PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 102(116) (2017), 155–174 DOI: https://doi.org/10.2298/PIM1716155G

EXPLICIT AND ASYMPTOTIC FORMULAE FOR VASYUNIN–COTANGENT SUMS

Mouloud Goubi, Abdelmejid Bayad, and Mohand Ouamar Hernane

Abstract. For coprime numbers p and q, we consider the Vasyunin–cotangent sum

V (q, p) = ∑_{k=1}^{p−1} \{ \frac{kq}{p}\} cot ( \frac{πk}{p} ) .

First, we prove explicit formula for the symmetric sum V (p, q) + V (q, p) which is a new reciprocity law for the sums above. This formula can be seen as a complement to the Bettin–Conrey result [13 , Theorem 1]. Second, we establish an asymptotic formula for V (p, q). Finally, by use of continued fraction theory, we give a formula for V (p, q) in terms of continued fraction of \frac{p}{q}.

1. Introduction and statement of results

1.1. Introduction. Let H = L^{2}([0, ∞]; t^{−2} dt) be the Hilbert space with the inner product

(1.1)  ⟨f, g⟩ = ∫_{0}^{∞} f(t) \overline{g(t)}t^{−2} dt, f, g ∈ H.

For any real number x, ⌊x⌋ is the integer part of x, and {x} = x − ⌊x⌋ is the fractional part of x. Let p be a positive integer. Denote by e_{p} the function in H given by e_{p}(t) = {t/p}. The properties of the subspace H_{n} = Vect(e_{1}, …, e_{n}) spanned by the functions e_{1}, …, e_{n} is studied in [6,8]. Consider the characteristic function

χ(t) := \{ \begin{matrix}1, & if t ∈ [1, ∞], \\ 0, & otherwise. \end{matrix}

2010 Mathematics Subject Classification: Primary: 11B99, 11F67, 11E45; Secondary: 11M26,11B68.

Key words and phrases: Vasyunin-cotangent sum, Estermann zeta function, fractional part function, Riemann Hypothesis.

Supported by Université d’Evry Val d’Essonne and PHC-Tassili program 14MDU914. Communicated by Gradimir Milovanović.

The distance d_{n} from χ to the subspace H_{n} is given by d_{n} = dist(χ, H_{n}) = inf_{h∈H_{n}} ∥χ − h∥. In [7] it conjectured that

(1.2)  d_{n}^{2} ∼ \frac{2 + γ − log4π + o(1)}{logn} , n → +∞.

Balazard and Roton proved in [9] that

d_{n}^{2} ⩾ \frac{2 + γ − log4π}{logn} , n → +∞.

It is well known that conjecture (1.2) implies the Riemann hypothesis [2–6,8]. For computation aspects of d_{n}, we refer to Landreau et al [18].

For fixed n ⩾ 1, from [10], we quote the formula

d_{n}^{2} = \frac{Gram(χ, e_{1}, ⋯, e_{n})}{Gram(e_{1}, ⋯, e_{n})} .

To compute this quantity, we need to evaluate two types of inner products, namely, ⟨χ, e_{p}⟩, ⟨e_{p}, e_{q}⟩. The first one is given in [6, §1] by

⟨χ, e_{p}⟩ = \frac{logp + 1 − γ}{p} .

On the other hand, for two coprime numbers p, q, the second inner product is given by the Vasyunin formula [8, 28]

(1.3)  ⟨e_{p}, e_{q}⟩ = \frac{log2π − γ}{2} ( \frac{1}{p} + \frac{1}{q}) + \frac{p − q}{2pq} log \frac{q}{p} − \frac{π}{2pq}[V (q, p) + V (p, q)]

where

(1.4)  V (q, p) = ∑_{k=1}^{p−1} \{ \frac{kq}{p}\} cot( \frac{πk}{p} ) .

The Vasyunin–cotangent sum V (p, q) is still curious. Recently, Bettin and Conrey [13, Theorem 1] proved a formula for

\frac{q̄}{p}V (q, p) + V (p̄, q̄)

where pp̄ ≡ 1(q) and qq̄ ≡ 1(p) with 1 ⩽ p̄ ⩽ q, 1 ⩽ q̄ ⩽ p.

For q = 1, the cotangent sum (1.4) is first studied by Vasyunin [28]. He proved the asymptotic formula (for large p)

V (1, p) = −\frac{p logp}{π} + \frac{p}{π}(log2π − γ) + O(logp).

This formula is improved by Rassias [26] and Maier and Rassias [20] as follows

V (1, p) = −\frac{p logp}{π} + \frac{p}{π}(log2π − γ) + O(1).

EXPLICIT AND ASYMPTOTIC FORMULAE FOR VASYUNIN–COTANGENT SUMS 157

Let p, n ∈ ℕ, p > 6N, with N = ⌊\frac{n}{2} ⌋ + 1. Maier and Rassias [21, Theorem 1.7] proved that there exist absolute real constants A_{1}, A_{2} ⩾ 1 and absolute real constants E_{l}, l ∈ ℕ with |E_{l}| ⩽ (A_{1}l)^{2l}, such that for each n, we have

(1.5)  V (1, p) = −\frac{p logp}{π} + \frac{p}{π}(log2π − γ) + \frac{1}{π} − ∑_{l=1}^{n} E_{l}p^{−l} − R_{n}^{∗}(p)

where |R_{n}^{∗}(p)| ⩽ (n A_{2})^{4n} p^{−(n+1)} .

The sum V (p, q) can be interpreted as the value of the Estermann zeta function [13, 17] at s = 0

E_{0}( s, \frac{p}{q}) = ∑_{k⩾1} \frac{τ(k)}{n^{s}} exp( \frac{2πikp}{q}) .

Recently, this sum is studied in [12, 13] and it is proved that V (p, q) satisfies the reciprocity formula for all positive coprime numbers p and q

(1.6)  \frac{q̄}{p}V (q, p) + V (p̄, q̄) = −\frac{1}{πp} − g( \frac{q̄}{p})

where g is an analytic function in ℂ ∖ ℝ_{−} , which has the following asymptotic expansion of order N ⩾ 2, x → 0

g(x) = −\frac{log2πx − γ}{π} + \frac{2}{π} ∑_{k=2}^{N} \frac{ζ(k)B_{k}}{k} x^{k} + O(x^{N+1} ),

B_{k} is the k^{th} Bernoulli number. Since ζ(2k) = (−1)^{k−1} 2^{2k−1} π^{2k}B_{2k} /(2k)!, we have

g(x) = −\frac{log2πx − γ}{π} − \frac{1}{2} ∑_{k=1}^{⌊N/2⌋} (−1)^{k} \frac{4^{k}π^{2k−1} B_{2k}^{2}}{k(2k)!} x^{k} + O(x^{N+1} ).

1.2. Statement of main results. Now consider the digamma function

ψ(z) = −γ − \frac{1}{z} + ∑_{k=1}^{∞} ( \frac{1}{k} − \frac{1}{k + z} )

and the symmetric function

(1.7)  G(p, q) = ∑_{r=1}^{pq−1} ( ψ( \frac{r + 1}{pq} ) − ψ( \frac{r}{pq}) ) \{ \frac{r}{p}\}\{ \frac{r}{q}\}.

We state a new reciprocity formula V (p, q).

Theorem 1.1. For p, q positive coprime numbers, we have

(1.8) V (p, q) + V (q, p) = \frac{1}{π} logp^{q−1} q^{p−1} − \frac{2}{π} − pg( \frac{1}{p}) − qg( \frac{1}{q}) − \frac{2}{π}G(p, q).

Corollary 1.1. For p, q positive coprime numbers, we have

⟨e_{p}, e_{q}⟩ = \frac{2 + (log2π − γ)(p + q)}{2pq} − \frac{1}{2pq}logp^{p−1} q^{q−1} (1.9)

+\frac{π}{2}( \frac{1}{q} g( \frac{1}{p}) + \frac{1}{p} g( \frac{1}{q}) ) + \frac{1}{pq}G(p, q) .

Next, we state an asymptotic formula for the sum V (ā, pa + r).

Theorem 1.2. Let a > r and p be integers such that (a, pa + r) = 1, āa ≡ 1 (mod ap + r) and r̄r ≡ 1 (mod a). Then for large p, we have

V (ā, pa + r) = −( p + \frac{r}{a}) V (r̄, a) − \frac{1}{πa} + \frac{1}{π}( p + \frac{r}{a}) ( log \frac{2πa}{pa + r} − γ) (1.10)

+\frac{1}{2} ∑_{k=1}^{⌊N/2⌋} (−1)^{k} \frac{4^{k}π^{2k−1} B_{2k}^{2}}{k(2k)!} ( \frac{a}{pa + r} )^{2k−1} + O( \frac{1}{p^{N}} ) .

We get the following corollary.

Corollary 1.2. Let a ⩾ 1. For large p, we have

V (p + 1, ap + a − 1) = (1.11)

− ( p + 1 − \frac{1}{a}) V (1, a) − \frac{1}{πa} + \frac{1}{π}( p + 1 − \frac{1}{a}) ( log \frac{2πa}{(p + 1)a − 1} − γ)

+ \frac{1}{2} ∑_{k=1}^{⌊ \frac{N}{2} ⌋}(−1)^{k} \frac{4^{k}π^{2k−1} B_{2k}^{2}}{k(2k)!} ( \frac{a}{(p + 1)a − 1} )^{2k−1} + O( \frac{1}{p^{N}} ) .

In the particular case a = 1, we have

(1.12)  V (1, p)= \frac{1}{π}( log \frac{2π}{p} − γ) p − \frac{1}{π} − \frac{π}{144p}

+ \frac{1}{2} ∑_{k=2}^{⌊ \frac{N}{2} ⌋}(−1)^{k} \frac{4^{k}π^{2k−1} B_{2k}^{2}}{k(2k)!} ( \frac{1}{p})^{2k−1} + O( \frac{1}{p^{N}} ) .

Relation (1.12) improves asymptotic formula (1.5) proved by Maier and Rassias in [21, Theorem 1.7].

In what follows, we give a few examples.

Examples 1.1. For large p we have

V (p + 1, 2p + 1) = − \frac{1}{2π} + \frac{1}{π}( p + \frac{1}{2}) ( log \frac{4π}{2p + 1} − γ)

+ \frac{1}{2} ∑_{k=1}^{⌊N/2⌋} (−1)^{k} \frac{4^{k}π^{2k−1} B_{2k}^{2}}{k(2k)!} ( \frac{2}{2p + 1} )^{2k−1} + O( \frac{1}{p^{N}} ) .

V (2p + 1, 3p + 1) = − \frac{p + \frac{1}{3}}{3\sqrt{3}} − \frac{1}{3π} + \frac{1}{π}( p + \frac{1}{3}) ( log \frac{3π}{2} − γ)

+ \frac{1}{2} ∑_{k=1}^{⌊N/2⌋} (−1)^{k} \frac{4^{k}π^{2k−1} B_{2k}^{2}}{k(2k)!} ( \frac{3}{3p + 1} )^{2k−1} + O( \frac{1}{p^{N}} ) .

V (p + 1, 3p + 2) = \frac{p + \frac{2}{3}}{3\sqrt{3}} − \frac{1}{3π} + \frac{1}{π}( p + \frac{2}{3}) ( log \frac{6π}{5} − γ)

+ \frac{1}{2} ∑_{k=1}^{⌊N/2⌋} (−1)^{k} \frac{4^{k}π^{2k−1} B_{2k}^{2}}{k(2k)!} ( \frac{3}{3p + 2} )^{2k−1} + O( \frac{1}{p^{N}} ) .

We can consider V (q, p) as a function of a single rational argument, by defining V (q/p) = V (q, p). That this function is well defined is clear from the conditions on q and p. By use of continued fractions [25, §7] and the reciprocity law [13, Theorem 1], we obtain

Theorem 1.3. Let 1 < q < p be coprime positive integers and p̄ denote the inverse of p modulo q. Write p̄/q = [a_{0}, a_{1}, …, a_{n}] for a simple finite continued fraction of p̄/q and (p_{k})_{0⩽k⩽n+1} for the finite sequence given by: p_{k−1} = a_{k−1} p_{k} + p_{k+1} with p_{0} = p̄, p_{1} = q and p_{n+1} = 0. Then we have

V (p, q)= \frac{1}{π(q[a_{0}, a_{1}, …, a_{n−2} ] − p̄)}

− ∑_{k=2}^{n} (−1)^{k}g([0, a_{k−1} , …, a_{n}]) ∏_{j=2}^{k} [a_{j−1} , …, a_{n}].

Corollary 1.3. Let \frac{q}{p} = [b_{0}, …, b_{n}] and (q_{k})_{k} the associated sequence: q_{k−1} =

b_{k−1} q_{k} + q_{k+1} , q_{0} = q, q_{1} = p, q_{n} = 1. Then we have

(1.13)  V (q, p) =− \frac{1}{π} − pg( \frac{1}{p}) − \frac{1}{π} logpq + \frac{1}{π}(logq + (−1)^{n} logq_{n−1} )

− \frac{1}{π} ∑_{k=1}^{n−2} (−1)^{k}(q_{k} logq_{k+1} + q_{k+1} logq_{k} − 2G(q_{k}, q_{k+1} )).

Moreover, we have

V (q, p) =− \frac{1}{π}(1 + logf_{1}) − q( f_{1}g( \frac{1}{qf_{1}} ) − \frac{1}{π} ∑_{k=1}^{n−2} (−1)^{k}(f_{k} logf_{k+1} ) + f_{k+1} logf_{k}) (1.14)

+ \frac{(−1)^{n}}{π} logf_{n−1} + \frac{(−1)^{n} − 1}{π} logq + ((−1)^{n}f_{n−1} − f_{1})\frac{q logq}{π}

+ \frac{2}{π} ∑_{k=1}^{n−2} (−1)^{k}G(qf_{k}, qf_{k+1} ), where f_{k} = ∏_{j=1}^{k} [0, b_{j−1} , …, b_{n}].

1.3. General case: p,q arbitrary. Let ω = gcd(p, q) ⩾ 1. Vasyunin formula in [28] is given by

⟨e_{p}, e_{q}⟩ = \frac{log2π − γ}{2} ( \frac{1}{p} + \frac{1}{q}) + \frac{p − q}{2pq} log \frac{q}{p} − \frac{πω}{2pq}( V ( \frac{p}{ω}, \frac{q}{ω}) + V ( \frac{q}{ω}, \frac{p}{ω}) ) .

As a consequence of Theorem 1.1, we find a new expression for ⟨e_{p}, e_{q}⟩:

⟨e_{p}, e_{q}⟩ = \frac{2ω + (log2π − γ)(p + q)}{2pq}

− \frac{1}{2pq}((p − ω) logp + (q − ω) logq − (p + q − 2ω) logω)

+ \frac{π}{2}( \frac{1}{p}g( \frac{ω}{q}) + \frac{1}{q}g( \frac{ω}{p}) ) + \frac{ω}{pq}G( \frac{p}{ω}, \frac{q}{ω}) .

2. Proof of Theorem 1.1 and Corollary 1.1

We start this section with some useful prelimary results.

2.1. Computation of ⟨v_{p}, v_{q}⟩. Let p be positive integer and v_{p} be the function given by v_{p}(t) = {⌊t⌋/p}; v_{p} is defined on ℝ_{+} and it can be regarded as a restriction of e_{p} to ℕ. Since

v_{p}(x) = \{ \begin{matrix}0, & if x ∈ [0, 1], \\ \{k/p\}, & x ∈ [k, k + 1] \end{matrix}

we have the relation

(2.1)  v_{p} = e_{p} − \frac{1}{p}e_{1}.

Then, from definition (1.1), we get

⟨v_{p}, v_{q}⟩ = ∑_{k⩾1} \frac{1}{k(k + 1)} \{ \frac{k}{p}\}\{ \frac{k}{q}\}.

Lemma 2.1. Let p, q be coprime numbers. Then we have

\frac{2pq}{π}⟨v_{p}, v_{q}⟩ = V (1, p) + V (1, q) − (V (p, q) + V (q, p)) + \frac{1}{π} logp^{q−1} q^{p−1} .

Proof. Using expression (2.1), we get

⟨v_{p}, v_{q}⟩ = ⟨e_{p} − \frac{1}{p}e_{1}, e_{q} − \frac{1}{q}e_{1}⟩ = ⟨e_{p}, e_{q}⟩ + \frac{1}{pq}⟨e_{1}, e_{1}⟩ − \frac{1}{p}⟨e_{1}, e_{q}⟩ − \frac{1}{q}⟨e_{1}, e_{p}⟩

On the other hand, from Vasyunin formula (1.3), we obtain

⟨e_{1}, e_{q}⟩ = \frac{log2π − γ}{2} ( 1 + \frac{1}{q}) + \frac{1 − q}{2q} logq− \frac{π}{2q}V (1, q),

⟨e_{1}, e_{p}⟩ = \frac{log2π − γ}{2} ( 1 + \frac{1}{p}) + \frac{1 − p}{2p} logp− \frac{π}{2p}V (1, p),

⟨e_{p}, e_{q}⟩ = \frac{log2π − γ}{2} ( \frac{1}{p} + \frac{1}{q}) + \frac{p − q}{2pq} log \frac{q}{p} − \frac{π}{2pq}[V (q, p) + V (p, q)].

Therefore we deduce

\frac{2pq}{π}⟨v_{p}, v_{q}⟩ = V (1, p) + V (1, q) − V (p, q) − V (q, p) + \frac{1}{π}(q logp + p logq − logpq). □

The series

(2.2)  ∑_{k⩾1} \frac{1}{k(k + 1)} \{ \frac{k}{p}\}\{ \frac{k}{q}\}

is convergent. Let us rewrite it in another form. For integer k we set k ≡ r(pq), 1 ⩽ r ⩽ pq − 1 and then \{ \frac{k}{p}\}\{ \frac{k}{q}\} = \{ \frac{r}{p}\}\{ \frac{r}{q}\}. Series (2.2) is equal to

(2.3)  ∑_{r=1}^{pq−1} ∑_{i⩾0} \frac{1}{(ipq + r)(ipq + r + 1)} \{ \frac{r}{p}\}\{ \frac{r}{q}\}.

Finally, to estimate ⟨v_{p}, v_{q}⟩ we need to compute the sums

∑_{i⩾0} \frac{1}{(ipq + r)(ipq + r + 1)} .

Lemma 2.2. For a, b two distinct positive numbers, we have

∑_{k=0}^{∞} \frac{1}{(k + a)(k + b)} = \frac{ψ(a) − ψ(b)}{a − b} .

Proof. We write

ψ(a) − ψ(b) = \frac{1}{b} − \frac{1}{a} + ∑_{k⩾1} ( \frac{1}{k + b} − \frac{1}{k + a} ) = (a − b) ∑_{k=0}^{∞} \frac{1}{(k + a)(k + b)} .  □

Corollary 2.1. Let α > 0. We have

(2.4)  ∑_{k=0}^{∞} \frac{1}{(αk + a)(αk + b)} = \frac{ψ(a/α) − ψ(b/α)}{α(a − b)} .

From [19], we have the integral representation

ψ(a) − ψ(b) = ∫_{0}^{∞} \frac{e^{−bt} − e^{−at}}{1 − e^{−t}} dt.

Thus, we can get

∑_{k=0}^{∞} \frac{a − b}{(k + a)(k + b)} = ∑_{k=0}^{∞} ( \frac{1}{k + b} − \frac{1}{k + a} ) = ∑_{k=0}^{∞} ∫_{0}^{1}(x^{k+b−1} − x^{x+a−1} )dx

= ∫_{0}^{1} \frac{x^{b−1} − x^{a−1}}{1 − x} dx = ∫_{0}^{∞} \frac{e^{−bt} − e^{−at}}{1 − e^{−t}} dt.

Hence, the special case α = pq of relations (1.7), (2.3), and (2.4) gives the result.

Corollary 2.2. For p, q arbitrary, we have

(2.5)  ∑_{k⩾1} \frac{pq}{k(k + 1)} \{ \frac{k}{p}\}\{ \frac{k}{q}\} = G(p, q).

2.2. Proof of Theorem 1.1 and Corollary 1.1. From Lemma 2.1 and Corollary 2.2, we deduce that

2G(p, q) = V (1, p) + V (1, q) − (V (p, q) + V (q, p)) + \frac{1}{π} logp^{q−1} q^{p−1} .

Therefore, we get (1.8) from the expression

(2.6)  V (1, p) = −\frac{1}{π} − pg( \frac{1}{p}) .

We can obtain formula (1.9) by use of reciprocity formula (1.8) and Vasyunin formula (1.3).

3. Proof of Theorem 1.2 and Corollary 1.2

From reciprocity law (1.6), we have

\frac{a}{pa + r} V (ā, pa + r) + V (r̄, a) = −\frac{1}{π(pa + r)} − g( \frac{a}{pa + r} )

and then we obtain

V (ā, pa + r) = −\frac{pa + r}{a} V (r̄, a) − \frac{1}{πa} − \frac{pa + r}{a} g( \frac{a}{pa + r} ) .

Moreover, for large p we have

g( \frac{a}{pa + r} ) = −\frac{1}{π}( log \frac{2πa}{pa + r} −γ) − \frac{1}{2} ∑_{k=1}^{⌊N/2⌋}(−1)^{k} \frac{4^{k}π^{2k−1} B_{2k}^{2}}{k(2k)!} ( \frac{a}{pa + r} )^{k}+O( \frac{1}{p^{N+1}} )

Taking r = a − 1, one remarks that for a ⩾ 1 (p + 1)a ≡ 1 (mod pa + a − 1) and (a − 1)^{2} ≡ 1(a). Then ā = p + 1 and r̄ = r = a − 1. Since V (a − 1, a) = V (1, a) and thanks to relation (1.10) of Theorem 1.2, we obtain relation (1.11). Finally from relation (1.11) we can get (1.12).

4. Proof of Theorem 1.3 and Corollary 1.3

4.1. Continued fractions – an overview. In this subsection we quote some facts about continued fractions, that will be useful later. For more details we refer to [25, §7]. For real number x let

x = a_{0} + \frac{1}{a_{1} + \frac{1}{a_{2} + \frac{1}{a_{3} + ⋯}}} = [a_{0}, a_{1}, a_{3}, …]

be its continued fraction expansion with partial quotients a_{0} ∈ ℤ, a_{k} ∈ ℕ ∖ {0}, k ⩾ 1. In fact, we can determine a_{0}, a_{1}, …, a_{n} via the algorithm:

• x = ⌊x⌋ + {x}, a_{0} = ⌊x⌋, ξ_{0} = {x}, if ξ_{0} = 0, then x is represented by

x = [a_{0}].

• if ξ_{0} ≠ 0 then ⌊\frac{1}{ξ_{0}} ⌋ > 1, r_{1} = \frac{1}{ξ_{0}} we obtain x = [a_{0}, r_{1}], a_{1} = ⌊r_{1}⌋ and

ξ_{1} = r_{1} − a_{1}; if ξ_{1} = 0 then x = a_{0} + \frac{1}{r_{1}} = a_{0} + \frac{1}{a_{1}} = [a_{0}, a_{1}].
• Otherwise we take r_{2} = \frac{1}{ξ_{1}} and iterate the process.

We then get the sequence a_{0}, a_{1}, a_{2}, . . .. This sequence is finite if and only if x is a rational number. In the rational case, this algorithm is the Euclidian algorithm. Let us express a/b as a continued fraction of the form a/b = [a_{0}, a_{1}, …, a_{n}], with a_{n+1} = 0. We can determine a_{0}, a_{1}, …, a_{n} by the Euclidean algorithm

(4.1)  p_{k−1} = a_{k−1} p_{k} + p_{k+1} , p_{0} = a, p_{1} = b, p_{n+1} = 0.

Then p_{n} = gcd(p, q) = 1. We quote from [25, §7] the following elementary properties of continued fractions which are needed in this paper. We omit their proofs.

EXPLICIT AND ASYMPTOTIC FORMULAE FOR VASYUNIN–COTANGENT SUMS 163

Lemma 4.1. Let [a_{0}, a_{1}, …, a_{n}] be continued fraction. Then

[a_{0}, a_{1}, …, a_{n}]= [a_{0}, a_{1}, …, a_{n−1} + \frac{1}{a_{n}} ],
[a_{0}, a_{1}, …, a_{n}]= a_{0} + \frac{1}{[a_{1}, ⋯, a_{n}]}.

We get this lemma from the continued fraction definition. Let s_{k}, t_{k} be two sequences of integers

s_{k} = \{ \begin{matrix}0, & if k = −2 \\ 1, & if k = −1 \\ a_{k}s_{k−1} + s_{k−2} , & if k ⩾ 0 \end{matrix} t_{k} = \{ \begin{matrix}1, & if k = −2 \\ 0, & if k = −1 \\ a_{k}t_{k−1} + t_{k−2} , & if k ⩾ 0 \end{matrix}

The sequences p_{k}, s_{k} and t_{k} have the following properties.

Lemma 4.2.

[a_{0}, a_{1}, …, a_{k}] = \frac{s_{k}}{t_{k}} , k ⩾ 0, (4.2)

[a_{k}, a_{k−1} , …, a_{1}] = \frac{t_{k}}{t_{k−1}} , k ⩾ 1,

[0, a_{k−1} , …, a_{n}] = \frac{p_{k}}{p_{k−1}} , k ⩾ 0, (4.3)

[a_{k}, a_{k−1} , …, a_{0}] = \frac{s_{k}}{s_{k−1}} , k ⩾ 0.

This lemma represents the classical properties of the continued fractions. The quotient \frac{s_{k}}{t_{k}} is called the k^{th}-convergent of the continued fraction of \frac{a}{b}.

Corollary 4.1. Let \frac{a}{b} = [a_{0}, a_{1}, …, a_{n}]; then

(4.4)  p_{k} = b ∏_{j=2}^{k} [0, a_{j−1} , …, a_{n}], k ⩾ 2,
b= p_{k}t_{k−1} + p_{k+1} t_{k−2} , k ⩾ 1.

Lemma 4.3. The sequences s_{k}, t_{k} and p_{k} satisfy

t_{k}s_{k−1} − t_{k−1} s_{k} = (−1)^{k},

p_{k} = (−1)^{k}(at_{k−2} − bs_{k−2} ). (4.5)

Proposition 4.1. Under the hypothesis of Corollary 4.1, we have

(4.6) ∑_{k=1}^{n−1} \frac{(−1)^{k}}{p_{k}p_{k+1}} = \frac{1}{b^{2}[a_{0}, a_{1}, …, a_{n−2} ] − ab}.

Proof. We have from Lemma 4.3

∑_{k=1}^{n−1} \frac{(−1)^{k}}{p_{k}p_{k+1}} = \frac{1}{b} ∑_{k=1}^{n−1} ( \frac{(−1)^{k}t_{k−1}}{p_{k+1}} + \frac{(−1)^{k}t_{k−2}}{p_{k}} )

= \frac{1}{b} ∑_{k=1}^{n−1} \frac{(−1)^{k}t_{k−1}}{p_{k+1}} + \frac{1}{b} ∑_{k=1}^{n−1} \frac{(−1)^{k}t_{k−2}}{p_{k}}

= \frac{1}{b} ∑_{k=2}^{n} \frac{(−1)^{k−1} t_{k−2}}{p_{k}} + \frac{1}{b} ∑_{k=1}^{n−1} \frac{(−1)^{k}t_{k−2}}{p_{k}}

= \frac{(−1)^{n−1} t_{n−2}}{p_{n} b} − \frac{t_{−1}}{b p_{1}} = \frac{(−1)^{n−1} t_{n−2}}{p_{n} b} .

From (4.5) we deduce p_{n} = (−1)^{n}(at_{n−2} − bs_{n−2} ) and then

(4.7)  \frac{(−1)^{n−1} p_{n}}{t_{n−2}} = ( b\frac{s_{n−2}}{t_{n−2}} − a) .

Relation (4.2) implies that

(4.8)  \frac{s_{n−2}}{t_{n−2}} = [a_{0}, a_{1}, …, a_{n−2} ]

From relations (4.7) and (4.8) we have

\frac{(−1)^{n−1} t_{n−2}}{p_{n}b} = \frac{1}{b^{2}[a_{0}, a_{1}, …, a_{n−2} ] − ab}.

Therefore we obtain

∑_{k=1}^{n−1} \frac{(−1)^{k}}{p_{k}p_{k+1}} = \frac{1}{b^{2}[a_{0}, a_{1}, …, a_{n−2} ] − ab}.  □

4.2. Application: Computation of V (p, q). Using reciprocity formula (1.6) and continued fractions properties, we can get an explicit formula for V (p, q).

Lemma 4.4. Let p and q be coprime positive numbers. We have

(4.9)  V (p, q) = \frac{q}{π} ∑_{k=1}^{n−1} \frac{(−1)^{k}}{p_{k}p_{k+1}} − q ∑_{k=2}^{n} \frac{(−1)^{k}}{p_{k}} g( \frac{p_{k}}{p_{k−1}} ) .

Proof. From the reciprocity law (1.6), with p̄ = a, q = b and the sequence (p_{k})_{k} defined in relation (4.1), we have

(4.10)  \frac{1}{p_{k−1}} V (p̄_{k}, p_{k−1} ) + \frac{1}{p_{k}}V (p̄_{k−1} , p_{k}) = \frac{1}{p_{k}}( \frac{1}{πp_{k−1}} + g( \frac{p_{k}}{p_{k−1}} ) ) .

Observe that

p_{k−1} = a_{k−1} p_{k} + p_{k+1} , p̄_{k−1} = p̄_{k+1} , V (p̄_{k−1} , p_{k}) = V (p̄_{k+1} , p_{k}),

then relation (4.10) becomes

\frac{1}{p_{k−1}} V (p̄_{k}, p_{k−1} ) + \frac{1}{p_{k}}V (p̄_{k+1} , p_{k}) = −\frac{1}{p_{k}}( \frac{1}{πp_{k−1}} + g( \frac{p_{k}}{p_{k−1}} ) ) .

We have the computation

∑_{k=1}^{n} [ \frac{(−1)^{k}}{p_{k−1}} V (p̄_{k}, p_{k−1} ) + \frac{(−1)^{k}}{p_{k}} V (p̄_{k+1} /p_{k})]

= ∑_{k=1}^{n} \frac{(−1)^{k}}{p_{k−1}} V (p̄_{k}, p_{k−1} ) + ∑_{k=1}^{n} \frac{(−1)^{k}}{p_{k}} V (p̄_{k+1} , p_{k})

= ∑_{k=0}^{n−1} \frac{(−1)^{k+1}}{p_{k}} V (p̄_{k+1} , p_{k}) + ∑_{k=1}^{n} \frac{(−1)^{k}}{p_{k}} V (p̄_{k+1} , p_{k})

= −\frac{1}{p_{0}}V (p̄_{1}, p_{0}) + \frac{(−1)^{n}}{p_{n}} V (p̄_{n+1} , p_{n}).

Since p_{n+1} = 0, the relation V (p̄_{n+1} , p_{n}) = 0 implies

∑_{k=1}^{n} [ \frac{(−1)^{k}}{p_{k−1}} V (p̄_{k}, p_{k−1} ) + \frac{(−1)^{k}}{p_{k}} V (p̄_{k+1} , p_{k})] = −\frac{1}{p̄}V (q̄, p̄)

= −\frac{1}{p̄}V (q̄, p̄) − \frac{1}{q}V (p, q) +∑_{k=2}^{n} [ \frac{(−1)^{k}}{p_{k−1}} V (p̄_{k}, p_{k−1} ) + \frac{(−1)^{k}}{p_{k}} V (p̄_{k+1} , p_{k})].

Moreover,

\frac{1}{q}V (p, q) = ∑_{k=2}^{n} [ \frac{(−1)^{k}}{p_{k−1}} V (p̄_{k}, p_{k−1} ) + \frac{(−1)^{k}}{p_{k}} V (p̄_{k+1} , p_{k})]

then we deduce

\frac{1}{q}V (p, q)=−∑_{k=2}^{n} \frac{(−1)^{k}}{p_{k}} ( \frac{1}{πp_{k−1}} +g( \frac{p_{k}}{p_{k−1}} )) = \frac{1}{π} ∑_{k=1}^{n−1} \frac{(−1)^{k}}{p_{k}p_{k+1}} −∑_{k=2}^{n} \frac{(−1)^{k}}{p_{k}} g( \frac{p_{k}}{p_{k−1}} ) . □

4.3. Proof of Theorem 1.3 and Corollary 1.3. First, we prove Theorem 1.3. Taking a = p̄ and b = q, by virtue of relation (4.6) we obtain

(4.11)  \frac{q}{π} ∑_{k=1}^{n−1} \frac{(−1)^{k}}{p_{k}p_{k+1}} = \frac{1}{q[a_{0}, a_{1}, …, a_{n−2} ] − p̄}.

From (4.4) we get

\frac{1}{p_{k}} = \frac{1}{q} ∏_{j=2}^{k} [a_{j−1} , …, a_{n}], k ⩾ 2,

and from (4.3) we obtain

(4.12)  ∑_{k=2}^{n} \frac{(−1)^{k}}{p_{k}} g( \frac{p_{k}}{p_{k−1}} ) = ∑_{k=2}^{n} \frac{1}{q}g ([a_{k−1} , ⋯, a_{n}]) ∏_{j=2}^{k} [a_{j−1} , ⋯, a_{n}].

Substituting quantities (4.11) and (4.12) into (4.9) we get Theorem 1.3. To prove Corollary 1.3, note that for p ≡ r(q) we have

(4.13)  V (p, q) = \{ \begin{matrix}0, & if r = 0 \\ V (r, q), & otherwise. \end{matrix}

166 GOUBI, BAYAD, AND HERNANE

Applying several times reciprocity formula (1.8) to the sequence q_{k}, we obtain

∑_{k=1}^{n−1} (−1)^{k}θ(q_{k−1} , q_{k}) = ∑_{k=1}^{n−1} (−1)^{k}[V (q_{k−1} , q_{k}) + V (q_{k}, q_{k−1} )]

where

θ(q_{k−1} , q_{k})= \frac{1}{π}(q_{k−1} logq_{k} + q_{k} logq_{k−1} − logq_{k−1} q_{k})

− \frac{2}{π} − q_{k−1} g( \frac{1}{q_{k−1}} ) − q_{k}g( \frac{1}{q_{k}}) − \frac{2}{π}G(q_{k−1} , q_{k}).

From (4.13) we have V (q_{k+1} , q_{k}) = V (q_{k−1} , q_{k}) and V (q_{n−1} , q_{n}) = 0, and then

∑_{k=1}^{n−1} (−1)^{k}θ(q_{k−1} , q_{k}) = ∑_{k=1}^{n−1} (−1)^{k}V (q_{k−1} , q_{k}) + ∑_{k=1}^{n−1} (−1)^{k}V (q_{k}, q_{k−1} )

= ∑_{k=1}^{n−1} (−1)^{k}V (q_{k+1} , q_{k}) + ∑_{k=0}^{n−2} (−1)^{k+1} V (q_{k+1} , q_{k})

= −V (p, q) − (−1)^{n}V (q_{n}, q_{n−1} ).

In addition we have

∑_{k=1}^{n−1} (−1)^{k}θ(q_{k−1} , q_{k}) = \frac{1}{π} ∑_{k=1}^{n−1} (−1)^{k}[(q_{k−1} logq_{k} + q_{k} logq_{k−1} − logq_{k−1} q_{k})]
= − \frac{2}{π} ∑_{k=1}^{n−1} (−1)^{k}[1 + G(q_{k−1} , q_{k})]

− ∑_{k=1}^{n−1} (−1)^{k}[q_{k−1} g( \frac{1}{q_{k−1}} ) + q_{k}g( \frac{1}{q_{k}}) ]

and we push this computation we arrive to

∑_{k=1}^{n−1} (−1)^{k}[q_{k−1} g( \frac{1}{q_{k−1}} ) + q_{k}g( \frac{1}{q_{k}}) ] =∑_{k=1}^{n−1} (−1)^{k}q_{k−1} g( \frac{1}{q_{k−1}} ) +∑_{k=1}^{n−1} (−1)^{k}q_{k}g( \frac{1}{q_{k}})

= ∑_{k=0}^{n−2} (−1)^{k+1} q_{k}g( \frac{1}{q_{k}}) + ∑_{k=1}^{n−1} (−1)^{k}q_{k}g( \frac{1}{q_{k}})

= −qg( \frac{1}{q}) + (−1)^{n−1} q_{n−1} g( \frac{1}{q_{n−1}} ) .

We have also

∑_{k=1}^{n−1} (−1)^{k} logq_{k−1} q_{k} = ∑_{k=1}^{n−1} (−1)^{k}(logq_{k−1} + logq_{k})

= ∑_{k=1}^{n−1} (−1)^{k} logq_{k−1} + ∑_{k=1}^{n−1} (−1)^{k} logq_{k}

= ∑_{k=0}^{n−2} (−1)^{k+1} logq_{k} + ∑_{k=1}^{n−1} (−1)^{k} logq_{k}

= − logq − (−1)^{n} logq_{n−1} .

Then we have

−V (p, q) − (−1)^{n}V (q_{n}, q_{n−1} ) = \frac{1}{π} ∑_{k=1}^{n−1} (−1)^{k}(q_{k−1} logq_{k} + q_{k} logq_{k−1} ) + qg( \frac{1}{q})

+(−1)^{n}q_{n−1} g( \frac{1}{q_{n−1}} ) − \frac{2}{π} ∑_{k=1}^{n−1} (−1)^{k}[1 + G(q_{k−1} , q_{k})]

+\frac{1}{π}(logq + (−1)^{n} logq_{n−1} )

and

−V (p, q) = (−1)^{n}V (q_{n}, q_{n−1} ) + (−1)^{n}q_{n−1} g( \frac{1}{q_{n−1}} ) − \frac{1}{π}(q logp + p logq)

+ qg( \frac{1}{q}) + \frac{2}{π}(1 + G(q, p)) + \frac{1}{π}(logq + (−1)^{n} logq_{n−1} )

− \frac{1}{π} ∑_{k=1}^{n−2} (−1)^{k}[q_{k} logq_{k+1} + q_{k+1} logq_{k} − 2G(q_{k}, q_{k+1} ) − 2]

= \frac{(−1)^{n+1}}{π} − V (p, q) − V (q, p) − \frac{1}{π} logpq − pg( \frac{1}{p})

+ \frac{1}{π}(logq + (−1)^{n} logq_{n−1} )

− \frac{1}{π} ∑_{k=1}^{n−2} (−1)^{k}[q_{k} logq_{k+1} + q_{k+1} logq_{k} − 2G(q_{k}, q_{k+1} ) − 2].

Then we complete the proof of relation (1.13)

V (q, p) =− \frac{1}{π} − pg( \frac{1}{p}) − \frac{1}{π} logpq + \frac{1}{π}(logq + (−1)^{n} logq_{n−1} )

− \frac{1}{π} ∑_{k=1}^{n−2} (−1)^{k}[q_{k} logq_{k+1} + q_{k+1} logq_{k} − 2G(q_{k}, q_{k+1} )].

We observe that for any integer 1 ⩽ k ⩽ n we have q_{k} = q ∏ _{j=1}^{k} [0, b_{j−1} , …, b_{n}], so that we can get (1.14) from (1.13).

5. Further identities on V (p, q) and computation

In this section we relate the sums V (p, q) to some interesting and well-known convergent series. The functions ψ and cotangent are related by the reflection formula [1, §6.3.7]

(5.1)  ψ(1 − z) − ψ(z) = π cot(πz).

168 GOUBI, BAYAD, AND HERNANE

Moreover, ψ can be written in terms of harmonic function H_{n}(z) = ∑ _{k=0}^{n} \frac{1}{(k+z)} and the n-th harmonic number H_{n} = ∑ _{k=1}^{n} \frac{1}{k} as follows.

Lemma 5.1. Let z ∈ ℂ such that Re(z) > 0. Then we have

ψ(z) = lim_{n→∞} (logn − H_{n}(z))

at z = n positive integer we have

(5.2)  ψ(n + 1) = −γ + H_{n}.

Proof. It is is easy to see that

ψ(z) = −γ − \frac{1}{z} + ∑_{k=1}^{∞} ( \frac{1}{k} − \frac{1}{k + z} ) = lim_{n→∞} ( logn − ∑_{k=0}^{n} \frac{1}{k + z} ) .

Then

ψ(n + 1) = lim_{m→∞} ( logm − ∑_{k=0}^{m} \frac{1}{k + n + 1} ) = lim_{m→∞} ( logm − ∑_{k=n}^{m+n} \frac{1}{k + 1} )

= lim_{m→∞} ( logm − ( ∑_{k=0}^{m+n} \frac{1}{k + 1} − ∑_{k=0}^{n−1} \frac{1}{k + 1} ) )

= lim_{m→∞} ( log(m + n) − ∑_{k=0}^{m+n} \frac{1}{k + 1} ) + H_{n} = −γ + H_{n}.  □

Proposition 5.1. For x = p̄/q with (p, q) = 1, we have

(5.3)  V (p, q) = \frac{1}{πq} ∑_{k⩾0} ∑_{r=1}^{q−1} \frac{r(1 − 2rx)}{(k + 1 − rx)(k + rx)} .

Proof. We have

V (p, q) = ∑_{r=1}^{q−1} \{ \frac{rp}{q}\} cot( \frac{πr}{q} ) = ∑_{r=1}^{q−1} \frac{r}{q} cot( \frac{πrp̄}{q}) .

By reflection formula (5.1) we have

cot( \frac{πrp̄}{q}) = \frac{1}{π}( ψ( \frac{q − rp̄}{q} ) − ψ( \frac{rp̄}{q}) ) .

From relations (1.7) and (2.5) we have

ψ( \frac{q − rp̄}{q} ) − ψ( \frac{rp̄}{q}) = q ∑_{k⩾0} \frac{q − 2rp̄}{(q(k + 1) − rp̄)(qk + rp̄)} .

Then

V (p, q) = \frac{1}{π} ∑_{r=1}^{q−1} ∑_{k⩾0} \frac{r(q − 2rp̄)}{(q(k + 1) − rp̄)(qk + rp̄)} ,

EXPLICIT AND ASYMPTOTIC FORMULAE FOR VASYUNIN–COTANGENT SUMS 169

and we obtain

V (p, q) = \frac{1}{πq} ∑_{r=1}^{q−1} ∑_{k⩾0} \frac{r(1 − 2rx)}{(k + 1 − rx)(k + rx)} .  □

As an immediate consequence we derive Corollary 5.1. Let q be a positive integer; we have

(5.4)  V (1, q)= −\frac{q(ψ(q) + γ − 2) + 2}{π} + \frac{1}{π} ∑_{k⩾1} ∑_{r=1}^{q−1} \frac{r(q − 2r)}{(q(k + 1) − r)(qk + r)} ,
V (1, q)= −\frac{q(ψ(q) + γ − 2) + 2}{π} − \frac{1}{π} ∑_{k⩾1} ∑_{r=1}^{⌊q/2⌋} \frac{(q − 2r)^{2}}{(q(k + 1) − r)(qk + r)} .

Proof. From (5.3), special case p = 1, we can get

V (1, q) = \frac{1}{πq} ∑_{k⩾0} ∑_{r=1}^{q−1} \frac{r(1 − 2r/q)}{(k + 1 − r/q)(k + r/q)}

then we have

V (1, q) = \frac{1}{π} ∑_{k⩾0} ∑_{r=1}^{q−1} \frac{r(q − 2r)}{(q(k + 1) − r)(qk + r)}

and

V (1, q)= \frac{1}{π} ∑_{r=1}^{q−1} \frac{q − 2r}{q − r} − \frac{1}{π} ∑_{k⩾1} ∑_{r=1}^{q−1} \frac{r(q − 2r)}{(q(k + 1) − r)(qk + r)}

= \frac{1}{π}(q(2 − H_{q−1} ) − 2) + \frac{1}{π} ∑_{k⩾1} ∑_{r=1}^{q−1} \frac{r(q − 2r)}{(q(k + 1) − r)(qk + r)} .

From (5.2) we obtain (5.4). To end the proof we use that for t + r = q, we have

\frac{1}{(q(k + 1) − r)(qk + r)} = \frac{1}{(q(k + 1) − t)(qk + t)} ,

r(q − 2r) + t(q − 2t) = (q − 2r)^{2}.  □

Proposition 5.2. The sum V (1, q) has the following integral representation.

(5.5)  V (1, q) = −\frac{1}{π} ∫_{0}^{1} \frac{(q − 2)x^{q} − qx^{q−1} + qx − q + 2}{(x − 1)^{2}(x^{q} − 1)} dx.

Remark 5.1. Note that from (5.5) and (2.6) we deduce that

g( \frac{1}{q}) = \frac{1}{πq} ∫_{0}^{1} ( \frac{(q − 2)x^{q} − qx^{q−1} + qx − q + 2}{(x − 1)^{2}(x^{q} − 1)} − 1) dx.

170 GOUBI, BAYAD, AND HERNANE

Proof. From the relation (5.4) we have

V (1, q) = \frac{1}{π} ∑_{r=1}^{q−1} r ∑_{k⩾0} \frac{q − 2r}{(q(k + 1) − r)(qk + r)} ,

\frac{q − 2r}{(q(k + 1) − r)(qk + r)} = \frac{1}{qk + r} − \frac{1}{qk + q − r} = ∫_{0}^{1}(x^{qk+r−1} − x^{qk+q−r−1} )dx.

Then

∑_{k⩾0} \frac{q − 2r}{(q(k + 1) − r)(qk + r)} = ∫_{0}^{1} \frac{x^{r−1} − x^{q−r−1}}{1 − x^{q}} dx

which gives

V (1, q) = −\frac{1}{π} ∫_{0}^{1} \frac{∑ _{r=1}^{q−1} r(x^{r−1} − x^{q−r−1} )}{x^{q} − 1} dx.

On the other hand

∑_{r=1}^{q−1} rx^{r−1} = ( \frac{x^{q} − 1}{x − 1} )′ = \frac{(q − 1)x^{q} − qx^{q−1} + 1}{(x − 1)^{2}} .  □

Thanks to Proposition 5.2 we give a few values of V (1, p) Example 5.1. V (1, 2) = 0,

V (1, 3)= −\frac{1}{π} ∫_{0}^{1} \frac{dx}{x^{2} + x + 1} = −\frac{1}{3\sqrt{3}}, V (1, 4) = −\frac{2}{π} ∫_{0}^{1} \frac{dx}{x^{2} + 1} = −\frac{1}{2},
V (1, 6)= −\frac{1}{π} ∫_{0}^{1} ( \frac{3}{x^{2} − x + 1} + \frac{1}{x^{2} + x + 1} ) dx = −\frac{7}{3\sqrt{3}}.

6. Another estimation of V (p, q) + V (q, p)

In this section we establish another asymptotic formula for the sum V (p, q) + V (q, p). For this we prove the lemma

Lemma 6.1. Let p be a positive integer. We have

\frac{logp}{p} = ∑_{k⩾1} \frac{1}{k(k + 1)} \{ \frac{k}{p}\}, (6.1)

∑_{r=1}^{pq−1} \{ \frac{r}{p}\}\{ \frac{r}{q}\} = \frac{(p − 1)(q − 1)}{4} .

Proof. We have

∑_{k⩾1} \frac{1}{k(k + 1)} \{ \frac{k}{p}\} = ∑_{k⩾1} \{ \frac{k}{p}\} ∫_{0}^{1}(x^{k−1} − x^{k})dx

= ∑_{r=1}^{p−1} ∑_{i⩾0} \{ \frac{r}{p}\} ∫_{0}^{1}(x^{ip+r−1} − x^{ip+r} )dx

EXPLICIT AND ASYMPTOTIC FORMULAE FOR VASYUNIN–COTANGENT SUMS 171

= \frac{1}{p} ∫_{0}^{1} \frac{∑ _{r=1}^{p−1} rx^{r−1}}{1 + x + ⋯+ x^{p−1}} dx = \frac{logp}{p}.

Thus we obtain (6.1). Let us write

∑_{r=1}^{pq−1} \{ \frac{r}{p}\}\{ \frac{r}{q}\} = ∑_{i=0}^{p−1} ∑_{t=1}^{q−1} \{ \frac{iq + t}{p} \}\{ \frac{t}{q}\} = \frac{1}{pq} ∑_{t=1}^{q−1} t( ∑_{i=0}^{p−1} p\{ \frac{iq + t}{p} \})

and we take r = p\{\frac{iq+t}{p} \}, observe that r is integer and runs from 1 to p − 1. Then, we have

∑_{r=1}^{pq−1} \{ \frac{r}{p}\}\{ \frac{r}{q}\} = \frac{1}{pq} ∑_{t=0}^{q−1} ∑_{r=1}^{p−1} tr = \frac{(p − 1)(q − 1)}{4} .  □

Remark 6.1. Another proof of relation (6.1) is given in [10].

The geometry of ψ(\frac{r}{p}) is studied in [14, 24]. In the next corollary we write logp as a linear combination of ψ(\frac{r}{p}) for 1 ⩽ r ⩽ p − 1.

Corollary 6.1. Let p be a positive integer. Then we have

logp = \frac{1}{p} ∑_{r=1}^{p−1} r( ψ( \frac{r + 1}{p} ) − ψ( \frac{r}{p}) ) .

Proof. We start with the equalities

\frac{logp}{p} = ∑_{k⩾1} \frac{1}{k(k + 1)} \{ \frac{k}{p}\} = ∑_{r=1}^{p−1} ( ∑_{i⩾0} \frac{1}{(ip + r)(ip + r + 1)} ) \{ \frac{r}{p}\}

and from (2.4) we have

∑_{i=0}^{∞} \frac{1}{(pi + r)(pi + r + 1)} = \frac{ψ(r + 1)/p − ψ(r/p)}{p}

thus the result follows. □

Proposition 6.1. Let p, q be two coprime integers and put

Δ(p, q) = \frac{2}{π}( 1 + max\{ \frac{(p − 1)(q − 1)}{4} ψ( \frac{1}{pq}) , min\{q − γ − ψ(q), p − γ − ψ(p)\}\}) .

Then we have

(6.2)  Δ(p, q) < \frac{1}{π} logp^{q−1} q^{p−1} − pg( \frac{1}{p}) − qg( \frac{1}{q}) − V (p, q) − V (q, p)

⩽ \frac{2 + 2\sqrt{pq logp logq}}{π}.

Proof. From equality (6.1) we remark that

⟨v_{p}, v_{q}⟩ ⩽ ∑_{k⩾1} \frac{1}{k(k + 1)} \{ \frac{k}{p}\} = \frac{logp}{p} and ⟨v_{p}, v_{q}⟩ ⩽ ∑_{k⩾1} \frac{1}{k(k + 1)} \{ \frac{k}{q}\} = \frac{logq}{q}

172 GOUBI, BAYAD, AND HERNANE

we deduce that ⟨v_{p}, v_{q}⟩ ⩽ \sqrt{logp logq/pq}. For q < p we have

⟨v_{p}, v_{q}⟩ > \frac{1}{pq} ∑_{k=1}^{q−1} \frac{k}{k + 1} .

Similarly if p < q we have ⟨v_{p}, v_{q}⟩ > \frac{1}{pq} ∑ _{k=1}^{p−1} \frac{k}{k+1} . Therefore we have

⟨v_{p}, v_{q}⟩ > \frac{1}{pq}( q − ∑_{k=1}^{q−1} \frac{1}{k + 1} ) or ⟨v_{p}, v_{q}⟩ > \frac{1}{pq}( p − ∑_{k=1}^{p−1} \frac{1}{k + 1} ) .

From (5.2) we have

⟨v_{p}, v_{q}⟩ > \frac{1}{pq}(q − γ − ψ(q)) or ⟨v_{p}, v_{q}⟩ > \frac{1}{pq}(p − γ − ψ(p))

and then ⟨v_{p}, v_{q}⟩ > \frac{1}{pq} min\{q − γ − ψ(q), p − γ − ψ(p)\}. Finally we obtain

\frac{1}{πpq} min\{q − γ − ψ(q), p − γ − ψ(p)\} < \frac{1}{π}⟨v_{p}, v_{q}⟩ ⩽ \sqrt{logp logq/pq}.

In addition we have

pq⟨v_{p}, v_{q}⟩ = G(p, q) = ∑_{r=1}^{pq−1} ( ψ( \frac{r + 1}{pq} ) − ψ( \frac{r}{pq}) ) \{ \frac{r}{p}\}\{ \frac{r}{q}\}

and for x > 0 and 0 < y < 1, from Sulaiman [27, Theorem 2.2] we have the inequality ψ(x + y) − ψ(x) ⩾ ψ(y). Taking x = \frac{r}{pq} and y = \frac{1}{pq}, from the above inequalities we obtain

G(p, q) ⩾ ψ( \frac{1}{pq}) ∑_{r=1}^{pq−1} \{ \frac{r}{p}\}\{ \frac{r}{q}\} ⩾ \frac{(p − 1)(q − 1)}{4} ψ( \frac{1}{pq}) .

Furthermore

\frac{1}{pq}max\{ \frac{(p − 1)(q − 1)}{4} ψ( \frac{1}{pq}) , min\{q − γ − ψ(q), p − γ − ψ(p)\}\}

< ⟨v_{p}, v_{q}⟩ ⩽ \sqrt{logp logq/pq}.

Then we have

\frac{2}{π} max\{ \frac{(p − 1)(q − 1)}{4} ψ( \frac{1}{pq}) , min\{q − γ − ψ(q), p − γ − ψ(p)\}\}

< \frac{2}{π}G(p, q) ⩽ \frac{2}{π} \sqrt{pq logp logq}.

From (1.8) we obtain

\frac{2}{π}G(p, q) + \frac{2}{π} = \frac{1}{π} logp^{q−1} q^{p−1} − (V (p, q) + V (q, p)) − pg( \frac{1}{p}) − qg( \frac{1}{q}) .

This implies relation (6.2). □

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Laboratoire d’Algèbre et Théorie des Nombres (Received 29 12 2015) Department of Mathematics (Revised 06 02 2016) University of UMMTO Tizi-ouzou Algeria

mouloud.ummto@hotmail.fr

LAMME, Univ. Evry Université Paris-Saclay Evry France

abayad@maths.univ-evry.fr

Département d’Algèbre et Théorie des Nombres Faculté de Mathématiques Université des Sciences et de la technologie Houari-Boumediène (USTHB) Alger Algérie

mhernane@usthb.dz