Appendix A
We give a proof of Lemma 4 using Cesàro means. Our proof provides a slightly better upper-bound than the one given in [2]. We begin with the following lemma:
Lemma 10
For all \(1\le j\le n\),
$$\begin{aligned} \sum _{p=j}^{n-1} \frac{A_{p-j}^{\theta -1}}{p A_p^\theta } = \varPsi _n (j) \left( \frac{1}{j} - \frac{1}{n} \right) . \end{aligned}$$
(34)
Proof
With the same notation as in the proof of Proposition 2, we observe that
$$\begin{aligned} J_n(Ms)_n&= \sum _{p=1}^{n-1} J_n M_{n,p} s_p + s_n = \sum _{p=1}^{n-1} \frac{\theta }{\theta +p} \frac{1}{p} \sum _{j=1}^{p} \varPsi _p (j) w_j + \frac{1}{n} \sum _{j=1}^n \varPsi _n (j) w_j \\&= \sum _{j=1}^{n-1} w_j \left[ \sum _{p=j}^{n-1} \frac{\theta }{\theta +p} \frac{\varPsi _p (j)}{p} + \frac{\varPsi _n (j)}{n} \right] \\&\quad +\,\frac{\varPsi _n (n)}{n} w_n \end{aligned}$$
and
$$\begin{aligned} J_n t_n = \sum _{j=1}^{n-1} w_j \frac{\varPsi _n (j)}{j} + \frac{\varPsi _n (n)}{n} w_n. \end{aligned}$$
Therefore, by identification, for all \(j\in [\![ 1 , n-1 ]\!]\),
$$\begin{aligned} \frac{\varPsi _n (j)}{j} = \sum _{p=j}^{n-1} \frac{\theta }{\theta +p} \frac{\varPsi _p (j)}{p} + \frac{\varPsi _n (j)}{n}. \end{aligned}$$
Finally, using the definition of Cesàro numbers it is clear that \(\frac{\theta }{\theta +p} \frac{\varPsi _p (j)}{p} = \frac{A_{p-j}^{\theta -1}}{p A_p^\theta }\) for all \(1\le j \le p\). \(\square \)
We are ready to prove Lemma 4.
We introduce independent Bernoulli variables \(\xi _r\), \(r\ge 1\), defined as
$$\begin{aligned} \mathbb {P} (\xi _r = 1)= \frac{\theta }{\theta + r -1} \quad , \quad \mathbb {P} (\xi _r =0) = \frac{r-1}{\theta +r-1}. \end{aligned}$$
The Feller Coupling characterizes the variables \(a_{n,j}\) and \(W_j\) on the same probability space in function of the \(\xi _r\), by the following equalities: For all \(1 \le j \le n\),
$$\begin{aligned} a_{n,j}&= \# \{j-\text {spacings between two consecutive 1 in the word } (1 \ \xi _2 \ \cdots \ \xi _n \ 1) \} \\&= \sum _{k=1}^{n-j} \xi _k (1-\xi _{k+1}) \cdots (1-\xi _{k+j-1})\xi _{k+j} + \xi _{n-j+1} (1- \xi _{n-j+1} ) \cdots (1- \xi _n ), \end{aligned}$$
and for all \(j \in \mathbb {N}^*\),
$$\begin{aligned} W_j = \sum _{k=1}^{+\infty } \xi _k (1-\xi _{k+1}) \cdots (1-\xi _{k+j-1})\xi _{k+j}. \end{aligned}$$
To begin with, it is easy to notice that
$$\begin{aligned} \left| a_{n,j} - W_j \right| \le \mathbb {1}_{J_n=j} + W_{j,n} + \mathbb {1}_{J_n+K_n = j+1} \end{aligned}$$
where
$$\begin{aligned} \left\{ \begin{array}{l} J_n := \min \{ k\ge 1 : \ \xi _{n-k+1} =1 \} \\ K_n:= \min \{ k\ge 1 : \ \xi _{n+k} =1\} \\ W_{j,n} := \sum _{k=n+1}^{+\infty } \xi _k (1-\xi _{k+1}) \cdots (1-\xi _{k+j-1})\xi _{k+j} \end{array} \right. \end{aligned}$$
therefore
$$\begin{aligned} \mathbb {E} \left| a_{n,j} - W_j \right| \le \mathbb {P} (J_n=j ) + \mathbb {E} (W_{j,n}) + \mathbb {P} (J_n+K_n = j+1 ). \end{aligned}$$
We look separately at the three right-hand side terms in this inequality.
$$\begin{aligned} \mathbb {P} (J_n = j) = \frac{\theta }{\theta + n - j} \frac{n-j+1}{\theta + n -j +1} \cdots \frac{n-1}{\theta + n-1} = \frac{\theta }{n} \varPsi _n (j). \end{aligned}$$
Hence, \(\sum \limits _{j=1}^n \mathbb {P} (J_n = j) = \frac{\theta }{n} \sum \limits _{j=1}^n \varPsi _n (j) = 1\).
$$\begin{aligned} \mathbb {E} (W_{j,n})&= \sum \limits _{k=n+1}^{+\infty } \frac{\theta }{\theta + k -1} \frac{k}{\theta + k} \cdots \frac{k+j-2}{\theta + k+j-2} \frac{\theta }{\theta + k +j -1} \\&= \sum \limits _{p=n+j+1}^{+\infty } \frac{\theta }{\theta + p-j -1} \frac{p-j}{\theta + p-j} \cdots \frac{p-2}{\theta + p-2} \frac{\theta }{\theta + p -1}. \end{aligned}$$
$$\begin{aligned}&\mathbb {P} (J_n+K_n = j+1 ) \\&\qquad = \sum \limits _{k=1}^j \frac{\theta }{\theta + n +k-j-1} \frac{n+k-j}{\theta + n+k-j} \cdots \frac{n+k-2}{\theta + n+k-2} \frac{\theta }{\theta + n + k-1} \\&\qquad = \sum \limits _{p=n+1}^{n+j} \frac{\theta }{\theta + p-j -1} \frac{p-j}{\theta + p-j} \cdots \frac{p-2}{\theta + p-2} \frac{\theta }{\theta + p -1}. \end{aligned}$$
In particular,
$$\begin{aligned}&\mathbb {E} (W_{j,n}) + \mathbb {P} (J_n+K_n = j+1 ) \\&\qquad = \sum \limits _{p=n+1}^{+\infty } \frac{\theta }{\theta + p-j -1} \frac{p-j}{\theta + p-j} \cdots \frac{p-2}{\theta + p-2} \frac{\theta }{\theta + p -1} \\&\qquad = \theta \sum _{p=n}^{+\infty } \frac{1}{p} \frac{A_{p-j}^{\theta -1}}{A_p^\theta } = \theta \left[ \sum _{p=j}^{+\infty } \frac{1}{p} \frac{A_{p-j}^{\theta -1}}{A_p^\theta } - \sum _{p=j}^{n-1} \frac{1}{p} \frac{A_{p-j}^{\theta -1}}{A_p^\theta } \right] . \end{aligned}$$
Since, by Lemma 10,
$$\begin{aligned} \sum _{p=j}^{n-1} \frac{1}{p} \frac{A_{p-j}^{\theta -1}}{A_p^\theta } = \varPsi _n (j) \left( \frac{1}{j} - \frac{1}{n} \right) \underset{n\rightarrow \infty }{\longrightarrow } \frac{1}{j} \end{aligned}$$
it follows
$$\begin{aligned} \mathbb {E} (W_{j,n}) + \mathbb {P} (J_n+K_n = j+1 ) = \theta \left[ \frac{1}{j} - \varPsi _n (j) \left( \frac{1}{j} - \frac{1}{n} \right) \right] . \end{aligned}$$
Then,
$$\begin{aligned} \sum _{j=1}^n \mathbb {E} (W_{j,n}) + \mathbb {P} (J_n+K_n = j+1 )&= \theta \sum _{j=1}^n \frac{1}{j} - \theta \sum _{j=1}^n \frac{\varPsi _n (j)}{j} + \frac{\theta }{n} \sum _{j=1}^n \varPsi _n (j) \\&= \theta \sum _{j=1}^n \frac{1}{j} - \theta \sum _{j=1}^n \frac{1}{\theta + j -1} \ + \ 1 \\&= \theta \sum _{j=1}^n \frac{\theta -1}{j(\theta + j -1)} \ + \ 1 \end{aligned}$$
using Lemma 3 for the second equality. We deduce
$$\begin{aligned} \begin{aligned} \mathbb {E} \left( \sum _{j=1}^n \left| a_{n,j} - W_j \right| \right)&\le \,2 + \theta (\theta -1) \sum _{j=1}^n \frac{1}{j(\theta + j-1)} \\&\le \,1 + \theta \mathbb {1}_{0<\theta <1} + (1 + \theta (\gamma + \psi (\theta )))\mathbb {1}_{\theta \ge 1} \end{aligned} \end{aligned}$$
(35)
where \(\gamma \) is the Euler–Mascheroni constant, and \(\psi \) is the digamma function. As a particular consequence,
$$\begin{aligned} \limsup \limits _{\theta \rightarrow 0^+} \ \sup \limits _{n \ge 1} \mathbb {E} \left( \sum _{j=1}^n \left| a_{n,j} - W_j \right| \right) \le 1. \end{aligned}$$
(36)
Appendix B
The following proposition gives the whole possible values for the constant \(c_2\) appearing in Proposition 4, in function of \(\alpha \) and \(\beta \).
Proposition 11
Let p, r be integers and q, s positive integers with p and q relatively prime, r and s relatively prime, \(r\ne 0\). Then
$$\begin{aligned}&c_2 \\&\quad = \left\{ \begin{array}{l@{\quad }l} \frac{1}{6} &{} \mathrm{{if}} \,\,\alpha \,\, \mathrm{{and}} \,\, \beta \,\, \mathrm{{are}}\, \mathrm{{irrational}}\, \mathrm{{and}} \\ &{} \mathrm{{linearly}} \,\, \mathrm{{independent}}\, \mathrm{{over}}\, \mathbb {Q} \\ \frac{1}{6} + \frac{1}{6q^2} &{} \mathrm{{if}} \,\, \alpha = \frac{p}{q}\,\, \mathrm{and} \,\,\beta \,\, \mathrm{{is}}\,\, \mathrm{{irrational }} \\ \frac{1}{6} + \frac{1}{6s^2} &{} \mathrm{{if}} \,\, \alpha \,\, \mathrm{{is}}\,\, \mathrm{{irrational}} \,\, \mathrm{{and }} \,\,\beta = \frac{r}{s} \\ \frac{(2q-1)(q-1)}{6q^2} + \frac{(2s-1)(s-1)}{6s^2} - \frac{2}{qs} \sum _{j=1}^{qs} \lbrace \frac{jp}{q} \rbrace \lbrace \frac{jr}{s} \rbrace &{} \mathrm{{if}}\,\, \alpha = \frac{p}{q} \,\, \mathrm{{and}} \,\,\beta = \frac{r}{s} \\ \frac{1}{6} - \frac{\gcd (s,q)^2 }{6srq^2} &{} \mathrm{{if}} \,\, \alpha \,\, \mathrm{{is}} \,\, \mathrm{{irrational}}~\mathrm{{and}} \\ &{} \beta = \frac{p}{q}+\frac{r}{s}\alpha . \end{array}\right. \end{aligned}$$
The four first cases have been shown by Wieand, as well as the last case for \(s=1\) by the same author in [13]. We complete her work in this appendix, treating the case s arbitrary.
Therefore, suppose \(\alpha \) and \(\beta \) irrational numbers which are linearly dependent over \(\mathbb {Q}\), say \(\beta = \frac{p}{q} + \frac{r}{s} \alpha \), with \(s\ge 2\). Let us recall that \(c_2 := \lim _{n \rightarrow \infty } \frac{1}{n} \sum _{j=1}^n (\{j\beta \} - \{j\alpha \})^2\). Since \(\lim _{n \rightarrow \infty } \frac{1}{n} \sum _{j=1}^n \{jx \}^2\) is easy to compute for all real numbers x, it remains to study what was denoted by \(s_3\) in [13], defined as
$$\begin{aligned} s_3 (\alpha , \beta ) := \lim _{n \rightarrow \infty } \frac{1}{n} \sum _{j=1}^n \{j\alpha \} \{j \beta \}. \end{aligned}$$
(37)
We first assume that we have \(r>0\). We are going to proceed in two steps.
\(\bullet \,\, {\hbox {Computation of} \,s_{3} (\alpha , \frac{r}{s}\alpha )}\) : Before starting the calculation, it is good to notice that if \(\psi := \frac{\alpha }{s}\) then \(\psi \) is still irrational and consequently the sequence \((n\psi )_{n\in \mathbb {N}^*}\) is still equidistributed. Thus \(s_3 (s\psi , r\psi ) = s_3 (s\alpha , r\alpha )\), i.e., \(s_3 (\alpha , \frac{r}{s}\alpha ) = s_3 (s\alpha , r\alpha ) \). Indeed, it is a direct consequence of Theorem 9 given in [13] that we recall here:
Theorem 7
Let f be a Riemann integrable function on [0, 1], and let \(t\in \mathbb {R}\setminus \mathbb {Q}\). Then for all real numbers b,
$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \sum _{j=1}^n f(\{jt + b \} ) = \int _0^1 f(x) \mathrm {d}x. \end{aligned}$$
(We take \(b=0\) and \(f(x)=\{sx\} \{rx\}\).) The expression of \(s_3 (s\alpha , r\alpha )\) is more convenient to handle in practice. We decompose for \(j\ge 1\),
$$\begin{aligned} \lbrace j s\alpha \rbrace \lbrace j r\alpha \rbrace&= \left( s \lbrace j \alpha \rbrace - \lfloor s \lbrace j \alpha \rbrace \rfloor \right) \left( r \lbrace j \alpha \rbrace - \lfloor r \lbrace j \alpha \rbrace \rfloor \right) \\&=\,rs \lbrace j \alpha \rbrace ^2 - r \sum _{l =1}^{s-1} \lbrace j \alpha \rbrace \mathbb {1}_{ \lbrace j \alpha \rbrace \ge \frac{l}{s}} - s \sum _{k=1}^{r-1} \lbrace j \alpha \rbrace \mathbb {1}_{ \lbrace j \alpha \rbrace \ge \frac{k}{r}} \\&\qquad + \,\sum \limits _{\begin{array}{c} 1\le k \le r-1 \\ 1\le l \le s-1 \end{array}} \mathbb {1}_{ \lbrace j \alpha \rbrace \ge \max \left( \frac{l}{s} , \frac{k}{r} \right) } \end{aligned}$$
The three first terms will not pose any difficulty since the limits of their respective means have already been evaluated in [13]. The novelty holds in the fourth term. For \(k \in [\![ 1 , r-1 ]\!]\) and \(l \in [\![ 1 , s-1 ]\!]\), let \(f_{k,l} (x) :=\mathbb {1}_{ x \ge \max \left( \frac{l}{s} , \frac{k}{r} \right) } \). These functions are clearly Riemann integrable on [0, 1], thus
$$\begin{aligned} \frac{1}{n} \sum _{j=1}^n \sum \limits _{\begin{array}{c} 1\le k \le r-1 \\ 1\le l \le s-1 \end{array}} \mathbb {1}_{ \lbrace j \alpha \rbrace \ge \max \left( \frac{l}{s} , \frac{k}{r} \right) }&\underset{n\rightarrow \infty }{\longrightarrow } \sum \limits _{\begin{array}{c} 1\le k \le r-1 \\ 1\le l \le s-1 \end{array}} \int _0^1 f_{k,l} (x) \mathrm {d}x \\&=\, (r-1)(s-1) - \sum \limits _{\begin{array}{c} 1\le k \le r-1 \\ 1\le l \le s-1 \end{array}} \max \left( \frac{l}{s} , \frac{k}{r} \right) \end{aligned}$$
with
$$\begin{aligned}&\sum \limits _{\begin{array}{c} 1\le k \le r-1 \\ 1\le l \le s-1 \end{array}} \max \left( \frac{l}{s} , \frac{k}{r} \right) \\&\qquad = \sum _{k=1}^{r-1} \left( \left( \sum _{l =1}^{\left\lfloor \frac{sk}{r}\right\rfloor } \frac{k}{r} \right) + \left( \sum _{l = \left\lfloor \frac{sk}{r}\right\rfloor + 1 }^{s-1} \frac{l}{s} \right) \right) \\&\qquad = \sum _{k=1}^{r-1} \left( \left( \frac{k}{r}\left\lfloor \frac{sk}{r}\right\rfloor \right) + \left( \frac{s-1}{2} -\frac{\left\lfloor \frac{sk}{r}\right\rfloor \left( \left\lfloor \frac{sk}{r}\right\rfloor +1\right) }{2s} \right) \right) \\&\qquad = \frac{(r-1)(s-1)}{2} + \frac{1}{2s} \sum _{k=1}^{r-1} \left( \frac{sk}{r} \right) ^2 - \frac{sk}{r} - \left\{ \frac{sk}{r} \right\} ^2 + \left\{ \frac{sk}{r} \right\} \\&\qquad = \frac{(r-1)(s-1)}{2} + \frac{1}{2s} \left( \frac{s^2-1}{r^2} \frac{r(r-1)(2r-1)}{6} - \frac{s-1}{r} \frac{r(r-1)}{2} \right) . \end{aligned}$$
Once expanded, we get
$$\begin{aligned} (r-1)(s-1) - \sum \limits _{\begin{array}{c} 1\le k \le r-1 \\ 1\le l \le s-1 \end{array}} \max \left( \frac{l}{s} , \frac{k}{r} \right) = \frac{1}{4} + \frac{rs}{3} - \frac{r}{4} - \frac{s}{4} - \frac{s}{12r} - \frac{r}{12s} + \frac{1}{12sr}. \end{aligned}$$
We deduce
$$\begin{aligned} s_3 (s\alpha , r\alpha )&= \frac{rs}{3} - r \left( \frac{s}{3} - \frac{1}{4} - \frac{1}{12s} \right) - s \left( \frac{r}{3} - \frac{1}{4} - \frac{1}{12r} \right) \\&\qquad + \,\left( \frac{1}{4} + \frac{rs}{3} - \frac{r}{4} - \frac{s}{4} - \frac{s}{12r} - \frac{r}{12s} + \frac{1}{12sr} \right) \\&=\, \frac{1}{4} + \frac{1}{12sr} \end{aligned}$$
\(\bullet \) Computation of \({s_3 (\alpha , \frac{p}{q} + \frac{r}{s}\alpha )}\) : First of all, the particular case \(q=1\) is trivial since \(s_3 (\alpha , p + \frac{r}{s}\alpha ) = s_3 (\alpha ,\frac{r}{s}\alpha )\) and then \(c_2= \frac{2}{3} - 2 \left( \frac{1}{4} + \frac{1}{12sr} \right) = \frac{1}{6} - \frac{\gcd (s,1)^2}{6sr\times 1^2}\). Thus, we assume in all the following that \(q\ge 2\). In the same manner as at the beginning of the first step, we notice that \(s_3 (\alpha , \frac{p}{q} + \frac{r}{s}\alpha ) = s_3 (s\alpha , \frac{p}{q} + r\alpha )\) (using Theorem 7 with \(b=\frac{p}{q}\) and for f a complicated expression that we do not precise here, which is bounded and piecewise continuous so Riemann integrable).
For all \(j \ge 1\),
$$\begin{aligned} \left\{ j s \alpha \right\} \left\{ j \frac{p}{q} + j r \alpha \right\} = \left\{ j s \alpha \right\} \left\{ j \frac{p}{q} \right\} + \lbrace j s \alpha \rbrace \lbrace j r \alpha \rbrace - \lbrace j s \alpha \rbrace \mathbb {1}_{ \lbrace j \frac{p}{q} \rbrace + \lbrace j r \alpha \rbrace \ge 1 }. \end{aligned}$$
The mean of the first term tends to \(s_3 (s \alpha , \frac{p}{q}) = \frac{1}{4} - \frac{1}{4q}\). The mean of the second term tends to \(s_3 (s\alpha , r\alpha ) = \frac{1}{4} + \frac{1}{12sr}\) by the first step. It remains to study the third term. In a similar way to what is done in [13], it is easy to check that
$$\begin{aligned} \frac{1}{q \left\lfloor \frac{n}{q} \right\rfloor } \sum _{j=1}^{q \left\lfloor \frac{n}{q} \right\rfloor } \lbrace j s \alpha \rbrace \mathbb {1}_{ \lbrace j \frac{p}{q} \rbrace + \lbrace j r \alpha \rbrace \ge 1 } \underset{n\rightarrow \infty }{=} \frac{1}{n} \sum _{j=1}^n \lbrace j s \alpha \rbrace \mathbb {1}_{ \lbrace j \frac{p}{q} \rbrace + \lbrace j r \alpha \rbrace \ge 1 } \ + \ o(1). \end{aligned}$$
Let \(n\in \mathbb {N}^*\). We use the periodicity of the sequence \((\lbrace \frac{kp}{q} \rbrace )\) to write that the left-hand side expression is equal to
$$\begin{aligned}&\frac{1}{q \left\lfloor \frac{n}{q} \right\rfloor } \sum _{j=0}^{\left\lfloor \frac{n}{q} \right\rfloor - 1} \sum _{k=1}^q \lbrace (k+jq)s\alpha \rbrace \mathbb {1}_{ \lbrace (k+jq)r\alpha \rbrace \ge 1 - \left\{ \frac{kp}{q} \right\} } \\&\qquad = \,s \cdot \frac{1}{q} \sum _{k=1}^q \frac{1}{\left\lfloor \frac{n}{q} \right\rfloor } \sum _{j=0}^{\left\lfloor \frac{n}{q} \right\rfloor - 1} \lbrace (k+jq)\alpha \rbrace \mathbb {1}_{ \lbrace (k+jq)r\alpha \rbrace \ge 1 - \left\{ \frac{kp}{q} \right\} } \\&\qquad \quad - \,\frac{1}{q} \sum _{k=1}^q \frac{1}{\left\lfloor \frac{n}{q} \right\rfloor } \sum _{j=0}^{\left\lfloor \frac{n}{q} \right\rfloor - 1} \sum _{m=1}^{s-1} \mathbb {1}_{ \left( \lbrace (k+jq) \alpha \rbrace \ge \frac{m}{s} \right) \cap \left( \lbrace (k+jq) r \alpha \rbrace \ge 1 - \left\{ \frac{kp}{q} \right\} \right) } \end{aligned}$$
with, (see [13]),
$$\begin{aligned} \frac{1}{q} \sum _{k=1}^q \frac{1}{\left\lfloor \frac{n}{q} \right\rfloor } \sum _{j=0}^{\left\lfloor \frac{n}{q} \right\rfloor - 1} \lbrace (k+jq)\alpha \rbrace \mathbb {1}_{ \lbrace (k+jq)r\alpha \rbrace \ge 1 - \left\{ \frac{kp}{q} \right\} } \underset{n\rightarrow \infty }{\longrightarrow } \frac{1}{4} + \frac{1}{12r} - \frac{1}{4q} - \frac{1}{12rq^2}. \end{aligned}$$
For \(k \in [\![ 1 , q]\!]\) and \(m \in [\![ 1 , s-1 ]\!]\), let
$$\begin{aligned} f_{k,m} (x) = \mathbb {1}_{ \left( x \ge \frac{m}{s} \right) \cap \left( rx - \sum _{l =1}^{r-1} \mathbb {1}_{x \ge \frac{l}{r} } \ge 1 - \left\{ \frac{kp}{q} \right\} \right) }. \end{aligned}$$
We compute its integral between 0 and 1:
$$\begin{aligned} \int _0^1 f_{k,m} (x) \mathrm {d}x&= \sum _{l = 0}^{r-1} \int _{\frac{1 - \left\{ \frac{kp}{q} \right\} + l}{r}}^{\frac{l+1}{r}} \mathbb {1}_{ \left( x \ge \frac{m}{s} \right) } \mathrm {d}x \\&= \,0 + \left( \frac{\left\lfloor \frac{rm}{s} \right\rfloor +1}{r} - \max \left( \frac{m}{s} , \frac{1 - \left\{ \frac{kp}{q} \right\} + \left\lfloor \frac{rm}{s} \right\rfloor }{r} \right) \right) \\&\quad + \,\sum _{l = \left\lfloor \frac{rm}{s} \right\rfloor +1 }^{r-1} \frac{l +1}{r} - \frac{1 - \left\{ \frac{kp}{q} \right\} + l}{r} \\&= \frac{1- \left\{ \frac{rm}{s} \right\} }{r} - \frac{1- \left\{ \frac{kp}{q} \right\} - \left\{ \frac{rm}{s} \right\} }{r} \mathbb {1}_{1- \left\{ \frac{kp}{q} \right\} - \left\{ \frac{rm}{s} \right\} \ge 0} \\&\quad + \,\left\{ \frac{kp}{q} \right\} \left( 1 - \frac{1}{r} - \frac{m}{s} + \frac{1}{r} \left\{ \frac{rm}{s} \right\} \right) \\&= \,\frac{1- \left\{ \frac{kp}{q} \right\} - \left\{ \frac{rm}{s} \right\} }{r} \mathbb {1}_{1- \left\{ \frac{kp}{q} \right\} - \left\{ \frac{rm}{s} \right\} < 0} \\&\quad + \left\{ \frac{kp}{q} \right\} \left( 1 - \frac{m}{s}+ \,\frac{1}{r} \left\{ \frac{rm}{s} \right\} \right) . \end{aligned}$$
Thus,
$$\begin{aligned} \sum _{k=1}^q \sum _{m=1}^{s-1} \int _0^1 f_{k,m} (x) \mathrm {d}x&= - \,\sum _{k=1}^q \left\{ \frac{kp}{q} \right\} + \sum _{k=1}^q \sum _{m=0}^{s-1} \int _0^1 f_{k,m} (x) \mathrm {d}x \\&=\, -\,\sum _{k=0}^{q-1} \frac{k}{q} + \sum _{k=0}^{q-1} \sum _{m=0}^{s-1} \frac{1 - \frac{k}{q} - \frac{m}{s}}{r} \mathbb {1}_{\frac{m}{s}> 1 - \frac{k}{q} }\\&\quad + \,\frac{k}{q} \left( 1- \frac{m}{s} + \frac{m}{sr} \right) \\&=\, \frac{q-1}{2} \left( s-1 - \frac{s-1}{2} + \frac{s-1}{2r} \right) \\&\quad +\, \sum _{k=0}^{q-1} \sum _{m=0}^{s-1} \frac{1 - \frac{k}{q} - \frac{m}{s}}{r} \mathbb {1}_{\frac{m}{s} > 1 - \frac{k}{q} } \end{aligned}$$
where we get the second equality noting that the numbers \(\left\{ \frac{kp}{q} \right\} \) (resp. \(\left\{ \frac{mr}{s} \right\} \) ) cycle through some rearrangement of the numbers \(0 , \frac{1}{q} ,\ldots , \frac{q-1}{q}\) (resp. \(0 , \frac{1}{s} ,\ldots , \frac{s-1}{s}\)). Now, it just remains to compute the second term in the last expression, which is equal to
$$\begin{aligned} \frac{1}{r} \sum _{k= \left\lfloor \frac{q}{s} \right\rfloor + 1}^{q-1} \sum _{m = \left\lfloor s- \frac{ks}{q} \right\rfloor + 1}^{s-1} \left( 1 - \frac{k}{q} - \frac{m}{s} \right) . \end{aligned}$$
Note that the condition \(\left\lfloor \frac{q}{s} \right\rfloor + 1 \le q-1\) is always satisfied when \(q\ge 3\) since \(s\ge 2 > \frac{3}{2} \ge \frac{q}{q-1}\). When \(q=2\), it only fails for \(s=2\) (in this case we have an empty sum). For now, assume we are not in the case \(q=s=2\). We will easily treat this remaining case at the end. We have:
$$\begin{aligned} \sum _{k= \left\lfloor \frac{q}{s} \right\rfloor + 1}^{q-1} \sum _{m = \left\lfloor s- \frac{ks}{q} \right\rfloor + 1}^{s-1} \left( 1 - \frac{k}{q} - \frac{m}{s} \right)&=\, \sum _{k= \left\lfloor \frac{q}{s} \right\rfloor + 1}^{q-1} \left( (s-1)- \left\lfloor s- \frac{ks}{q} \right\rfloor \right) \left( 1 - \frac{k}{q} \right) \\&\quad - \,\frac{1}{s} \sum _{m = \left\lfloor s- \frac{ks}{q} \right\rfloor + 1}^{s-1} m \\&= \,\sum _{k= \left\lfloor \frac{q}{s} \right\rfloor + 1}^{q-1} \frac{1}{2} \left( \frac{k}{q} - s \left( \frac{k}{q} \right) ^2 \right) \\&\quad + \,\frac{1}{2s} \left( \left\{ \frac{-ks}{q} \right\} ^2 - \left\{ \frac{-ks}{q} \right\} \right) . \end{aligned}$$
In addition, using the facts that for all \(x \in \mathbb {R}\), \(\{-x\}(1-\{-x\})= \{ x\}(1-\{ x\})\), and for all \(k \in [\![ 0 , \left\lfloor \frac{q}{s} \right\rfloor ]\!]\), \(0\le \frac{ks}{q} <1\), it comes
$$\begin{aligned}&\sum _{k=0}^{\left\lfloor \frac{q}{s} \right\rfloor } \frac{1}{2} \left( \frac{k}{q} - s \left( \frac{k}{q} \right) ^2 \right) + \frac{1}{2s} \left( \left\{ \frac{-ks}{q} \right\} ^2 - \left\{ \frac{-ks}{q} \right\} \right) \\&\qquad \qquad = \sum _{k=0}^{\left\lfloor \frac{q}{s} \right\rfloor } \frac{1}{2} \left( \frac{k}{q} - s \left( \frac{k}{q} \right) ^2 \right) + \frac{1}{2s} \left( \left( \frac{ks}{q} \right) ^2 - \frac{ks}{q} \right) = 0. \end{aligned}$$
Furthermore, denoting \(d:=\gcd (s,q)\) and \(s=d\tilde{s}\), \(q=d \tilde{q}\),
$$\begin{aligned} \sum _{k=0}^{q-1} \left\{ \frac{ks}{q} \right\} \left( 1 - \left\{ \frac{ks}{q} \right\} \right)&=\, d \sum _{k=0}^{\tilde{q}-1} \left\{ \frac{k\tilde{s}}{\tilde{q}} \right\} \left( 1 - \left\{ \frac{k\tilde{s}}{\tilde{q}} \right\} \right) \\&=\, d \sum _{k=0}^{\tilde{q}-1} \frac{k}{\tilde{q}} \left( 1 - \frac{k}{\tilde{q}} \right) \\&= \frac{q}{6} - \frac{d^2}{6q}. \end{aligned}$$
Thus,
$$\begin{aligned} \sum _{k= \left\lfloor \frac{q}{s} \right\rfloor + 1}^{q-1} \sum _{m = \left\lfloor s- \frac{ks}{q} \right\rfloor + 1}^{s-1} \left( 1 - \frac{k}{q} - \frac{m}{s} \right)&= \,-\, \frac{q}{12s} + \frac{d^2}{12qs} + \frac{1}{2} \sum _{k=0}^{q-1} \left( \frac{k}{q} - s \left( \frac{k}{q} \right) ^2 \right) . \end{aligned}$$
Expanding the expressions it follows
$$\begin{aligned} \frac{1}{n} \sum _{j=1}^n \lbrace js \alpha \rbrace \mathbb {1}_{ \lbrace j \frac{p}{q} \rbrace + \lbrace j r \alpha \rbrace \ge 1 }&\underset{n\rightarrow \infty }{\longrightarrow } s \left( \frac{1}{4} + \frac{1}{12r} - \frac{1}{4q} - \frac{1}{12rq^2} \right) \\&\qquad - \,\frac{1}{q} \left( \frac{1}{4} + \frac{qs}{4} - \frac{q}{4} + \frac{qs}{4r} - \frac{q}{4r} - \frac{s}{4} - \frac{s}{4r} + \frac{1}{4r} \right) \\&\qquad - \,\frac{1}{qr} \left( - \frac{q}{12s} + \frac{d^2}{12sq} -\frac{qs}{6} + \frac{s}{4} - \frac{s}{12q} + \frac{q}{4} - \frac{1}{4} \right) \\&\quad = \frac{1}{4} - \frac{1}{4q} + \frac{1}{12sr} - \frac{d^2}{12srq^2} \end{aligned}$$
Consequently,
$$\begin{aligned} s_3 \left( \alpha , \frac{p}{q} + \frac{r}{s} \alpha \right)&= \left( \frac{1}{4} - \frac{1}{4q} \right) + \left( \frac{1}{4} + \frac{1}{12sr} \right) - \left( \frac{1}{4} - \frac{1}{4q} + \frac{1}{12sr} - \frac{d^2}{12srq^2} \right) \\&= \frac{1}{4} + \frac{d^2}{12srq^2}. \end{aligned}$$
It is easy to check that this formula remains true for \(q=s=2\), since in this case we have
$$\begin{aligned} \sum _{k=1}^q \sum _{m=1}^{s-1} \int _0^1 f_{k,m} (x) \mathrm {d}x = \frac{1}{4}+ \frac{1}{4r} \end{aligned}$$
and thus
$$\begin{aligned} s_3 \left( \alpha , \frac{p}{q} + \frac{r}{s} \alpha \right)&= \left( \frac{1}{4} - \frac{1}{4q} \right) + \left( \frac{1}{4} + \frac{1}{12sr} \right) \\&\quad - \,s \left( \frac{1}{4} + \frac{1}{12sr} - \frac{1}{4q} - \frac{1}{12rq^2} \right) + \frac{1}{q} \left( \frac{1}{4}+ \frac{1}{4r} \right) \\&= \frac{1}{4} + \frac{1}{24r} = \frac{1}{4} + \frac{d^2}{12srq^2}. \end{aligned}$$
Finally, we assume \(r<0\).
As we have seen above, the result of \(s_3 \left( \alpha , \frac{p}{q} + \frac{r}{s} \alpha \right) \) is not depending on the choice of the irrational \(\alpha \); thus, one can replace it by \((-s)\alpha \), and then
$$\begin{aligned} s_3 \left( \alpha , \frac{p}{q} + \frac{r}{s} \alpha \right)&= s_3 \left( s (-\alpha ) , \frac{p}{q} + \vert r \vert \alpha \right) \\&= \lim _{n \rightarrow \infty } \frac{1}{n} \sum _{j=1}^n (1- \{ js\alpha \} ) \left\{ j \left( \frac{p}{q} + \vert r \vert \alpha \right) \right\} \\&= \lim _{n \rightarrow \infty } \frac{1}{n} \sum _{j=1}^n \left\{ j \left( \frac{p}{q} + \vert r \vert \alpha \right) \right\} - s_3 \left( s\alpha , \frac{p}{q} + \vert r \vert \alpha \right) \\&= \frac{1}{2} - \left( \frac{1}{4} + \frac{d^2}{12s \vert r \vert q^2} \right) \\&= \frac{1}{4} + \frac{d^2}{12srq^2}. \end{aligned}$$