Series of Error Terms for Rational Approximations of Irrational Numbers

Carsten Elsner Fachhochschule für die Wirtschaft Hannover Freundallee 15 D-30173 Hannover Germany

                Carsten.Elsner@fhdw.de

Abstract

Let p_{n}/q_{n} be the n-th convergent of a real irrational number α, and let ε_{n} = αq_{n}−p_{n}. In this paper we investigate various sums of the type ∑_{m} ε_{m}, ∑_{m} |ε_{m}|, and ∑_{m} ε_{m}x^{m}. The main subject of the paper is bounds for these sums. In particular, we investigate the behaviour of such sums when α is a quadratic surd. The most significant properties of the error sums depend essentially on Fibonacci numbers or on related numbers.

1 Statement of results for arbitrary irrationals

Given a real irrational number α and its regular continued fraction expansion

α = ⟨ a_{0}; a_{1}, a_{2}, … ⟩ (a_{0} ∈ ℤ , a_{ν} ∈ ℕ for ν ≥ 1) ,

the convergents p_{n}/q_{n} of α form a sequence of best approximating rationals in the following sense: for any rational p/q satisfying 1 ≤ q < q_{n} we have

| α − \frac{p_{n}}{q_{n}} | < | α − \frac{p}{q} | .

The convergents p_{n}/q_{n} of α are defined by finite continued fractions

\frac{p_{n}}{q_{n}} = ⟨ a_{0}; a_{1}, …, a_{n} ⟩ .

The integers p_{n} and q_{n} can be computed recursively using the initial values p_{−1} = 1, p_{0} = a_{0}, q_{−1} = 0, q_{0} = 1, and the recurrence formulae

p_{n} = a_{n}p_{n−1} + p_{n−2} ,  q_{n} = a_{n}q_{n−1} + q_{n−2}  (1)

with n ≥ 1. Then p_{n}/q_{n} is a rational number in lowest terms satisfying the inequalities

\frac{1}{q_{n} + q_{n+1}} < |q_{n}α − p_{n}| < \frac{1}{q_{n+1}}  (n ≥ 0) .  (2)

The error terms q_{n}α−p_{n} alternate, i.e., sgn (q_{n}α−p_{n}) = (−1)^{n}. For basic facts on continued fractions and convergents see [4, 5, 8]. Throughout this paper let

ρ = \frac{1 + \sqrt{5}}{2} and \overline{ρ} = −\frac{1}{ρ} = \frac{1 − \sqrt{5}}{2} .

The Fibonacci numbers F_{n} are defined recursively by F_{−1} = 1, F_{0} = 0, and F_{n} = F_{n−1} +F_{n−2} for n ≥ 1. In this paper we shall often apply Binet’s formula,

F_{n} = \frac{1}{\sqrt{5}} (ρ^{n} − ( − \frac{1}{ρ})^{n}) (n ≥ 0) .  (3)

While preparing a talk on the subject of so-called leaping convergents relying on the papers [2, 6, 7], the author applied results for convergents to the number α = e = exp(1). He found two identities which are based on formulas given by Cohn [1]:

∑_{n=0}^{∞}(q_{n}e − p_{n}) = 2 ∫_{0}^{1} exp(t^{2}) dt − 2e + 3 = 0.4887398 …,

∑_{n=0}^{∞} |q_{n}e − p_{n}| = 2e ∫_{0}^{1} exp(−t^{2}) dt − e = 1.3418751 ….

These identities are the starting points of more generalized questions concerning error series of real numbers α.

1.) What is the maximum size M of the series ∑_{m=0}^{∞} |q_{m}α − p_{m}| ? One easily concludes that M ≥ (1 + \sqrt{5})/2, because ∑_{m=0}^{∞} |q_{m}(1 + \sqrt{5})/2 − p_{m}| = (1 + \sqrt{5})/2.

2.) Is there a method to compute ∑_{m=0}^{∞} |q_{m}α − p_{m}| explicitly for arbitrary real quadratic irrationals ?

The series ∑_{m=0}^{∞} |q_{m}α−p_{m}| ∈ [0, M] measures the approximation properties of α on average. The smaller this series is, the better rational approximations α has. Nevertheless, α can be a Liouville number and ∑_{m=0}^{∞} |q_{m}α − p_{m}| takes a value close to M. For example, let us consider the numbers

α_{n} = ⟨ 1; \underbrace{1, …, 1}_{n}, a_{n+1}, a_{n+2}, … ⟩

for even positive integers n, where the elements a_{n+1}, a_{n+2}, … are defined recursively in the following way. Let p_{k}/q_{k} = ⟨1; \underbrace{1, …, 1}_{k}⟩ for k = 0, 1, …, n and set

\begin{aligned}a_{n+1} & := & q_{n}^{n} , & q_{n+1} & = & a_{n+1}q_{n} + q_{n−1} & = & q_{n}^{n}(q_{n} + q_{n−1}) , \\ a_{n+2} & := & q_{n+1}^{n+1} , & q_{n+2} & = & a_{n+2}q_{n+1} + q_{n} & = & q_{n+1}^{n+1}(q_{n}^{n+1} + q_{n−1}) , \\ a_{n+3} & := & q_{n+2}^{n+2} , & q_{n+3} & = & a_{n+3}q_{n+2} + q_{n+1} & = & … \end{aligned}

and so on. In the general case we define a_{k+1} by a_{k+1} = q_{k}^{k} for k = n, n + 1, . . .. Then we have with (1) and (2) that

0 < | α_{n} − \frac{p_{k}}{q_{k}} | < \frac{1}{q_{k}q_{k+1}} < \frac{1}{a_{k+1}q_{k}^{2}} = \frac{1}{q_{k}^{k+2}} (k ≥ n) .

Hence α_{n} is a Liouville number. Now it follows from (9) in Theorem 2 below with 2k = n and n_{0} = (n/2) − 1 that

∑_{m=0}^{∞} |q_{m}α_{n} − p_{m}| > ∑_{m=0}^{n−1} |q_{m}α_{n} − p_{m}| = (F_{n−1} − 1)(ρ − α_{n}) + ρ − ρ^{1−n} ≥ ρ − \frac{1}{ρ^{n−1}} .

We shall show by Theorem 2 that M = ρ, such that the error sums of the Liouville numbers α_{n} tend to this maximum value ρ for increasing n. We first treat infinite sums of the form ∑_{n} |q_{n}α − p_{n}| for arbitrary real irrational numbers α = ⟨1; a_{1}, a_{2}, …⟩, when we may assume without loss of generality that 1 < α < 2.

Proposition 1. Let α = ⟨1; a_{1}, a_{2}, …⟩ be a real irrational number. Then for every integer m ≥ 0, the following two inequalities hold: Firstly,

|q_{2m}α − p_{2m}| + |q_{2m+1}α − p_{2m+1}| < \frac{1}{ρ^{2m}} ,  (4)

provided that either

a_{2m}a_{2m+1} > 1  or  (a_{2m} = a_{2m+1} = 1 and a_{1}a_{2} ⋯ a_{2m−1} > 1) .  (5)

Secondly,

|q_{2m}α − p_{2m}| + |q_{2m+1}α − p_{2m+1}| = \frac{1}{ρ^{2m}} + F_{2m}(ρ − α)  (0 ≤ m ≤ k) ,  (6)

provided that

a_{1} = a_{2} = … = a_{2k+1} = 1 .  (7)

In the second term on the right-hand side of (6), ρ − α takes positive or negative values according to the parity of the smallest subscript r ≥ 1 with a_{r} > 1: For odd r we have ρ > α, otherwise, ρ < α. Next, we introduce a set M of irrational numbers, namely

M := { α ∈ ℝ \ ℚ | ∃ k ∈ ℕ : α = ⟨1; 1, …, 1, a_{2k+1}, a_{2k+2}, …⟩ ∧ a_{2k+1} > 1} .

Note that ρ > α for α ∈ M. Our main result for real irrational numbers is given by the subsequent theorem.

Theorem 2. Let 1 < α < 2 be a real irrational number and let g, n ≥ 0 be integers with n ≥ 2g. Set n_{0} := ⌊n/2⌋. Then the following inequalities hold. 1.) For α ∉ M we have

∑_{ν=2g}^{n} |q_{ν}α − p_{ν}| ≤ ρ^{1−2g} − ρ^{−2n_{0}−1} ,  (8)

with equality for α = ρ and every odd n ≥ 0. 2.) For α ∈ M, say α = ⟨1; 1, …, 1, a_{2k+1}, a_{2k+2}, …⟩ with a_{2k+1} > 1, we have

∑_{ν=2g}^{n} |q_{ν}α − p_{ν}| ≤ (F_{2k−1} − F_{2g−1})(ρ − α) + ρ^{1−2g} − ρ^{−2n_{0}−1} ,  (9)

with equality for n = 2k − 1. 3.) We have

∑_{ν=2g}^{∞} |q_{ν}α − p_{ν}| ≤ ρ^{1−2g} ,  (10)

with equality for α = ρ.

In particular, for any positive ε and any even integer n satisfying

n ≥ \frac{log(ρ/ε)}{log ρ} ,

it follows that

∑_{ν=n}^{∞} |q_{ν}α − p_{ν}| ≤ ε .

For ν ≥ 1 we know by q_{2} ≥ 2 and by (2) that |q_{ν}α − p_{ν}| < 1/q_{ν+1} ≤ 1/q_{2} ≤ 1/2, which implies |q_{ν}α −p_{ν}| = ∥q_{ν}α∥, where ∥β∥ denotes the distance of a real number β to the nearest integer. For α = ⟨a_{0}; a_{1}, a_{2}, …⟩, |q_{0}α − p_{0}| = α − a_{0} = {α} is the fractional part of α. Therefore, we conclude from Theorem 2 that

∑_{ν=1}^{∞} ∥q_{ν}α∥ ≤ ρ − {α} .

In particular, we have for α = ρ that

∑_{ν=1}^{∞} ∥q_{ν}ρ∥ = 1 .

The following theorem gives a simple bound for ∑_{m}(q_{m}α − p_{m}). Theorem 3. Let α be a real irrational number. Then the series ∑_{m=0}^{∞}(q_{m}α − p_{m})x^{m} converges absolutely at least for |x| < ρ, and

0 < ∑_{m=0}^{∞}(q_{m}α − p_{m}) < 1 .

Both the upper bound 1 and the lower bound 0 are best possible.

The proof of this theorem is given in Section 3. We shall prove Proposition 1 and Theorem 2 in Section 4, using essentially the properties of Fibonacci numbers.

2 Statement of Results for Quadratic Irrationals

In this section we state some results for error sums involving real quadratic irrational numbers α. Any quadratic irrational α has a periodic continued fraction expansion,

α = ⟨ a_{0}; a_{1}, …, a_{ω}, T_{1}, …, T_{r}, T_{1}, …, T_{r}, … ⟩ = ⟨ a_{0}; a_{1}, …, a_{ω}, \overline{T_{1}, …, T_{r}} ⟩ ,

say. Then there is a linear three-term recurrence formula for z_{n} = p_{rn+s} and z_{n} = q_{rn+s} (s = 0, 1, …, r − 1), [3, Corollary 1]. This recurrence formula has the form

z_{n+2} = Gz_{n+1} ± z_{n} (rn > ω) .

Here, G denotes a positive integer, which depends on α and r, but not on n and s. The number G can be computed explicitly from the numbers T_{1}, …, T_{r} of the continued fraction expansion of α. This is the basic idea on which the following theorem relies.

Theorem 4. Let α be a real quadratic irrational number. Then

∑_{m=0}^{∞}(q_{m}α − p_{m})x^{m} ∈ ℚ[α](x) .

It is not necessary to explain further technical details of the proof. Thus, the generating function of the sequence (q_{m}α − p_{m})_{m≥0} is a rational function with coefficients from ℚ[α].

Example 5. Let α = \sqrt{7} = ⟨2; \overline{1, 1, 1, 4}⟩. Then

∑_{m=0}^{∞}(q_{m}\sqrt{7} − p_{m})x^{m} = \frac{x^{3} − (2 + \sqrt{7})x^{2} + (3 + \sqrt{7})x − (5 + 2\sqrt{7})}{x^{4} − (8 + 3\sqrt{7})} .  (11)

In particular, for x = 1 and x = −1 we obtain

∑_{m=0}^{∞}(q_{m}\sqrt{7} − p_{m}) = \frac{21 − 5\sqrt{7}}{14} = 0.555088817 …,

∑_{m=0}^{∞} |q_{m}\sqrt{7} − p_{m}| = \frac{7 + 5\sqrt{7}}{14} = 1.444911182 ….

Next, we consider the particular quadratic surds

α = \frac{n + \sqrt{4 + n^{2}}}{2} = ⟨ n; n, n, n, … ⟩

and compute the generating function of the error terms q_{m}α − p_{m}.

Corollary 6. Let n ≥ 1 and α = (n + \sqrt{4 + n^{2}})/2. Then

∑_{m=0}^{∞}(q_{m}α − p_{m})x^{m} = \frac{1}{x + α} ,

particularly

∑_{m=0}^{∞}(q_{m}α − p_{m}) = \frac{1}{α + 1} , ∑_{m=0}^{∞} |q_{m}α − p_{m}| = \frac{1}{α − 1} , ∑_{m=0}^{∞} \frac{q_{m}α − p_{m}}{m + 1} = log (1 + \frac{1}{α}) .

For the number ρ = (1 + \sqrt{5})/2 we have p_{m} = F_{m+2} and q_{m} = F_{m+1}. Hence, using 1/(ρ + 1) = (3 − \sqrt{5})/2 = 1 + \overline{ρ}, 1/(ρ − 1) = ρ, and 1 + 1/ρ = ρ, we get from Corollary 6

∑_{m=0}^{∞}(F_{m+1}ρ − F_{m+2}) = 1 + \overline{ρ} , ∑_{m=0}^{∞} |F_{m+1}ρ − F_{m+2}| = ρ , ∑_{m=0}^{∞} \frac{F_{m+1}ρ − F_{m+2}}{m + 1} = log ρ .

(12)

Similarly, we obtain for the number α = \sqrt{7} from (11):

∑_{m=0}^{∞} \frac{q_{m}\sqrt{7} − p_{m}}{m + 1} = ∫_{0}^{1} \frac{x^{3} − (2 + \sqrt{7})x^{2} + (3 + \sqrt{7})x − (5 + 2\sqrt{7})}{x^{4} − (8 + 3\sqrt{7})} dx = 0.5568649708 …

3 Proof of Theorem 3

Throughout this paper we shall use the abbreviations ε_{m}(α) = ε_{m} := q_{m}α − p_{m} and ε(α) = ∑_{m=0}^{∞} |ε_{m}(α)|. The sequence (|ε_{m}|)_{m≥0} converges strictly decreasing to zero. Since ε_{0} > 0 and ε_{m}ε_{m+1} < 0, we have

ε_{0} + ε_{1} < ∑_{m=0}^{∞} ε_{m} < ε_{0} .

Put a_{0} = ⌊α⌋, θ := ε_{0} = α − a_{0}, so that 0 < θ < 1. Moreover,

ε_{0} + ε_{1} = θ + a_{1}α − (a_{0}a_{1} + 1) = θ + a_{1}θ − 1 = θ + ⌊ \frac{1}{θ} ⌋ θ − 1 .

Choosing an integer k ≥ 1 satisfying

\frac{1}{k + 1} < θ < \frac{1}{k} ,

we get

θ + ⌊ \frac{1}{θ} ⌋ θ − 1 > \frac{1}{k + 1} + \frac{k}{k + 1} − 1 = 0 ,

which proves the lower bound for ∑ ε_{m}. In order to estimate the radius of convergence for the series ∑ ε_{m}x^{m} we first prove the inequality

q_{m} ≥ F_{m+1} (m ≥ 0) ,  (13)

which follows inductively. We have q_{0} = 1 = F_{1}, q_{1} = a_{1} ≥ 1 = F_{2}, and

q_{m} = a_{m}q_{m−1} + q_{m−2} ≥ q_{m−1} + q_{m−2} ≥ F_{m} + F_{m−1} = F_{m+1} (m ≥ 2) ,

provided that (13) is already proven for q_{m−1} and q_{m−2}. With Binet’s formula (3) and (13) we conclude that

q_{m+1} ≥ \frac{1}{\sqrt{5}}(ρ^{m+2} − ( − \frac{1}{ρ})^{m+2}) ≥ \frac{1}{\sqrt{5}}ρ^{m} (m ≥ 0) .  (14)

Hence, we have

|ε_{m}|x^{m} = |q_{m}α − p_{m}|x^{m} < \frac{x^{m}}{q_{m+1}} ≤ \sqrt{5}(\frac{x}{ρ})^{m} (m ≥ 0) .

It follows that the series ∑ ε_{m}x^{m} converges absolutely at least for |x| < ρ. In order to prove that the upper bound 1 is best possible, we choose 0 < ε < 1 and a positive integer n satisfying

\frac{1}{n} (1 + \frac{ρ\sqrt{5}}{ρ − 1}) < ε .

Put

α_{n} := ⟨ 0; 1, \overline{n} ⟩ = \frac{1}{2} − \frac{1}{n} + \frac{1}{2}\sqrt{1 + \frac{4}{n^{2}}} > 1 − \frac{1}{n} .

With p_{0} = 0 and q_{0} = 1 we have by (1), (2), and (14),

∑_{m=0}^{∞}(q_{m}α_{n} − p_{m}) ≥ α_{n} − ∑_{m=1}^{∞} |q_{m}α_{n} − p_{m}|

> 1 − \frac{1}{n} − ∑_{m=1}^{∞} \frac{1}{q_{m+1}} ≥ 1 − \frac{1}{n} − ∑_{m=1}^{∞} \frac{1}{nq_{m}}

≥ 1 − \frac{1}{n} − \frac{\sqrt{5}}{n} ∑_{m=1}^{∞} \frac{1}{ρ^{m−1}}

= 1 − \frac{1}{n} (1 + \frac{ρ\sqrt{5}}{ρ − 1}) > 1 − ε .

For the lower bound 0 we construct quadratic irrational numbers β_{n} := ⟨ 0; \overline{n} ⟩ and complete the proof of the theorem by similar arguments.

4 Proofs of Proposition 1 and Theorem 2

Lemma 7. Let α = ⟨a_{0}; a_{1}, a_{2}, …⟩ be a real irrational number with convergents p_{m}/q_{m}. Let n ≥ 1 be a subscript satisfying a_{n} > 1. Then

q_{n+k} ≥ F_{n+k+1} + F_{k+1}F_{n}  (k ≥ 0) .  (15)

In the case n ≡ k + 1 ≡ 0 (mod 2) we additionally assume that n ≥ 4, k ≥ 3. Then

F_{n+k+1} + F_{k+1}F_{n} > ρ^{n+k} .  (16)

When α − ρ ∉ ℤ, the inequality (15) with m = n + k is stronger than (13).

Proof. We prove (15) by induction on k. Using (1) and (13), we obtain for k = 0 and k = 1, respectively,

q_{n} = a_{n}q_{n−1} + q_{n−2} ≥ 2F_{n} + F_{n−1} = (F_{n} + F_{n−1}) + F_{n} = F_{n+1} + F_{1}F_{n} ,
q_{n+1} = a_{n+1}q_{n} + q_{n−1} ≥ q_{n} + q_{n−1} ≥ (F_{n+1} + F_{n}) + F_{n} = F_{n+2} + F_{2}F_{n} .

Now, let k ≥ 0 and assume that (15) is already proven for q_{n+k} and q_{n+k+1}. Then

q_{n+k+2} ≥ q_{n+k+1} + q_{n+k}
≥ (F_{n+k+2} + F_{k+2}F_{n}) + (F_{n+k+1} + F_{k+1}F_{n})
= F_{n+k+3} + F_{k+3}F_{n} .

This corresponds to (15) with k replaced by k + 2. In order to prove (16) we express the Fibonacci numbers F_{m} by Binet’s formula (3). Hence, we have

F_{n+k+1} + F_{k+1}F_{n}
= ρ^{n+k} (ρ (\frac{1}{5} + \frac{1}{\sqrt{5}}) + (−1)^{n+k}(\frac{1}{\sqrt{5}} − \frac{1}{5}) \frac{1}{ρ^{2n+2k+1}} + \frac{1}{5} (\frac{(−1)^{n+1}}{ρ^{2n−1}} + \frac{(−1)^{k}}{ρ^{2k+1}} )) .

Case 1: Let n ≡ k ≡ 1 (mod 2). In particular, we have k ≥ 1. Then

F_{n+k+1} + F_{k+1}F_{n}
= ρ^{n+k} (ρ (\frac{1}{5} + \frac{1}{\sqrt{5}}) + (\frac{1}{\sqrt{5}} − \frac{1}{5}) \frac{1}{ρ^{2n+2k+1}} + \frac{1}{5} (\frac{1}{ρ^{2n−1}} − \frac{1}{ρ^{2k+1}} ))
> ρ^{n+k} (ρ (\frac{1}{5} + \frac{1}{\sqrt{5}}) − \frac{1}{5ρ^{3}} ) = ρ^{n+k} .

Case 2: Let n ≡ 1 (mod 2), k ≡ 0 (mod 2). In particular, we have n ≥ 1 and k ≥ 0. First, we assume that k ≥ 2. Then, by similar computations as in Case 1, we obtain

F_{n+k+1} + F_{k+1}F_{n}
= ρ^{n+k} (ρ (\frac{1}{5} + \frac{1}{\sqrt{5}}) − (\frac{1}{\sqrt{5}} − \frac{1}{5}) \frac{1}{ρ^{2n+2k+1}} + \frac{1}{5} (\frac{1}{ρ^{2n−1}} + \frac{1}{ρ^{2k+1}} ) ) > ρ^{n+k} .

For k = 0 and some odd n ≥ 1 we get

F_{n+k+1} + F_{k+1}F_{n} > ρ^{n} (ρ (\frac{1}{5} + \frac{1}{\sqrt{5}}) − (\frac{1}{\sqrt{5}} − \frac{1}{5}) \frac{1}{ρ^{3}} + \frac{1}{5ρ}) > ρ^{n} .

Case 3: Let n ≡ 0 (mod 2), k ≡ 1 (mod 2). By the assumption of the lemma, we have n ≥ 4 and k ≥ 3. Then

F_{n+k+1} + F_{k+1}F_{n}
= ρ^{n+k} (ρ (\frac{1}{5} + \frac{1}{\sqrt{5}}) − (\frac{1}{\sqrt{5}} − \frac{1}{5}) \frac{1}{ρ^{2n+2k+1}} + \frac{1}{5} ( − \frac{1}{ρ^{2n−1}} − \frac{1}{ρ^{2k+1}} )) > ρ^{n+k} .

Case 4: Let n ≡ k ≡ 0 (mod 2). In particular, we have n ≥ 2. Then

F_{n+k+1} + F_{k+1}F_{n}
= ρ^{n+k} (ρ (\frac{1}{5} + \frac{1}{\sqrt{5}}) + (\frac{1}{\sqrt{5}} − \frac{1}{5}) \frac{1}{ρ^{2n+2k+1}} + \frac{1}{5} ( − \frac{1}{ρ^{2n−1}} + \frac{1}{ρ^{2k+1}} )) > ρ^{n+k} .

This completes the proof of Lemma 7.

Lemma 8. Let m be an integer. Then

\frac{ρ^{2m}}{F_{2m+2}} < 1  (m ≥ 1) ,  (17)
ρ^{2m}(\frac{1}{F_{2m+3}} + \frac{1}{F_{2m+3} + F_{2m+1}} ) < 1  (m ≥ 0) .  (18)

Proof. For m ≥ 1 we estimate Binet’s formula (3) for F_{2m+2} using 4m + 2 ≥ 6:

F_{2m+2} = \frac{ρ^{2m}}{\sqrt{5}} (ρ^{2} − \frac{1}{ρ^{4m+2}} ) ≥ \frac{ρ^{2m}}{\sqrt{5}} (ρ^{2} − \frac{1}{ρ^{6}} ) > ρ^{2m} .

Similarly, we prove (18) by

F_{2n+1} = \frac{1}{\sqrt{5}} (ρ^{2n+1} + \frac{1}{ρ^{2n+1}} ) > \frac{ρ^{2n+1}}{\sqrt{5}} (n ≥ 0) .

Hence,

ρ^{2m}(\frac{1}{F_{2m+3}} + \frac{1}{F_{2m+3} + F_{2m+1}} ) < ρ^{2m} (\frac{\sqrt{5}}{ρ^{2m+3}} + \frac{\sqrt{5}}{ρ^{2m+3} + ρ^{2m+1}} ) < 1 .

The lemma is proven.

Proof of Proposition 1: Firstly, we assume the hypotheses in (5) and prove (4). As in the proof of Theorem 3, put a_{0} = ⌊α⌋, θ := α − a_{0}, a_{1} = ⌊1/θ⌋ with 0 < θ < 1 and ε_{0} = θ < 1. Then

|ε_{0}| + |ε_{1}| = θ + (a_{0}a_{1} + 1) − a_{1}α = θ + 1 − a_{1}θ = θ + 1 − ⌊ \frac{1}{θ} ⌋ θ .

We have 0 < θ < 1/2, since otherwise for θ > 1/2, we obtain a_{1} = ⌊1/θ⌋ = 1. With a_{0} = a_{1} = 1 the conditions in (5) are unrealizable both. Hence, there is an integer k ≥ 2 with

\frac{1}{k + 1} < θ < \frac{1}{k} .

Obviously, it follows that [1/θ] = k, and therefore

θ + 1 − ⌊ \frac{1}{θ} ⌋ θ < \frac{1}{k} + 1 − \frac{k}{k + 1} = \frac{2k + 1}{k(k + 1)} ≤ \frac{5}{6} (k ≥ 2) .

Altogether, we have proven that

|ε_{0}| + |ε_{1}| ≤ \frac{5}{6} < 1 .  (19)

Therefore we already know that the inequality (4) holds for m = 0. Thus, we assume m ≥ 1 in the sequel. Noting that ε_{2m} > 0 and ε_{2m+1} < 0 hold for every integer m ≥ 0, we may rewrite (4) as follows:

(0 < ) (p_{2m+1} − p_{2m}) − α(q_{2m+1} − q_{2m}) < \frac{1}{ρ^{2m}}  (m ≥ 0) .  (20)

We distinguish three cases according to the conditions in (5).

Case 1: Let a_{2m+1} ≥ 2. Additionally, we apply the trivial inequality a_{2m+2} ≥ 1. Then, using (2), (13), and (18),

|ε_{2m}| + |ε_{2m+1}| < \frac{1}{q_{2m+1}} + \frac{1}{q_{2m+2}}
≤ \frac{1}{2q_{2m} + q_{2m−1}} + \frac{1}{q_{2m+1} + q_{2m}}
≤ \frac{1}{2q_{2m} + q_{2m−1}} + \frac{1}{3q_{2m} + q_{2m−1}}
≤ \frac{1}{2F_{2m+1} + F_{2m}} + \frac{1}{3F_{2m+1} + F_{2m}}
< \frac{1}{ρ^{2m}}  (m ≥ 0) .

Case 2: Let a_{2m+1} = 1 and a_{2m} ≥ 2. Here, we have p_{2m+1} − p_{2m} = p_{2m} + p_{2m−1} − p_{2m} = p_{2m−1}, and similarly q_{2m+1} − q_{2m} = q_{2m−1}.

Therefore, by (20), it suffices to show that 0 < p_{2m−1} − αq_{2m−1} < ρ^{−2m} for m ≥ 1. This follows with (2), (13), and (17) from

0 < p_{2m−1} − αq_{2m−1} < \frac{1}{q_{2m}}
≤ \frac{1}{2q_{2m−1} + q_{2m−2}} ≤ \frac{1}{2F_{2m} + F_{2m−1}}
= \frac{1}{F_{2m+2}} < \frac{1}{ρ^{2m}}  (m ≥ 1) .

Case 3: Let a_{2m} = a_{2m+1} = 1 ∧ a_{1}a_{2} ⋯ a_{2m−1} > 1. Since a_{2m+1} = 1, we again have (as in Case 2):

0 < |ε_{2m}| + |ε_{2m+1}| = p_{2m−1} − αq_{2m−1} < \frac{1}{q_{2m}} .  (21)

By the hypothesis of Case 3, there is an integer n satisfying 1 ≤ n ≤ 2m − 1 and a_{n} ≥ 2. We define an integer k ≥ 1 by setting 2m = n + k. Then we obtain using (15) and (16),

q_{2m} = q_{n+k} ≥ F_{n+k+1} + F_{k+1}F_{n} > ρ^{n+k} = ρ^{2m} .

From the identity n + k = 2m it follows that the particular condition n ≡ k + 1 ≡ 0 (mod 2) in Lemma 7 does not occur. Thus, by (21), we conclude that the desired inequality (4). In order to prove (6), we now assume the hypothesis (7), i.e., a_{1}a_{2} ⋯ a_{2k+1} = 1 and 0 ≤ m ≤ k. From 2m − 1 ≤ 2k − 1 and a_{0} = a_{1} = … = a_{2k−1} = 1 it is clear that p_{2m−1} = F_{2m+1} and q_{2m−1} = F_{2m}. Since a_{2k+1} = 1 and 0 ≤ m ≤ k, we have

|q_{2m}α − p_{2m}| + |q_{2m+1}α − p_{2m+1}|
= p_{2m−1} − αq_{2m−1} = F_{2m+1} − αF_{2m} = F_{2m+1} − ρF_{2m} + (ρ − α)F_{2m} .

From Binet’s formula (3) we conclude that

F_{2m+1} − ρF_{2m} = \frac{1}{\sqrt{5}} (\frac{1}{ρ^{2m+1}} + \frac{1}{ρ^{2m−1}} ) = ρ^{−2m} ,

which finally proves the desired identity (6) in Proposition 1.

Lemma 9. Let k ≥ 1 be an integer, and let α := ⟨ 1; 1, …, 1, a_{2k+1}, a_{2k+2} … ⟩ be a real irrational number with partial quotients a_{2k+1} > 1 and a_{μ} ≥ 1 for μ ≥ 2k + 2. Then we have the inequalities

(F_{2k−1} − 1)(ρ − α) < \frac{1}{ρ^{2k}} − |ε_{2k}| − |ε_{2k+1}|  (22)

for a_{2k+1} ≥ 3, and

(F_{2k−1} − 1)(ρ − α) < \frac{1}{ρ^{2k}} + \frac{1}{ρ^{2k+2}} − |ε_{2k}| − |ε_{2k+1}| − |ε_{2k+2}| − |ε_{2k+3}|  (23)

for a_{2k+1} = 2.

One may conjecture that (22) also holds for a_{2k+1} = 2.

Example 10. Let α = ⟨1; 1, 1, 1, 1, 2, \overline{1}⟩ = (21ρ + 8)/(13ρ + 5) = (257 − \sqrt{5})/158. With k = 2 and a_{5} = 2, we have on the one side

ρ − α = F_{2}(ρ − α) = \frac{40\sqrt{5} − 89}{79} = 0.005604 …,

on the other side,

\frac{1}{ρ^{4}} − |ε_{4}| − |ε_{5}| = \frac{1}{ρ^{4}} − \frac{4\sqrt{5} − 1}{79} = 0.045337 …

Proof of Lemma 9:

Case 1: Let n := a_{2k+1} ≥ 3. Then there is a real number η satisfying 0 < η < 1 and

r_{2k+1} := ⟨a_{2k+1}; a_{2k+2}, …⟩ = n + η =: 1 + β .

It is clear that n − 1 < β < n. From the theory of regular continued fractions (see [5, formula (16)]) it follows that

α = ⟨ 1; 1, …, 1, a_{2k+1}, a_{2k+2} … ⟩ = \frac{F_{2k+2}r_{2k+1} + F_{2k+1}}{F_{2k+1}r_{2k+1} + F_{2k}}
= \frac{F_{2k+2}(1 + β) + F_{2k+1}}{F_{2k+1}(1 + β) + F_{2k}} = \frac{βF_{2k+2} + F_{2k+3}}{βF_{2k+1} + F_{2k+2}} .

Similarly, we have

ρ = \frac{F_{2k+2}ρ + F_{2k+1}}{F_{2k+1}ρ + F_{2k}} ,

hence, by some straightforward computations,

ρ − α = \frac{1 + β − ρ}{(ρF_{2k+1} + F_{2k})(βF_{2k+1} + F_{2k+2})} < \frac{n}{(ρF_{2k+1} + F_{2k})(βF_{2k+1} + F_{2k+2})} .  (24)

Here, we have applied the identities

F_{2k+2}^{2} − F_{2k+1}F_{2k+3} = −1 , F_{2k+1}^{2} − F_{2k}F_{2k+2} = 1 ,

and the inequality 1 + β − ρ < 1 + n − ρ < n. Since β > n − 1 and, by (2),

|ε_{2k}| < \frac{1}{q_{2k+1}} = \frac{1}{nF_{2k+1} + F_{2k}} ,
|ε_{2k+1}| < \frac{1}{q_{2k+2}} = \frac{1}{a_{2k+2}q_{2k+1} + F_{2k+1}} ≤ \frac{1}{(n + 1)F_{2k+1} + F_{2k}} .

(22) follows from

\frac{n(F_{2k−1} − 1)}{(ρF_{2k+1} + F_{2k})((n − 1)F_{2k+1} + F_{2k+2})} < \frac{1}{ρ^{2k}} − \frac{1}{nF_{2k+1} + F_{2k}} − \frac{1}{(n + 1)F_{2k+1} + F_{2k}} . (25)

In order to prove (25), we need three inequalities for Fibonacci numbers, which rely on Binet’s formula. Let δ := 1/ρ^{4}. Then, for all integers s ≥ 1, we have

\frac{ρ^{2s+1}}{\sqrt{5}} < F_{2s+1} < \frac{(1 + δ)ρ^{2s+1}}{\sqrt{5}}  and  \frac{(1 − δ)ρ^{2s}}{\sqrt{5}} ≤ F_{2s} .  (26)

We start to prove (25) by observing that

\sqrt{5} (\frac{1 + δ}{ρ^{2}(ρ^{2} + 1 − δ)} + \frac{1}{3ρ + 1 − δ} + \frac{1}{4ρ + 1 − δ} ) < 1 .

Here, the left-hand side can be diminished by noting that

\frac{1}{ρ} > \frac{n}{(n − 1)ρ + (1 − δ)ρ^{2}} .

By n ≥ 3 we get

\sqrt{5} (\frac{(1 + δ)n}{ρ(ρ^{2} + 1 − δ)((n − 1)ρ + (1 − δ)ρ^{2})} + \frac{1}{nρ + 1 − δ} + \frac{1}{(n + 1)ρ + 1 − δ} ) < 1 ,

or, equivalently,

\frac{(1 + δ)nρ^{2k−1}/\sqrt{5}}{(ρ · ρ^{2k+1}/\sqrt{5} + (1 − δ)ρ^{2k}/\sqrt{5})((n − 1)ρ^{2k+1}/\sqrt{5} + (1 − δ)ρ^{2k+2}/\sqrt{5})}

< \frac{1}{ρ^{2k}} − \frac{1}{nρ^{2k+1}/\sqrt{5} + (1 − δ)ρ^{2k}/\sqrt{5}} − \frac{1}{(n + 1)ρ^{2k+1}/\sqrt{5} + (1 − δ)ρ^{2k}/\sqrt{5}} .

From this inequality, (25) follows easily by applications of (26) with s ∈ {2k − 1, 2k, 2k + 1, 2k + 2}.

Case 2: Let a_{2k+1} = 2.

Case 2.1: Let k ≥ 2. We first consider the function

f(β) := \frac{1 − ρ + β}{βF_{2k+1} + F_{2k+2}} (1 ≤ β ≤ 2) .

The function f increases monotonically with β, therefore we have

f(β) ≤ f(2) = \frac{3 − ρ}{2F_{2k+1} + F_{2k+2}} ,

and consequently we conclude from the identity stated in (24) that

ρ − α ≤ \frac{3 − ρ}{(ρF_{2k+1} + F_{2k})(2F_{2k+1} + F_{2k+2})} .

Hence, (23) follows from the inequality

\frac{(3 − ρ)F_{2k−1}}{(ρF_{2k+1} + F_{2k})(2F_{2k+1} + F_{2k+2})} + \frac{1}{q_{2k+1}} + \frac{1}{q_{2k+2}} + \frac{1}{q_{2k+3}} + \frac{1}{q_{2k+4}} < \frac{1}{ρ^{2k}} + \frac{1}{ρ^{2k+2}} . (27)

On the left-hand side we now replace the q’s by certain smaller terms in Fibonacci numbers. For q_{2k+2}, q_{2k+3}, and q_{2k+4}, we find lower bounds by (15) in Lemma 7:

q_{2k+1} = a_{2k+1}q_{2k} + q_{2k−1} = 2F_{2k+1} + F_{2k} ,
q_{2k+2} ≥ F_{2k+3} + F_{2}F_{2k+1} = F_{2k+3} + F_{2k+1} ,
q_{2k+3} ≥ F_{2k+4} + F_{3}F_{2k+1} = F_{2k+4} + 2F_{2k+1} ,
q_{2k+4} ≥ F_{2k+5} + F_{4}F_{2k+1} = F_{2k+5} + 3F_{2k+1} .

Substituting these expressions into (27), we then conclude that (23) from

\frac{(3 − ρ)F_{2k−1}}{(ρF_{2k+1} + F_{2k})(2F_{2k+1} + F_{2k+2})} + \frac{1}{2F_{2k+1} + F_{2k}} + \frac{1}{F_{2k+3} + F_{2k+1}}

+ \frac{1}{F_{2k+4} + 2F_{2k+1}} + \frac{1}{F_{2k+5} + 3F_{2k+1}} < \frac{1}{ρ^{2k}} (1 + \frac{1}{ρ^{2}} ) .  (28)

We apply the inequalities in (26) for all s ≥ 2 when δ is replaced by δ := 1/ρ^{8}. Using this redefined number δ, we have

\sqrt{5} (\frac{(3 − ρ)(1 + δ)}{ρ(ρ^{2} + 1 − δ)(2ρ + (1 − δ)ρ^{2})} + \frac{1}{2ρ + 1 − δ} + \frac{1}{ρ^{3} + ρ} + \frac{1}{(1 − δ)ρ^{4} + 2ρ} + \frac{1}{ρ^{5} + 3ρ}) − \frac{1}{ρ^{2}}

< 1 ,

or, equivalently,

\frac{(3 − ρ)(1 + δ)ρ^{2k−1}/\sqrt{5}}{(ρ · ρ^{2k+1}/\sqrt{5} + (1 − δ)ρ^{2k}/\sqrt{5})(2ρ^{2k+1}/\sqrt{5} + (1 − δ)ρ^{2k+2}/\sqrt{5})}

+ \frac{1}{2ρ^{2k+1}/\sqrt{5} + (1 − δ)ρ^{2k}/\sqrt{5}} + \frac{1}{ρ^{2k+3}/\sqrt{5} + ρ^{2k+1}/\sqrt{5}}
+ \frac{1}{(1 − δ)ρ^{2k+4}/\sqrt{5} + 2ρ^{2k+1}/\sqrt{5}} + \frac{1}{ρ^{2k+5}/\sqrt{5} + 3ρ^{2k+1}/\sqrt{5}}
< \frac{1}{ρ^{2k}} (1 + \frac{1}{ρ^{2}} ) .

From this inequality, (28) follows by applications of (26) with s ∈ {2k − 1, 2k, 2k + 1, 2k + 2, 2k + 3, 2k + 4, 2k + 5} for k ≥ 2 (which implies s ≥ 3).

Case 2.2: Let k = 1. From the hypotheses we have a_{2k+1} = a_{3} = 2. To prove (23) it suffices to check the inequality in (28) for k = 1. We have

F_{2k−1} = F_{1} = 1 ,  F_{2k} = F_{2} = 1 , F_{2k+1} = F_{3} = 2 ,
F_{2k+2} = F_{4} = 3 , F_{2k+3} = F_{5} = 5 ,
F_{2k+4} = F_{6} = 8 , F_{2k+5} = F_{7} = 13 .

Then (28) is satisfied because

ρ^{2} (\frac{3 − ρ}{7(1 + 2ρ)} + \frac{1}{5} + \frac{1}{7} + \frac{1}{12} + \frac{1}{19}) − \frac{1}{ρ^{2}} < 1 .

This completes the proof of Lemma 9.

Proof of Theorem 2: In the sequel we shall use the identity

F_{2g} + F_{2g+2} + F_{2g+4} + … + F_{2n} = F_{2n+1} − F_{2g−1} (n ≥ g ≥ 0) ,  (29)

which can be proven by induction by applying the recurrence formula of Fibonacci numbers. Note that F_{−1} = 1. Next, we prove (8).

Case 1: Let α ∉ M, α = ⟨1; a_{1}, a_{2}, …⟩ = ⟨1; 1, …, 1, a_{2k}, a_{2k+1}, …⟩ with a_{2k} > 1 for some subscript k ≥ 1. This implies α > ρ.

Case 1.1: Let 0 ≤ n < 2k. Then n_{0} = ⌊n/2⌋ ≤ k − 1. In order to treat |ε_{2m}| + |ε_{2m+1}|, we apply (6) with k replaced by k − 1 in Proposition 1. For α the condition (7) with k replaced by k − 1 is fulfilled. Note that the term F_{2m}(ρ − α) in (6) is negative. Therefore, we have

S(n) := ∑_{ν=2g}^{n} |ε_{ν}| ≤ ∑_{m=g}^{⌊n/2⌋} (|ε_{2m}| + |ε_{2m+1}|)

< ∑_{m=g}^{n_{0}} \frac{1}{ρ^{2m}} = \frac{ρ^{2−2g} − ρ^{−2n_{0}}}{ρ^{2} − 1} = ρ^{1−2g} − ρ^{−2n_{0}−1} .

Case 1.2: Let n ≥ 2k.

Case 1.2.1: Let k ≥ g. Here, we get

S(n) ≤ ∑_{m=g}^{k−1} (|ε_{2m}| + |ε_{2m+1}|) + (|ε_{2k}| + |ε_{2k+1}|) + ∑_{m=k+1}^{n_{0}} (|ε_{2m}| + |ε_{2m+1}|) . (30)

When n_{0} ≤ k, the right-hand sum is empty and becomes zero. The same holds for the left-hand sum for k = g.

a) We estimate the left-hand sum as in the preceding case applying (6), ρ − α < 0, and the hypothesis a_{1}a_{2} ⋯ a_{2k−1} = 1:

∑_{m=g}^{k−1} (|ε_{2m}| + |ε_{2m+1}|) < ∑_{m=g}^{k−1} \frac{1}{ρ^{2m}} .

b) Since a_{2k} > 1, the left-hand condition in (5) allows us to apply (4) for m = k:

|ε_{2k}| + |ε_{2k+1}| < \frac{1}{ρ^{2k}} .

c) We estimate the right-hand sum in (30) again by (4). To check the conditions in (5), we use a_{1}a_{2} ⋯ a_{2m−1} > 1, which holds by m ≥ k + 1 and a_{2k} > 1. Hence,

∑_{m=k+1}^{n_{0}} (|ε_{2m}| + |ε_{2m+1}|) < ∑_{m=k+1}^{n_{0}} \frac{1}{ρ^{2m}} .

Altogether, we find with (30) that

S(n) < ∑_{m=g}^{n_{0}} \frac{1}{ρ^{2m}} = ρ^{1−2g} − ρ^{−2n_{0}−1} .  (31)

Case 1.2.2: Let k < g. In order to estimate |ε_{2m}| + |ε_{2m+1}| for g ≤ m ≤ n_{0}, we use k + 1 ≤ g and the arguments from c) in Case 1.2.1. Again, we obtain the inequality (31). The results from Case 1.1 and Case 1.2 prove (8) for a_{2k} > 1 with k ≥ 1. It remains to investigate the following case.

Case 2: Let α ∉ M, α = ⟨1; a_{1}, a_{2}, …⟩ with a_{1} > 1. For m = 0 (provided that g = 0) the first condition in (5) is fulfilled by a_{2m}a_{2m+1} = a_{0}a_{1} = a_{1} > 1. For m ≥ 1 we know that a_{1}a_{2} ⋯ a_{2m−1} > 1 always satisfies one part of the second condition. Therefore, we apply the inequality from (4):

S(n) < ∑_{m=g}^{n_{0}} \frac{1}{ρ^{2m}} = ρ^{1−2g} − ρ^{−2n_{0}−1} .

Next, we prove (9). Let α ∈ M, α = ⟨1; a_{1}, a_{2}, …⟩ = ⟨1; 1, …, 1, a_{2k+1}, a_{2k+2}, …⟩ with a_{2k+1} > 1 for some subscript k ≥ 1. This implies ρ > α.

Case 3.1: Let 0 ≤ n < 2k. Then n_{0} = ⌊n/2⌋ ≤ k − 1. In order to treat |ε_{2m}| + |ε_{2m+1}|, we apply (6) with k replaced by k − 1 in Proposition 1. For α the condition (7) with k replaced by k − 1 is fulfilled. Note that the term F_{2m}(ρ − α) in (6) is positive. Therefore we have, using (29),

S(n) ≤ ∑_{m=g}^{n_{0}} (\frac{1}{ρ^{2m}} + (ρ − α)F_{2m})

= ρ^{1−2g} − ρ^{−2n_{0}−1} + (ρ − α) ∑_{m=g}^{n_{0}} F_{2m}
= ρ^{1−2g} − ρ^{−2n_{0}−1} + (ρ − α)(F_{2n_{0}+1} − F_{2g−1})
≤ ρ^{1−2g} − ρ^{−2n_{0}−1} + (F_{2k−1} − F_{2g−1})(ρ − α) .

Here we have used that 2n_{0} + 1 ≤ 2k − 1.

Case 3.2: Let n ≥ 2k. Our arguments are similar to the proof given in Case 1.2, using a_{1}a_{2} ⋯ a_{2k−1} = 1 and a_{2k+1} > 1.

Case 3.2.1: Let k ≥ g. Applying (29) again, we obtain

S(n) ≤ ∑_{m=g}^{k−1} (|ε_{2m}| + |ε_{2m+1}|) + (|ε_{2k}| + |ε_{2k+1}|) + ∑_{m=k+1}^{n_{0}} (|ε_{2m}| + |ε_{2m+1}|)

< ∑_{m=g}^{k−1} (\frac{1}{ρ^{2m}} + (ρ − α)F_{2m}) + \frac{1}{ρ^{2k}} + ∑_{m=k+1}^{n_{0}} \frac{1}{ρ^{2m}}

= ∑_{m=g}^{n_{0}} \frac{1}{ρ^{2m}} + (ρ − α) ∑_{m=g}^{k−1} F_{2m}
= ρ^{1−2g} − ρ^{−2n_{0}−1} + (F_{2k−1} − F_{2g−1})(ρ − α) .

Case 3.2.2: Let k < g. From g ≥ k + 1 we get

S(n) ≤ ∑_{m=g}^{n_{0}} \frac{1}{ρ^{2m}} = ρ^{1−2g} − ρ^{−2n_{0}−1} .

The results of Case 3.1 and Case 3.2 complete the proof of (9). For the inequality (10) we distinguish whether α belongs to M or not.

Case 4.1: Let α ∉ M. Then (10) is a consequence of the inequality in (8):

∑_{ν=2g}^{∞} |ε_{ν}| ≤ lim_{n_{0}→∞} (ρ^{1−2g} − ρ^{−2n_{0}−1}) = ρ^{1−2g} .

Case 4.2: Let α ∈ M. There is a subscript k ≥ 1 satisfying α = ⟨1; 1, …, 1, a_{2k+1}, a_{2k+2}, …⟩ and a_{2k+1} > 1. To simplify arguments, we introduce the function χ(k, g) defined by χ(k, g) = 1 (if k > g), and χ(k, g) = 0 (if k ≤ g). We have

S := ∑_{ν=2g}^{∞} |ε_{ν}| = ∑_{ν=2g}^{2k−1} |ε_{ν}| + ∑_{ν=max{2k,2g}}^{∞} |ε_{ν}|

= χ(k, g) ((F_{2k−1} − F_{2g−1})(ρ − α) + ρ^{1−2g} − ρ^{−2k+1}) + ∑_{m=max{k,g}}^{∞} (|ε_{2m}| + |ε_{2m+1}|)

≤ (F_{2k−1} − F_{2g−1})(ρ − α) + ρ^{1−2g} − ρ^{−2k+1} + ∑_{m=k}^{∞} (|ε_{2m}| + |ε_{2m+1}|)

≤ (F_{2k−1} − 1)(ρ − α) + ρ^{1−2g} − ρ^{−2k+1} + ∑_{m=k}^{∞} (|ε_{2m}| + |ε_{2m+1}|) ,  (32)

where we have used (9) with n = 2k − 1 and n_{0} = ⌊n/2⌋ = k − 1.

Case 4.2.1: Let a_{2k+1} ≥ 3. The conditions in Lemma 9 for (22) are satisfied. Moreover, the terms |ε_{2m}| + |ε_{2m+1}| of the series in (32) for m ≥ k + 1 can be estimated using (4), since a_{1}a_{2} ⋯ a_{2k+1} > 1. Therefore, we obtain

S < \frac{1}{ρ^{2k}} − \frac{1}{ρ^{2k−1}} + ρ^{1−2g} + ∑_{m=k+1}^{∞} (|ε_{2m}| + |ε_{2m+1}|)
< \frac{1}{ρ^{2k}} − \frac{1}{ρ^{2k−1}} + ρ^{1−2g} + \frac{1}{ρ^{2k+1}} = ρ^{1−2g} .

Case 4.2.2: Let a_{2k+1} = 2. Now the conditions in Lemma 9 for (23) are satisfied. Thus, from (32) and (4) we have

S < \frac{1}{ρ^{2k}} + \frac{1}{ρ^{2k+2}} − \frac{1}{ρ^{2k−1}} + ρ^{1−2g} + ∑_{m=k+2}^{∞} (|ε_{2m}| + |ε_{2m+1}|)

< \frac{1}{ρ^{2k}} + \frac{1}{ρ^{2k+2}} − \frac{1}{ρ^{2k−1}} + ρ^{1−2g} + ∑_{m=k+2}^{∞} \frac{1}{ρ^{2m}}
= \frac{1}{ρ^{2k}} + \frac{1}{ρ^{2k+2}} − \frac{1}{ρ^{2k−1}} + ρ^{1−2g} + \frac{1}{ρ^{2k+3}} = ρ^{1−2g} .

This completes the proof of Theorem 2.

5 Concluding remarks

In this section we state some additional identities for error sums ε(α). For this purpose let α = ⟨a_{0}; a_{1}, a_{2}, … ⟩ be the continued fraction expansion of a real irrational number. Then the numbers α_{n} are defined by

α = ⟨a_{0}; a_{1}, a_{2}, …, a_{n−1}, α_{n}⟩ (n = 0, 1, 2, …) .

Proposition 11. For every real irrational number α we have

ε(α) = ∑_{n=1}^{∞} ∏_{k=1}^{n} \frac{1}{α_{k}}

and

( \begin{matrix}ε(α) \\ · \end{matrix} ) = ∑_{n=0}^{∞} (−1)^{n} ( \begin{matrix}a_{n} & 1 \\ 1 & 0 \end{matrix} ) ( \begin{matrix}a_{n−1} & 1 \\ 1 & 0 \end{matrix} ) ⋯ ( \begin{matrix}a_{0} & 1 \\ 1 & 0 \end{matrix} ) ( \begin{matrix}−1 \\ α \end{matrix} ) .

Next, let α = ⟨a_{0}; a_{1}, a_{2}, … ⟩ with a_{0} ≥ 1 be a real number with convergents p_{m}/q_{m} (m ≥ 0), where p_{−1} = 1, q_{−1} = 0. Then the convergents \overline{p}_{m}/\overline{q}_{m} of the number 1/α = ⟨0; a_{0}, a_{1}, a_{2}, … ⟩ satisfy the equations \overline{q}_{m} = p_{m−1} and \overline{p}_{m} = q_{m−1} for m ≥ 0, since we know that \overline{p}_{−1} = 1, \overline{p}_{0} = 0 and \overline{q}_{−1} = 0, \overline{q}_{0} = 1. Therefore we obtain a relation between ε(α) and ε(1/α):

ε(1/α) = ∑_{m=0}^{∞} |\frac{\overline{q}_{m}}{α} − \overline{p}_{m}| = ∑_{m=0}^{∞} |\frac{p_{m−1}}{α} − q_{m−1}| = \frac{1}{α} ∑_{m=0}^{∞} |q_{m−1}α − p_{m−1}|

= \frac{1}{α}(|q_{−1}α − p_{−1}| + ∑_{m=0}^{∞} |q_{m}α − p_{m}|) = \frac{1}{α}(1 + ε(α)) .

This proves

Proposition 12. For every real number α > 1 we have

ε(1/α) = \frac{1 + ε(α)}{α} .

6 Acknowledgment

The author would like to thank Professor Iekata Shiokawa for helpful comments and useful hints in organizing the paper. I am very much obliged to the anonymous referee, who suggested the results stated in Section 5. Moreover, the presentation of the paper was improved by following the remarks of the referee.

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[8] O. Perron, Die Lehre von den Kettenbrüchen, Chelsea Publishing Company, New York, 1929.

2010 Mathematics Subject Classification: Primary 11J04; Secondary 11J70, 11B39. Keywords: continued fractions, convergents, approximation of real numbers, error terms.

(Concerned with sequences \underline{A000045}, \underline{A001519}, \underline{A007676}, \underline{A007677}, \underline{A041008}, and \underline{A041009}.)

Received July 12 2010; revised version received January 10 2011. Published in Journal of Integer Sequences, January 28 2011.

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