Series of Error Terms for Rational Approximations of Irrational Numbers

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1 Journal of Integer Sequences, Vol. 4 20, Article..4 Series of Error Terms for Rational Approximations of Irrational Numbers Carsten Elsner Fachhochschule für die Wirtschaft Hannover Freundallee D-3073 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 mx 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. Statement of results for arbitrary irrationals Given a real irrational number α and its regular continued fraction expansion α = a 0 ;a,a 2,... a 0 Z, a ν N for ν, 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 q < q n we have α p n < α p q. The convergents p n /q n of α are defined by finite continued fractions q n p n q n = a 0 ;a,...,a n.

2 The integers p n and q n can be computed recursively using the initial values p =, p 0 = a 0, q = 0, q 0 =, and the recurrence formulae p n = a n p n + p n 2, q n = a n q n + q n 2 with n. Then p n /q n is a rational number in lowest terms satisfying the inequalities q n + q n+ < q n α p n < q n+ n 0. 2 The error terms q n α p n alternate, i.e., sgnq n α p n = n. For basic facts on continued fractions and convergents see [4,, 8]. Throughout this paper let ρ = + 2 and ρ = ρ = 2. The Fibonacci numbers F n are defined recursively by F =, F 0 = 0, and F n = F n +F n 2 for n. In this paper we shall often apply Binet s formula, F n = ρ n n ρ 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. He found two identities which are based on formulas given by Cohn []: q n e p n = 2 n=0 q n e p n = 2e n=0 0 expt 2 dt 2e + 3 = , 0 exp t 2 dt e = These identities are the starting points of more generalized questions concerning error series of real numbers α.. What is the maximum size M of the series q mα p m? One easily concludes that M + /2, because qm + /2 p m = + /2. 2. Is there a method to compute q mα p m explicitly for arbitrary real quadratic irrationals? The series q mα 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 q mα p m takes a value close to M. For example, let us consider the numbers α n = ;,...,,a }{{} n+,a n+2,... n 2

3 for even positive integers n, where the elements a n+,a n+2,... are defined recursively in the following way. Let p k /q k = ;,..., for k = 0,,...,n and set }{{} k a n+ := qn n, q n+ = a n+ q n + q n = qn n qn + q n, a n+2 := qn+, q n+2 = a n+2 q n+ + q n = qn+ q n+ n + q n, a n+3 := qn+2, q n+3 = a n+3 q n+2 + q n+ =... and so on. In the general case we define a k+ by a k+ = qk k for k = n,n +,... Then we have with and 2 that 0 < α n p k < < = k n. q k q k q k+ a k+ qk 2 q k+2 k Hence α n is a Liouville number. Now it follows from 9 in Theorem 2 below with 2k = n and n 0 = n/2 that q m α n p m > n q m α n p m = F n ρ α n + ρ ρ n ρ ρ n. 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 nα p n for arbitrary real irrational numbers α = ;a,a 2,..., when we may assume without loss of generality that < α < 2. Proposition. Let α = ;a,a 2,... be a real irrational number. Then for every integer m 0, the following two inequalities hold: Firstly, provided that either Secondly, provided that a 2m a 2m+ > q 2m α p 2m + q 2m+ α p 2m+ <, 4 ρ2m or a2m = a 2m+ = and a a 2 a 2m >. q 2m α p 2m + q 2m+ α p 2m+ = ρ 2m + F 2mρ α 0 m k, 6 a = a 2 =... = a 2k+ =. 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 with a r > : 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 = ;,...,,a 2k+,a 2k+2,... a 2k+ > }. Note that ρ > α for α M. Our main result for real irrational numbers is given by the subsequent theorem. 3

4 Theorem 2. Let < 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.. For α M we have n q ν α p ν ρ 2g ρ 2n0, 8 ν=2g with equality for α = ρ and every odd n For α M, say α = ;,...,,a 2k+,a 2k+2,... with a 2k+ >, we have n q ν α p ν F 2k F 2g ρ α + ρ 2g ρ 2n0, 9 ν=2g with equality for n = 2k. 3. We have with equality for α = ρ. q ν α p ν ρ 2g, 0 ν=2g In particular, for any positive ε and any even integer n satisfying it follows that n logρ/ε log ρ q ν α p ν ε. ν=n For ν we know by q 2 2 and by 2 that q ν α p ν < /q ν+ /q 2 /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,a 2,..., q 0 α p 0 = α a 0 = {α} is the fractional part of α. Therefore, we conclude from Theorem 2 that q ν α ρ {α}. ν= In particular, we have for α = ρ that q ν ρ =. The following theorem gives a simple bound for m q mα p m. ν= Theorem 3. Let α be a real irrational number. Then the series q mα p m x m converges absolutely at least for x < ρ, and 0 < q m α p m <. Both the upper bound and the lower bound 0 are best possible. The proof of this theorem is given in Section 3. We shall prove Proposition and Theorem 2 in Section 4, using essentially the properties of Fibonacci numbers. 4,

5 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,...,a ω,t,...,t r,t,...,t r,... = a 0 ;a,...,a ω,t,...,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,,..., r, [3, Corollary ]. This recurrence formula has the form z n+2 = Gz n+ ± 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,...,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 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. Let α = 7 = 2;,,, 4. Then q m 7 pm x m = x x x x In particular, for x = and x = we obtain q m 7 pm = q m 7 pm = Next, we consider the particular quadratic surds α = n n 2 2 = , = = n;n,n,n,... and compute the generating function of the error terms q m α p m.

6 Corollary 6. Let n and α = n n 2 /2. Then particularly q m α p m = α +, q m α p m x m = x + α, q m α p m = α, q m α p m m + = log +. α For the number ρ = + /2 we have p m = F m+2 and q m = F m+. Hence, using /ρ + = 3 /2 = + ρ, /ρ = ρ, and + /ρ = ρ, we get from Corollary 6 F m+ ρ F m+2 = + ρ, F m+ ρ F m+2 = ρ, Similarly, we obtain for the number α = 7 from : q m 7 pm m + = 3 Proof of Theorem 3 0 F m+ ρ F m+2 m + = log ρ. x x x x dx = Throughout this paper we shall use the abbreviations ε m α = ε m := q m α p m and εα = ε mα. The sequence ε m m 0 converges strictly decreasing to zero. Since ε 0 > 0 and ε m ε m+ < 0, we have ε 0 + ε < ε m < ε 0. Put a 0 = α, θ := ε 0 = α a 0, so that 0 < θ <. Moreover, ε 0 + ε = θ + a α a 0 a + = θ + a θ = θ + θ Choosing an integer k satisfying k + < θ < k, θ. we get θ + θ > θ k + + k k + = 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+ m 0, 3 6 2

7 which follows inductively. We have q 0 = = F, q = a = F 2, and q m = a m q m + q m 2 q m + q m 2 F m + F m = F m+ m 2, provided that 3 is already proven for q m and q m 2. With Binet s formula 3 and 3 we conclude that q m+ ρ m+2 ρ m+2 ρ m m 0. 4 Hence, we have ε m x m = q m α p m x m < xm m x m 0. q m+ ρ It follows that the series ε m x m converges absolutely at least for x < ρ. In order to prove that the upper bound is best possible, we choose 0 < ε < and a positive integer n satisfying + ρ < ε. n ρ Put α n := 0;,n = 2 n + 2 With p 0 = 0 and q 0 = we have by, 2, and 4, q m α n p m α n q m α n p m m= + 4 n 2 > n. > n q m= m+ n nq m= m n n ρ m m= = + ρ > ε. n ρ For the lower bound 0 we construct quadratic irrational numbers β n := 0;n and complete the proof of the theorem by similar arguments. 4 Proofs of Proposition and Theorem 2 Lemma 7. Let α = a 0 ;a,a 2,... be a real irrational number with convergents p m /q m. Let n be a subscript satisfying a n >. Then q n+k F n+k+ + F k+ F n k 0. 7

8 In the case n k + 0 mod 2 we additionally assume that n 4, k 3. Then F n+k+ + F k+ F n > ρ n+k. 6 When α ρ Z, the inequality with m = n + k is stronger than 3. Proof. We prove by induction on k. Using and 3, we obtain for k = 0 and k =, respectively, q n = a n q n + q n 2 2F n + F n = F n + F n + F n = F n+ + F F n, q n+ = a n+ q n + q n q n + q n F n+ + F n + F n = F n+2 + F 2 F n. Now, let k 0 and assume that is already proven for q n+k and q n+k+. Then q n+k+2 q n+k+ + q n+k F n+k+2 + F k+2 F n + F n+k+ + F k+ F n = F n+k+3 + F k+3 F n. This corresponds to with k replaced by k + 2. In order to prove 6 we express the Fibonacci numbers F m by Binet s formula 3. Hence, we have F n+k+ + F k+ F n = ρ ρ n+k + Case : Let n k mod 2. In particular, we have k. Then + n+k ρ + n+ + k. 2n+2k+ ρ 2n ρ 2k+ F n+k+ + F k+ F n = ρ ρ n+k + + > ρ ρ n+k + = ρ n+k. ρ 3 ρ + 2n+2k+ ρ 2n ρ2k+ Case 2: Let n mod 2, k 0 mod 2. In particular, we have n and k 0. First, we assume that k 2. Then, by similar computations as in Case, we obtain F n+k+ + F k+ F n = ρ ρ n+k + ρ + 2n+2k+ ρ + 2n ρ 2k+ > ρ n+k. 8

9 For k = 0 and some odd n we get F n+k+ + F k+ F n > ρ ρ n + ρ + 3 ρ > ρ n. Case 3: Let n 0 mod 2, k mod 2. By the assumption of the lemma, we have n 4 and k 3. Then F n+k+ + F k+ F n = ρ ρ n+k + ρ + 2n+2k+ ρ 2n ρ 2k+ > ρ n+k. Case 4: Let n k 0 mod 2. In particular, we have n 2. Then F n+k+ + F k+ F n = ρ ρ n+k + + ρ + 2n+2k+ ρ + 2n ρ 2k+ > ρ n+k. This completes the proof of Lemma 7. Lemma 8. Let m be an integer. Then ρ 2m F 2m+3 + F 2m+3 + F 2m+ ρ 2m F 2m+2 < m, 7 < m 0. 8 Proof. For m we estimate Binet s formula 3 for F 2m+2 using 4m + 2 6: Similarly, we prove 8 by F 2m+2 = ρ2m ρ 2 ρ 4m+2 F 2n+ = ρ 2n+ + ρ 2n+ ρ2m ρ 2 ρ 6 > ρ 2m. > ρ2n+ n 0. Hence, ρ 2m + F 2m+3 F 2m+3 + F 2m+ < ρ 2m ρ 2m+3 + ρ 2m+3 + ρ 2m+ <. The lemma is proven. 9

10 Proof of Proposition : Firstly, we assume the hypotheses in and prove 4. As in the proof of Theorem 3, put a 0 = α, θ := α a 0, a = /θ with 0 < θ < and ε 0 = θ <. Then ε 0 + ε = θ + a 0 a + a α = θ + a θ = θ + θ. θ We have 0 < θ < /2, since otherwise for θ > /2, we obtain a = /θ =. With a 0 = a = the conditions in are unrealizable both. Hence, there is an integer k 2 with k + < θ < k. Obviously, it follows that [/θ] = k, and therefore θ + θ Altogether, we have proven that θ < k + k k + = 2k + kk + 6 k 2. ε 0 + ε 6 <. 9 Therefore we already know that the inequality 4 holds for m = 0. Thus, we assume m in the sequel. Noting that ε 2m > 0 and ε 2m+ < 0 hold for every integer m 0, we may rewrite 4 as follows: 0 < p 2m+ p 2m αq 2m+ q 2m < ρ 2m m We distinguish three cases according to the conditions in. Case : Let a 2m+ 2. Additionally, we apply the trivial inequality a 2m+2. Then, using 2, 3, and 8, ε 2m + ε 2m+ < < q 2m+ + q 2m+2 2q 2m + q 2m + q 2m+ + q 2m + 2q 2m + q 2m 3q 2m + q 2m + 2F 2m+ + F 2m 3F 2m+ + F 2m m 0. ρ 2m Case 2: Let a 2m+ = and a 2m 2. Here, we have p 2m+ p 2m = p 2m +p 2m p 2m = p 2m, and similarly q 2m+ q 2m = q 2m. 0

11 Therefore, by 20, it suffices to show that 0 < p 2m αq 2m < ρ 2m for m. This follows with 2, 3, and 7 from 0 < p 2m αq 2m < = q 2m 2q 2m + q 2m 2 2F 2m + F 2m < m. F 2m+2 ρ 2m Case 3: Let a 2m = a 2m+ = a a 2 a 2m >. Since a 2m+ =, we again have as in Case 2: 0 < ε 2m + ε 2m+ = p 2m αq 2m < q 2m. 2 By the hypothesis of Case 3, there is an integer n satisfying n 2m and a n 2. We define an integer k by setting 2m = n + k. Then we obtain using and 6, q 2m = q n+k F n+k+ + F k+ F n > ρ n+k = ρ 2m. From the identity n+k = 2m it follows that the particular condition n k + 0 mod 2 in Lemma 7 does not occur. Thus, by 2, we conclude that the desired inequality 4. In order to prove 6, we now assume the hypothesis 7, i.e., a a 2 a 2k+ = and 0 m k. From 2m 2k and a 0 = a =... = a 2k = it is clear that p 2m = F 2m+ and q 2m = F 2m. Since a 2k+ = and 0 m k, we have q 2m α p 2m + q 2m+ α p 2m+ = p 2m αq 2m = F 2m+ αf 2m = F 2m+ ρf 2m + ρ αf 2m. From Binet s formula 3 we conclude that F 2m+ ρf 2m = ρ 2m+ + ρ 2m which finally proves the desired identity 6 in Proposition. = ρ 2m, Lemma 9. Let k be an integer, and let α := ;,...,,a 2k+,a 2k+2... be a real irrational number with partial quotients a 2k+ > and a µ for µ 2k + 2. Then we have the inequalities F 2k ρ α < ρ 2k ε 2k ε 2k+ 22 for a 2k+ 3, and for a 2k+ = 2. F 2k ρ α < ρ 2k + ρ 2k+2 ε 2k ε 2k+ ε 2k+2 ε 2k+3 23

12 One may conjecture that 22 also holds for a 2k+ = 2. Example 0. Let α = ;,,,, 2, = 2ρ + 8/3ρ + = 27 /8. With k = 2 and a = 2, we have on the one side on the other side, Proof of Lemma 9: ρ α = F 2 ρ α = ρ ε 4 ε 4 = ρ = , = Case : Let n := a 2k+ 3. Then there is a real number η satisfying 0 < η < and r 2k+ := a 2k+ ;a 2k+2,... = n + η =: + β. It is clear that n < β < n. From the theory of regular continued fractions see [, formula 6] it follows that Similarly, we have α = ;,...,,a 2k+,a 2k+2... = F 2k+2r 2k+ + F 2k+ F 2k+ r 2k+ + F 2k = F 2k+2 + β + F 2k+ F 2k+ + β + F 2k = βf 2k+2 + F 2k+3 βf 2k+ + F 2k+2. ρ = F 2k+2ρ + F 2k+ F 2k+ ρ + F 2k, hence, by some straightforward computations, ρ α = + β ρ ρf 2k+ + F 2k βf 2k+ + F 2k+2 < n ρf 2k+ + F 2k βf 2k+ + F 2k Here, we have applied the identities F 2 2k+2 F 2k+ F 2k+3 =, F 2 2k+ F 2k F 2k+2 =, and the inequality + β ρ < + n ρ < n. Since β > n and, by 2, ε 2k < ε 2k+ < = q 2k+ = q 2k+2 nf 2k+ + F 2k, a 2k+2 q 2k+ + F 2k+ n + F 2k+ + F 2k. 2

13 22 follows from nf 2k < ρf2k+ + F 2k n F2k+ + F 2k+2 ρ. 2 2k nf 2k+ + F 2k n + F 2k+ + F 2k In order to prove 2, we need three inequalities for Fibonacci numbers, which rely on Binet s formula. Let δ := /ρ 4. Then, for all integers s, we have ρ 2s+ < F 2s+ < + δρ2s+ and δρ 2s F 2s. 26 We start to prove 2 by observing that + δ ρ 2 ρ 2 + δ + 3ρ + δ + 4ρ + δ Here, the left-hand side can be diminished by noting that ρ > n n ρ + δρ 2. <. By n 3 we get + δn ρ ρ 2 + δ n ρ + δρ 2 + nρ + δ + n + ρ + δ <, or, equivalently, + δnρ 2k / ρ ρ 2k+ / + δρ 2k / n ρ 2k+ / + δρ 2k+2 / < ρ 2k nρ 2k+ / + δρ 2k / n + ρ 2k+ / + δρ 2k /. From this inequality, 2 follows easily by applications of 26 with s {2k, 2k, 2k +, 2k + 2}. Case 2: Let a 2k+ = 2. Case 2.: Let k 2. We first consider the function fβ := ρ + β βf 2k+ + F 2k+2 β 2. The function f increases monotonically with β, therefore we have fβ f2 = 3 ρ 2F 2k+ + F 2k+2, 3

14 and consequently we conclude from the identity stated in 24 that ρ α Hence, 23 follows from the inequality 3 ρ ρf 2k+ + F 2k 2F 2k+ + F 2k+2. 3 ρf 2k ρf 2k+ + F 2k 2F 2k+ + F 2k < q 2k+ q 2k+2 q 2k+3 q 2k+4 ρ k ρ2k+2 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 in Lemma 7: q 2k+ = a 2k+ q 2k + q 2k = 2F 2k+ + F 2k, q 2k+2 F 2k+3 + F 2 F 2k+ = F 2k+3 + F 2k+, q 2k+3 F 2k+4 + F 3 F 2k+ = F 2k+4 + 2F 2k+, q 2k+4 F 2k+ + F 4 F 2k+ = F 2k+ + 3F 2k+. Substituting these expressions into 27, we then conclude that 23 from 3 ρf 2k ρf 2k+ + F 2k 2F 2k+ + F 2k+2 + 2F 2k+ + F 2k + + F 2k+3 + F 2k+ + < F 2k+4 + 2F 2k+ F 2k+ + 3F 2k+ ρ 2k ρ 2 We apply the inequalities in 26 for all s 2 when δ is replaced by δ := /ρ 8. Using this redefined number δ, we have 3 ρ + δ ρ ρ 2 + δ 2ρ + δρ 2 + 2ρ + δ + ρ 3 + ρ + δρ 4 + 2ρ + ρ + 3ρ ρ 2 or, equivalently, + + <, 3 ρ + δρ 2k / ρ ρ 2k+ / + δρ 2k / 2ρ 2k+ / + δρ 2k+2 / 2ρ 2k+ / + δρ 2k / + ρ 2k+3 / + ρ 2k+ / δρ 2k+4 / + 2ρ 2k+ / + ρ 2k+ / + 3ρ 2k+ / < ρ 2k + ρ 2. From this inequality, 28 follows by applications of 26 with s {2k, 2k, 2k +, 2k + 2, 2k + 3, 2k + 4, 2k + } for k 2 which implies s 3. 4

15 Case 2.2: Let k =. From the hypotheses we have a 2k+ = a 3 = 2. To prove 23 it suffices to check the inequality in 28 for k =. We have F 2k = F =, F 2k = F 2 =, F 2k+ = F 3 = 2, F 2k+2 = F 4 = 3, F 2k+3 = F =, F 2k+4 = F 6 = 8, F 2k+ = F 7 = 3. Then 28 is satisfied because This completes the proof of Lemma 9. 3 ρ ρ ρ ρ <. 2 Proof of Theorem 2: In the sequel we shall use the identity F 2g + F 2g+2 + F 2g F 2n = F 2n+ F 2g n g 0, 29 which can be proven by induction by applying the recurrence formula of Fibonacci numbers. Note that F =. Next, we prove 8. Case : Let α M, α = ;a,a 2,... = ;,...,,a 2k,a 2k+,... with a 2k > for some subscript k. This implies α > ρ. Case.: Let 0 n < 2k. Then n 0 = n/2 k. In order to treat ε 2m + ε 2m+, we apply 6 with k replaced by k in Proposition. For α the condition 7 with k replaced by k is fulfilled. Note that the term F 2m ρ α in 6 is negative. Therefore, we have Sn := < n ε ν ν=2g n 0 n/2 ε2m + ε 2m+ ρ = ρ2 2g ρ 2n0 2m ρ 2 = ρ 2g ρ 2n 0. Case.2: Let n 2k. Case.2.: Let k g. Here, we get Sn k ε2m + ε 2m+ + ε 2k + ε 2k+ + n 0 m=k+ ε2m + ε 2m+. 30

16 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 a 2 a 2k = : k ε2m + ε 2m+ < k ρ 2m. b Since a 2k >, the left-hand condition in allows us to apply 4 for m = k: ε 2k + ε 2k+ < ρ 2k. c We estimate the right-hand sum in 30 again by 4. To check the conditions in, we use a a 2 a 2m >, which holds by m k + and a 2k >. Hence, n 0 m=k+ ε2m + ε 2m+ < n 0 m=k+ ρ 2m. Altogether, we find with 30 that Sn < n 0 ρ 2m = ρ 2g ρ 2n 0. 3 Case.2.2: Let k < g. In order to estimate ε 2m + ε 2m+ for g m n 0, we use k + g and the arguments from c in Case.2.. Again, we obtain the inequality 3. The results from Case. and Case.2 prove 8 for a 2k > with k. It remains to investigate the following case. Case 2: Let α M, α = ;a,a 2,... with a >. For m = 0 provided that g = 0 the first condition in is fulfilled by a 2m a 2m+ = a 0 a = a >. For m we know that a a 2 a 2m > always satisfies one part of the second condition. Therefore, we apply the inequality from 4: Sn < n 0 ρ 2m = ρ 2g ρ 2n 0. Next, we prove 9. Let α M, α = ;a,a 2,... = ;,...,,a 2k+,a 2k+2,... with a 2k+ > for some subscript k. This implies ρ > α. Case 3.: Let 0 n < 2k. Then n 0 = n/2 k. In order to treat ε 2m + ε 2m+, we apply 6 with k replaced by k in Proposition. For α the condition 7 with k replaced by k is fulfilled. Note 6

17 that the term F 2m ρ α in 6 is positive. Therefore we have, using 29, Sn n 0 ρ + ρ αf 2m 2m = ρ 2g ρ 2n 0 + ρ α Here we have used that 2n 0 + 2k. n 0 F 2m = ρ 2g ρ 2n 0 + ρ αf 2n0 + F 2g ρ 2g ρ 2n 0 + F 2k F 2g ρ α. Case 3.2: Let n 2k. Our arguments are similar to the proof given in Case.2, using a a 2 a 2k = and a 2k+ >. Case 3.2.: Let k g. Applying 29 again, we obtain Sn < = k k n 0 ε2m + ε 2m+ + ε 2k + ε 2k+ + ρ + ρ αf 2m 2m + ρ + 2k ρ + ρ α k 2m F 2m n0 m=k+ = ρ 2g ρ 2n 0 + F 2k F 2g ρ α. n 0 m=k+ ρ 2m ε2m + ε 2m+ Case 3.2.2: Let k < g. From g k + we get Sn n 0 ρ 2m = ρ 2g ρ 2n 0. The results of Case 3. and Case 3.2 complete the proof of 9. For the inequality 0 we distinguish whether α belongs to M or not. Case 4.: Let α M. Then 0 is a consequence of the inequality in 8: ν=2g ε ν lim ρ 2g ρ 2n 0 = ρ 2g. n 0 Case 4.2: Let α M. There is a subscript k satisfying α = ;,...,,a 2k+,a 2k+2,... and a 2k+ >. To 7

18 simplify arguments, we introduce the function χk,g defined by χk,g = if k > g, and χk,g = 0 if k g. We have S := ε ν = ν=2g 2k ν=2g ε ν + ν=max{2k,2g} ε ν = χk,g F 2k F 2g ρ α + ρ 2g ρ 2k+ + F 2k F 2g ρ α + ρ 2g ρ 2k+ + F 2k ρ α + ρ 2g ρ 2k+ + m=max{k,g} ε2m + ε 2m+ m=k ε2m + ε 2m+ ε2m + ε 2m+, 32 m=k where we have used 9 with n = 2k and n 0 = n/2 = k. Case 4.2.: Let a 2k+ 3. The conditions in Lemma 9 for 22 are satisfied. Moreover, the terms ε 2m + ε 2m+ of the series in 32 for m k + can be estimated using 4, since a a 2 a 2k+ >. Therefore, we obtain S < ρ 2k ρ 2k + ρ 2g + m=k+ ε2m + ε 2m+ < ρ 2k ρ 2k + ρ 2g + ρ 2k+ = ρ 2g. Case 4.2.2: Let a 2k+ = 2. Now the conditions in Lemma 9 for 23 are satisfied. Thus, from 32 and 4 we have S < ρ + 2k ρ 2k+2 ρ + 2k ρ 2g + ε2m + ε 2m+ < ρ 2k + ρ 2k+2 ρ 2k + ρ 2g + m=k+2 m=k+2 ρ 2m = ρ 2k + ρ 2k+2 ρ 2k + ρ 2g + ρ 2k+3 = ρ 2g. This completes the proof of Theorem 2. Concluding remarks In this section we state some additional identities for error sums εα. For this purpose let α = a 0 ;a,a 2,... be the continued fraction expansion of a real irrational number. Then the numbers α n are defined by α = a 0 ;a,a 2,...,a n,α n n = 0,, 2,... 8

19 Proposition. For every real irrational number α we have εα = n α n= k k= and εα = n=0 n an 0 an 0 a0 0 α. Next, let α = a 0 ;a,a 2,... with a 0 be a real number with convergents p m /q m m 0, where p =, q = 0. Then the convergents p m /q m of the number /α = 0;a 0,a,a 2,... satisfy the equations q m = p m and p m = q m for m 0, since we know that p =, p 0 = 0 and q = 0, q 0 =. Therefore we obtain a relation between εα and ε/α: This proves ε/α = q m α p m = = q α p + α p m α q m q m α p m Proposition 2. For every real number α > we have ε/α = + εα α. = α q m α p m = α + εα. 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. Moreover, the presentation of the paper was improved by following the remarks of the referee. References [] H. Cohn, A short proof of the simple continued fraction expansion of e, Amer. Math. Monthly , [2] C. Elsner, On arithmetic properties of the convergents of Euler s number, Colloq. Math ,

20 [3] C. Elsner and T. Komatsu, On the residue classes of integer sequences satisfying a linear three term recurrence formula, Linear Algebra Appl , [4] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Clarendon Press, Oxford, 984. [] A. Khintchine, Kettenbrüche, Teubner, Leipzig, 949. [6] T. Komatsu, Arithmetical properties of the leaping convergents of e /s, Tokyo J. Math , 2. [7] T. Komatsu, Some combinatorial properties of the leaping convergents, Integers , A2. Available electronically at [8] O. Perron, Die Lehre von den Kettenbrüchen, Chelsea Publishing Company, New York, Mathematics Subject Classification: Primary J04; Secondary J70, B39. Keywords: continued fractions, convergents, approximation of real numbers, error terms. Concerned with sequences A00004, A009, A007676, A007677, A04008, and A Received July 2 200; revised version received January Published in Journal of Integer Sequences, January Return to Journal of Integer Sequences home page. 20