N}/N\ll\log N+a_{K+1}^{1/2}+\log s_{K+1}$. Therefore

$\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
1/(2n)\end{subarray}}^{N}\frac{1}{n\|n\alpha\|}=(\log
N)^{2}+O\left(\sum_{k=1}^{K+1}a_{k}^{1/2}\log a_{k}+(\log s_{K+1})\log
N\right),$

which together with (8) proves the claim. ∎

###  2.2 Corollaries

Theorems 3 and 4 establish the asymptotics of the sums in (1) for a large
class of irrational $\alpha$. For instance, we immediately obtain

$\begin{split}\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}&\sim 2N\log N\quad\textrm{if
}a_{k}=o(k^{2}),\\\ \sum_{n=1}^{N}\frac{1}{n\|n\alpha\|}&\sim(\log
N)^{2}\quad\textrm{if }s_{k}=o(k^{2}).\end{split}$

As a further example, consider Euler’s number
$e=[2;1,2,1,\ldots,1,2n,1,\ldots]$. The convergent denominators grow at the
rate $\log q_{k}=(k/3)\log k+O(k)$, and $s_{K}=K^{2}/9+O(K)$. Theorems 3 and 4
thus give

$\begin{split}\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|ne\|}&=2N\log N+O\left(\frac{N(\log
N)^{1/2}}{(\log\log N)^{1/2}}\right),\\\
\sum_{n=1}^{N}\frac{1}{n\|ne\|}&=(\log
N)^{2}+\frac{\pi^{2}}{6}\cdot\frac{(\log N)^{2}}{(\log\log
N)^{2}}+O\left(\frac{(\log N)^{3/2}}{(\log\log N)^{1/2}}\right).\end{split}$

We will need certain basic facts from the metric theory of continued fractions
in order to deduce the a.e. asymptotics from Theorems 3 and 4. Khinchin and
Lévy showed that $\log q_{k}\sim\frac{\pi^{2}}{12\log 2}k$ for a.e. $\alpha$,
whereas Borel and Bernstein proved that given a positive function $\varphi$,
for a.e. $\alpha$ we have $a_{k}\geq\varphi(k)$ for infinitely many
$k\in\mathbb{N}$ if and only if $\sum_{k=1}^{\infty}1/\varphi(k)=\infty$. A
theorem of Diamond and Vaaler [14] on trimmed sums of partial quotients states
that

$\lim_{K\to\infty}\frac{\sum_{k=1}^{K}a_{k}-\max_{1\leq k\leq K}a_{k}}{K\log
K}=\frac{1}{\log 2}\qquad\textrm{for a.e. }\alpha.$ (9)

We refer to the monograph [25] for the proof of all these results and for more
context.

###### Proof of Theorem 2.

Assume first that $\sum_{k=1}^{\infty}1/\varphi(k)<\infty$. By the
Khinchin–Lévy and Borel–Bernstein theorems mentioned above, we have
$a_{K+1}\leq\varphi(K/100)\leq\varphi(\log N)$ for all but finitely many $K$,
and also $\log s_{K+1}\ll\log K\ll\log\log N$. Theorem 3 thus yields

$\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}=2N\log N+O(N\varphi(\log
N)^{1/2}+N\log\log N).$

Since $\varphi$ is nondecreasing and $\sum_{k=1}^{\infty}1/\varphi(k)<\infty$,
we have $k/\varphi(k)\to 0$. In particular, the $N\log\log N$ error term in
the previous formula is negligible compared to $N\varphi(\log N)^{1/2}$, as
claimed.

By the Diamond–Vaaler theorem (9),

$s_{K+1}\ll\max_{1\leq k\leq K+1}a_{k}+K\log K\ll\varphi(K/100)+K\log
K\ll\varphi(\log N)+\log N\log\log N.$

Theorem 4 thus yields

$\sum_{n=1}^{N}\frac{1}{n\|n\alpha\|}=(\log N)^{2}+O\left(\varphi(\log N)+\log
N\log\log N\right),$

as claimed.

Assume next that $\sum_{k=1}^{\infty}1/\varphi(k)=\infty$. Fix a small
$\varepsilon>0$. The function
$\varphi^{*}(x)=4\varepsilon^{-2}\varphi(100x)+x$ is also positive and
nondecreasing, and satisfies $\sum_{k=1}^{\infty}1/\varphi^{*}(k)=\infty$. By
the Borel–Bernstein theorem, for a.e. $\alpha$ we have
$a_{K+1}\geq\varphi^{*}(K)\geq K$ for infinitely many $K$. Theorem 3 gives
that for infinitely many $K$ and any $4(ca_{K+1})^{1/2}q_{K}\leq
N\leq\varepsilon^{-1}a_{K+1}^{1/2}q_{K}$,

$\begin{split}\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}&\geq 2N\log N+q_{K+1}-O((\log
s_{K+1})N)\\\ &\geq 2N\log N+\varepsilon a_{K+1}^{1/2}N-O(N\log\log N)\\\
&\geq 2N\log N+\frac{\varepsilon}{2}a_{K+1}^{1/2}N\\\ &\geq 2N\log
N+N\varphi(\log N)^{1/2}.\end{split}$

In particular, the upper asymptotic density of the set

$\Bigg{\\{}N\in\mathbb{N}\,:\,\sum_{\begin{subarray}{c}n=1\\\ \|n\alpha\|\geq
c/N\end{subarray}}^{N}\frac{1}{\|n\alpha\|}\geq 2N\log N+N\varphi(\log
N)^{1/2}\Bigg{\\}}$

is at least $1-4c^{1/2}\varepsilon$. Since $\varepsilon>0$ was arbitrary, the
upper asymptotic density is 1, as claimed.

Similarly, for a.e. $\alpha$ we have $a_{K+1}=\max_{1\leq k\leq K+1}a_{k}\geq
2\varphi(100K)+K\log K\log\log K$ for infinitely many $K$. Theorem 4 gives
that for infinitely many $K$ and any $q_{K}\leq N<q_{K+1}$,

$\begin{split}\sum_{n=1}^{N}\frac{1}{n\left\|n\alpha\right\|}&\geq(\log
N)^{2}+a_{K+1}-O(a_{K+1}^{1/2}\log a_{K+1}+\log N\log\log N)\\\ &\geq(\log
N)^{2}+\frac{1}{2}a_{K+1}\geq(\log N)^{2}+\varphi(\log N).\end{split}$

In particular, the set

$\Bigg{\\{}N\in\mathbb{N}\,:\,\sum_{n=1}^{N}\frac{1}{n\|n\alpha\|}\geq(\log
N)^{2}+\varphi(\log N)\Bigg{\\}}$

has upper asymptotic density 1, as claimed. ∎

###  2.3 Badly approximable irrationals

In the special case of a badly approximable $\alpha$ Theorems 3 and 4 yield
the value of the sums in (1) up to an error $O(N\log\log N)$ resp. $O(\log
N\log\log N)$. We now show how to modify the proof to remove the factor
$\log\log N$ from the error terms.

###### Proof of Theorem 1.

We have $\|n\alpha\|\geq c/N$ for all $1\leq n\leq N$ with a suitably small
constant $0<c<1/2$ depending only on $\alpha$, hence

$\sum_{n=1}^{N}\frac{1}{\|n\alpha\|}=\int_{c/N}^{\infty}\frac{1}{t^{2}}|\\{1\leq
n\leq N\,:\,\|n\alpha\|\leq t\\}|\,\mathrm{d}t.$

The contribution of the integral on $[1/2,\infty)$ is negligible:

$\int_{1/2}^{\infty}\frac{1}{t^{2}}|\\{1\leq n\leq N\,:\,\|n\alpha\|\leq
t\\}|\,\mathrm{d}t\leq\int_{1/2}^{\infty}\frac{N}{t^{2}}\,\mathrm{d}t=2N.$

To estimate the integral on $[c/N,1/2]$, we use the local discrepancy
estimates [30, Theorem 2]

$\begin{split}\max_{1\leq N<q_{K+1}}\left(|\\{1\leq n\leq
N\,:\,\\{n\alpha\\}\leq t\\}|-tN\right)&=\sum_{\begin{subarray}{c}k=1\\\
k\textrm{
even}\end{subarray}}^{K}\\{q_{k}t\\}\left(a_{k+1}(1-\\{q_{k}t\\})+\\{q_{k+1}t\\}-\\{q_{k-1}t\\}\right)+O(1),\\\
\min_{1\leq N<q_{K+1}}\left(|\\{1\leq n\leq N\,:\,\\{n\alpha\\}\leq
t\\}|-tN\right)&=-\sum_{\begin{subarray}{c}k=1\\\ k\textrm{
odd}\end{subarray}}^{K}\\{q_{k}t\\}\left(a_{k+1}(1-\\{q_{k}t\\})+\\{q_{k+1}t\\}-\\{q_{k-1}t\\}\right)+O(1)\end{split}$

with universal implied constants, where $\\{\cdot\\}$ is the fractional part
function. In the special case of a badly approximable $\alpha$ these estimates
immediately show that for all $1\leq N<q_{K+1}$,

$\left||\\{1\leq n\leq N\,:\,\\{n\alpha\\}\leq
t\\}|-tN\right|\ll\sum_{k=1}^{K}\\{q_{k}t\\}+1\ll\sum_{\begin{subarray}{c}k=1\\\
q_{k}\leq 1/t\end{subarray}}^{K}q_{k}t+\sum_{\begin{subarray}{c}k=1\\\
q_{k}>1/t\end{subarray}}^{K}1+1\ll\log(tN)+1.$

In the last step we used the fact that $\sum_{j=1}^{k}q_{j}\leq 3q_{k}$, and
that there are $\ll\log(B/A)+1$ convergent denominators $q_{k}$ that fall in
any given interval $[A,B]$. Since $-\alpha$ is also badly approximable, we can
similarly estimate the number of $1\leq n\leq N$ such that $\\{-n\alpha\\}\leq
t$, that is, $\\{n\alpha\\}\geq 1-t$. In particular,

$|\\{1\leq n\leq N\,:\,\|n\alpha\|\leq
t\\}|=2tN+O\left(\log(tN)+1\right)\quad\textrm{uniformly in }0<t\leq 1/2.$
(10)

We mention that (10) can also be easily deduced from an explicit formula for
the local discrepancy due to T. Sós [32]. Hence

$\int_{c/N}^{1/2}\frac{1}{t^{2}}|\\{1\leq n\leq N\,:\,\|n\alpha\|\leq
t\\}|\,\mathrm{d}t=2N\log N+O(N),$

and the claim $\sum_{n=1}^{N}1/\|n\alpha\|=2N\log N+O(N)$ follows. Summation
by parts then yields $\sum_{n=1}^{N}1/(n\|n\alpha\|)=(\log N)^{2}+O(\log N)$,
as claimed. ∎

##  3 Related Diophantine sums

###  3.1 Sums with fractional parts

Theorems 1, 2, 3, 4 have perfect analogues with the distance from the nearest
integer function $\|\cdot\|$ replaced by the fractional part function
$\\{\cdot\\}$.

######  Theorem 5.

Let $c>0$ be an arbitrary constant. For any $K\geq 0$ and $q_{K}\leq
N<q_{K+1}$,

$\sum_{\begin{subarray}{c}n=1\\\ \\{n\alpha\\}\geq
c/N\end{subarray}}^{N}\frac{1}{\\{n\alpha\\}}=N\log
N+O\left((\mathds{1}_{\\{K+1\textrm{ odd}\\}}a_{K+1}^{1/2}+\log
s_{K+1})N\right).$

If in addition $4(ca_{K+1})^{1/2}q_{K}\leq N$ and $K+1$ is odd, then

$\sum_{\begin{subarray}{c}n=1\\\ \\{n\alpha\\}\geq
c/N\end{subarray}}^{N}\frac{1}{\\{n\alpha\\}}\geq N\log N+q_{K+1}-O((\log
s_{K+1})N).$

The implied constants depend only on $c$. Further, for any $K\geq 0$ and
$q_{K}\leq N<q_{K+1}$,

$\begin{split}\sum_{n=1}^{N}\frac{1}{n\\{n\alpha\\}}=&\frac{1}{2}(\log
N)^{2}+\frac{\pi^{2}}{6}\sum_{\begin{subarray}{c}k=1\\\ k\textrm{
odd}\end{subarray}}^{K}a_{k}+\mathds{1}_{\\{K+1\textrm{
odd}\\}}a_{K+1}\sum_{1\leq j\leq N/q_{K}}\frac{1}{j^{2}}\\\
&+O\Bigg{(}\sum_{\begin{subarray}{c}k=1\\\ k\textrm{
odd}\end{subarray}}^{K+1}a_{k}^{1/2}\log a_{k}+(\log s_{K+1})\log
N\Bigg{)}\end{split}$

with a universal implied constant.

###### Proof.

This is a straightforward modification of the proof of Theorems 3 and 4. The
parity conditions follow from the fact that
$p_{2k}/q_{2k}<\alpha<p_{2k+1}/q_{2k+1}$ for all $k\geq 0$. ∎

The same holds for the sums

$\sum_{\begin{subarray}{c}n=1\\\ 1-\\{n\alpha\\}\geq
c/N\end{subarray}}^{N}\frac{1}{1-\\{n\alpha\\}}\qquad\textrm{and}\qquad\sum_{n=1}^{N}\frac{1}{n(1-\\{n\alpha\\})}$

with “odd” replaced by “even”. Theorem 2 also remains true with $\|n\alpha\|$
replaced either by $\\{n\alpha\\}/2$ or by $(1-\\{n\alpha\\})/2$. The proof is
identical to that of Theorem 2; note that the Borel–Bernstein theorem holds
without any monotonicity assumption on $\varphi$, so the parity conditions do
not cause any difficulty.

A straightforward modification of the proof of Theorem 1 similarly shows that
for any badly approximable $\alpha$,

$\sum_{n=1}^{N}\frac{1}{\\{n\alpha\\}}=N\log
N+O(N)\qquad\textrm{and}\qquad\sum_{n=1}^{N}\frac{1}{n\\{n\alpha\\}}=\frac{1}{2}(\log
N)^{2}+O(\log N),$

and the same holds with $\\{n\alpha\\}$ replaced by $1-\\{n\alpha\\}$.

###  3.2 Shifted sums

Some of our methods apply to shifted Diophantine sums as well. Here we only
focus on the case of a badly approximable $\alpha$, for which we find the
precise asymptotics. All previously known results on shifted Diophantine sums
have a constant factor gap between the upper and the lower estimates [4, 5].

######  Theorem 6.

Let $\alpha$ be a badly approximable irrational, and let $\beta\in\mathbb{R}$.
For any $N\geq 1$,

$\sum_{\begin{subarray}{c}n=1\\\ n\neq
n_{0}\end{subarray}}^{N}\frac{1}{\|n\alpha+\beta\|}=2N\log N+O(N\log\log N)$

with an implied constant depending only on $\alpha$, where
$n_{0}=n_{0}(\alpha,\beta,N)\in[1,N]$ is an integer for which $\min_{1\leq
n\leq N}\|n\alpha+\beta\|=\|n_{0}\alpha+\beta\|$. If in addition
$\inf_{n\in\mathbb{N}}(n\log\log n)\|n\alpha+\beta\|>0$, then

$\sum_{n=1}^{N}\frac{1}{\|n\alpha+\beta\|}=2N\log N+O(N\log\log
N),\qquad\sum_{n=1}^{N}\frac{1}{n\|n\alpha+\beta\|}=(\log N)^{2}+O(\log
N\log\log N)$

with implied constants depending only on $\alpha$ and $\beta$.

###### Proof.

For any integers $1\leq n<n^{\prime}\leq N$, we have
$\|(n\alpha+\beta)-(n^{\prime}\alpha+\beta)\|\geq 2c/N$ with a suitably small
constant $c>0$ depending only on $\alpha$. In particular, for all $1\leq n\leq
N$, $n\neq n_{0}$ we have $\|n\alpha+\beta\|\geq c/N$. One readily checks that
formulas (3), (4), (5) and (7) in the proof of Theorem 3 remain true with an
arbitrary shift $\beta$, hence

$\sum_{\begin{subarray}{c}n=1\\\ n\neq
n_{0}\end{subarray}}^{N}\frac{1}{\|n\alpha+\beta\|}=2N\log N+O(N\log\log N),$

as claimed.

Under the additional assumption $\inf_{n\in\mathbb{N}}(n\log\log
n)\|n\alpha+\beta\|>0$ the contribution of the $n=n_{0}$ term is $O(N\log\log
N)$, thus $\sum_{n=1}^{N}1/\|n\alpha+\beta\|=2N\log N+O(N\log\log N)$, as
claimed. Summation by parts then yields
$\sum_{n=1}^{N}1/(n\|n\alpha+\beta\|)=(\log N)^{2}+O(\log N\log\log N)$, as
claimed. ∎

Similarly, for any badly approximable $\alpha$ and any $\beta\in\mathbb{R}$,

$\sum_{\begin{subarray}{c}n=1\\\ n\neq
n_{0}^{\prime}\end{subarray}}^{N}\frac{1}{\\{n\alpha+\beta\\}}=N\log
N+O(N\log\log N)$

with an implied constant depending only on $\alpha$, where
$n_{0}^{\prime}=n_{0}^{\prime}(\alpha,\beta,N)\in[1,N]$ is an integer for
which $\min_{1\leq n\leq
N}\\{n\alpha+\beta\\}=\\{n_{0}^{\prime}\alpha+\beta\\}$. If in addition
$\inf_{n\in\mathbb{N}}(n\log\log n)\\{n\alpha+\beta\\}>0$, then

$\sum_{n=1}^{N}\frac{1}{\\{n\alpha+\beta\\}}=N\log N+O(N\log\log
N),\qquad\sum_{n=1}^{N}\frac{1}{n\\{n\alpha+\beta\\}}=\frac{1}{2}(\log
N)^{2}+O(\log N\log\log N)$

with implied constants depending only on $\alpha$ and $\beta$. The same holds
with $\\{n\alpha+\beta\\}$ replaced by $1-\\{n\alpha+\beta\\}$.

###  3.3 A higher dimensional generalization

There are several natural higher dimensional generalizations of the sums in
(1), which have been studied using Fourier analysis [1, 29], geometry of
numbers [4] and lattices [19, 20]. In this setting, a vector
$\alpha\in\mathbb{R}^{d}$ is called badly approximable if
$\inf_{n\in\mathbb{Z}^{d}\backslash\\{0\\}}\|n\|_{\infty}^{d}\cdot\|n_{1}\alpha_{1}+\cdots+n_{d}\alpha_{d}\|>0$,
where $\|n\|_{\infty}=\max_{1\leq k\leq d}|n_{k}|$. Here we only comment on a
result of Fregoli [19], who showed that for any badly approximable vector
$\alpha\in\mathbb{R}^{d}$,

$N^{d}\log
N\ll\sum_{n\in[-N,N]^{d}\backslash\\{0\\}}\frac{1}{\|n_{1}\alpha_{1}+\cdots+n_{d}\alpha_{d}\|}\ll
N^{d}\log N.$

His main result in fact yields the precise asymptotics of the previous sum, in
particular giving an alternative proof of Theorem 1 based on lattices.

######  Theorem 7.

Let $\alpha\in\mathbb{R}^{d}$ be a badly approximable vector. For any $N\geq
1$,

$\begin{split}\sum_{n\in[-N,N]^{d}\backslash\\{0\\}}\frac{1}{\|n_{1}\alpha_{1}+\cdots+n_{d}\alpha_{d}\|}&=d2^{d+1}N^{d}\log
N+O(N^{d}),\\\
\sum_{n\in[-N,N]^{d}\backslash\\{0\\}}\frac{1}{\|n\|_{\infty}^{d}\cdot\|n_{1}\alpha_{1}+\cdots+n_{d}\alpha_{d}\|}&=d^{2}2^{d}(\log
N)^{2}+O(\log N)\end{split}$

with implied constants depending only on $\alpha$.

###### Proof.

The main result [19, Proposition 1.3] states that

$|\\{n\in[-N,N]^{d}\backslash\\{0\\}\,:\,\|n_{1}\alpha_{1}+\cdots+n_{d}\alpha_{d}\|\leq
t\\}|=2^{d+1}tN^{d}+O\left(t^{d/(d+1)}N^{d^{2}/(d+1)}\right)$

uniformly in $0<t\leq 1/2$. Using this instead of formula (10), the rest of
the proof is identical to that of Theorem 1. Writing

$\sum_{n\in[-N,N]^{d}\backslash\\{0\\}}\frac{1}{\|n\|_{\infty}^{d}\cdot\|n_{1}\alpha_{1}+\cdots+n_{d}\alpha_{d}\|}=\sum_{\ell=1}^{N}\frac{1}{\ell^{d}}\sum_{\|n\|_{\infty}=\ell}\frac{1}{\|n_{1}\alpha_{1}+\cdots+n_{d}\alpha_{d}\|},$

the second claim follows from summation by parts. ∎

## Acknowledgments

The author is supported by the Austrian Science Fund (FWF) project M 3260-N.

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