AN APPROXIMATION FORMULA FOR EULER ... - inmabb

REVISTA DE LA ´ MATEMATICA ´ UNION ARGENTINA Vol. 57, No. 1, 2016, Pages 9–22 Published online: October 26, 2015

AN APPROXIMATION FORMULA FOR EULER–MASCHERONI’S CONSTANT PABLO A. PANZONE

Abstract. A fast approximation formula for Euler–Mascheroni’s constant based on a certain integral due to Peter Borwein is given. This asymptotic formula for γ involves harmonic numbers, ln 2 and rational numbers whose denominators are powers of 2.

1. Introduction and results There are many results concerning the arithmetic nature of Euler–Mascheroni’s constant γ := limN →∞ 1 + 1/2 + 1/3 + · · · + 1/N − ln N (or approximations of it). P. Bundschuh (see [7]) proved that either γ is transcendental or certain roots of transcendental equations involving the gamma function and its derivative are irrational. J. Sondow (see [12]) proved some neat criteria for the irrationality of γ; many formulae for γ were discovered by him and can be found online in his personal homepage. Extensions and refinements of these results are given in [8] and [9]. A. Aptekarev was the first to prove that either γ or the so called Gompertz’s constant is irrational (see [1]). T. Rivoal has recently extended this result in [11]. T. Rivoal and A. Aptekarev constructed rapidly converging sequences to γ or combinations of γ and logarithms (see [2, 10, 11]). S. Ramanujan has given many formulae for γ which can be found in [4]; these have been proved thanks to B. Berndt’s work. Also Wolfram’s website contains many other formulae. Sondow’s construction is based on the famous integrals introduced by F. Beukers for a simplified proof of Apery’s result that ζ(3) is irrational (see also [13]). One of Sondow’s criteria is the following: If Sn :=

n min(k−1,n−k) n−i Y Y Y k=1

i=0

(n + k)2

2

LCM(1,...,2n)/j (n i)

,

j=i+1

and {ln Sn } ≥ 1/2n infinitely often, then γ is irrational. Here {x} means the fractional part of x. Numerical computations suggest that {ln Sn } is dense in (0, 1). 2010 Mathematics Subject Classification. 41-99, 11Jxx. Key words and phrases. Euler’s constant, Approximation, Linear form, Irrationality. The author was supported in part by Universidad Nacional del Sur and CONICET.

9

10

PABLO A. PANZONE

In this note we prove, in Theorem A below, a fast approximation formula for γ. The idea is to have an approximation formula involving the logarithms of fixed numbers. We show in Theorem B, which follows immediately from Theorem A, that this is possible in some sense (with a price to be paid). In Theorem B a formula depending on a parameter n of the form: X Z1 γ + Z2 ln 2 + r, tending to zero as n → ∞, is given. Here P Z1 , Z2 are integer numbers (tending to infinity as n → ∞) and the expression rP means a finite sum of certain rational n numbers involving harmonic numbers, i.e. j=1 1/j, and rational numbers with denominators a power of two. In the present paper we use ideas from P. Borwein [6], where a very nice proof of the irrationality of certain series is given using a certain integral which we modify slightly to construct an approximation to Euler’s constant. Specifically we define the following function for n ≥ 2: Fn (z) :=

1 2πi

n−1 Y

Z

|t|=1 k=1

∞ (1 − 2k /t) (−1/t) X 1 dt, k n (1 − 2 t) (1 − 2 t) (z + 2h /t)

(1)

h=θ(n)

where θ(n) is a function from the set of positive integers to itself such that n ≤ θ(n). Our proof is based on the integral Z 1 Z −z 1 τ pn (τ ) Fn (z)dzdτ, (2) 2πi 0 β (−z)n where the rational function pn (τ ) is defined recursively by p0 (τ ) = 1/(1 + τ ),

(τ pr−1 (τ ))0 = pr (τ ),

and β is a curve which, roughly speaking, encloses the negative integers and goes from −∞ + i∞ to −∞ − i∞. Specifically we take β = β1 ∪ β2 ∪ β3 , where with α := 3π/4 one defines β2 (resp. −β1 ) := {exp(−iα)t (resp. exp(iα)t), 1/2 ≤ t ≤ ∞}. The curve β3 is the arc of the circle given by the equation {exp(it)/2; α ≤ t ≤ 2π − α}. Recall the notation for the q-binomial coefficients hni Πn (1 − q s ) := s=n−m+1 , if 0 < m < n, m m q Πs=1 (1 − q s ) [ n0 ]q = [ nn ]q := 1, and defined to be zero otherwise (see [5]). Definition 1. Define, for 2 ≤ n, qm,n := (−1)m−1 2m(m−1)/2 an := (−1)n

n X

h n − 1 i hn + m − 1i , m−1 2 m 2

qm,n ,

m=1

Rev. Un. Mat. Argentina, Vol. 57, No. 1 (2016)

APPROXIMATION FORMULA FOR EULER–MASCHERONI’S CONSTANT

bn := −

n X

θ(n)+m−1

m=1

2hn

h=1

cn := (−1)n+1

n−1 1 X

X

qm,n n X

11

(−1)n−1−i 2h(n−1−i) pi (1),

i=0

qm,n θ(n) + m − 1 ,

m=1

dn :=

n X

(−1)k

k=1

en := (−1)

n

pk−1 (1) , 2k − 1

n X

qm,n

m=1

2θ(n)+m−1 X j=1

1 . j

Observe that an , cn ∈ Z, en is a sum of harmonic numbers, and by Lemma 3 (a) the coefficient pn (1) can be written in closed form: n DnE 1 X pn (1) = n+1 (−1)i , 2 i i=0

n for n ≥ 1, and p0 (1)

n = 1/2, where i are the Eulerian numbers. Recall that the Eulerian number k gives the number of permutations of {1, 2, . . . , n} having k permutation ascents, and one has the formula D n E k+1 X j n+1 (−1) (k − j + 1)n . = k j j=0 Our main result is the following explicit approximation formula. Theorem A. Let θ(n) be any function from the set of positive integers to itself such that n ≤ θ(n). Then − γan + an dn + bn + en + cn ln 2 0.792(n + 1) n+1 θ(n)2n(n+1)/2 1 n! max , =O . ln(n + 2) 2nθ(n)+θ(n) 2θ(n)n+n Note: the constant implied in the O-symbol does not depend on n nor the function θ(n). For example taking θ(n) = n the above O-term is 1 O n2 , 2 2 −cn ln n where c > 0 is some positive constant. The proof of Theorem A is given in Section 2. Definition 2. For fixed 0 < ρ < 18 , set gn := 2[n

2

ρ]

Πnk=[n/2] (2k − 1),

b0n := (−1)n+1

n X m=1

qm,n

n−1 X i=0

(−1)i pi (1) . (2i+1 − 1)2(i+1)(n+m−1)


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PABLO A. PANZONE

Next we specialize θ(n) = n. Indeed one has the following theorem. Theorem B. Assume θ(n) = n, 0 < ρ < 18 . For some constant 0 < c0 , − γan gn + b0n gn + en gn + cn gn ln 2 1

= γZ1 + b0n gn + en gn + Z2 ln 2 = O(2−( 8 −ρ)n

2

+c0 n ln n

),

where Zi are integer numbers. Moreover en gn is a sum of harmonic numbers and b0n gn is a rational number whose denominator is a power of 2. Theorem B is proved in Section 3. We make some remarks concerning the last theorem. One has the following corollary, where {x} means the fractional part of x. Corollary. If γ is a rational number then: a) The sequence {b0n gn + en gn + cn gn ln 2} cannot be dense in [A, B] ⊂ [0, 1], A < B. b) The sequence {n!(b0n gn + en gn + cn gn ln 2)} has {0, 1} as its only possible limit points. The proof, which follows from Theorem B, is given in Section 3. 2

3

Remark 1. Notice that |an | ≈ 2( 2 )(n (iii) gives

−n)

15

2

|an gn | ≈ 2( 8 +ρ)n Also if [n2 ρ] ≥ n, then pi (1)2[n Z.

2

ρ]

as n → ∞, which together with Lemma 7 +O(n)

as n → ∞.

∈ Z, i = 1, . . . , n − 1, and therefore b0n gn 22n

2

−n

∈

Remark 2. The coefficient an has some similitude with a particular case of the so called q-little Legendre polynomials defined by: Pn (x, q) =

n X

(−1)m q m(m+1)/2−mn

m=0

h n i hn + mi xm . m q m q

Notice that an+1 = (−1)n+1

n X

(−1)m 2m(m+1)/2

m=0

h n i hn + mi 2n − 1 2n + m+1 . m 2 m 2 −1 2

2. Outline and proof of Theorem A The proof of Theorem A is rather technical so we give first an outline and the ideas behind it. In Lemma 1 we decompose the function (1) as Fn (z) = An (z) + Bn (z) + Cn (z), Rev. Un. Mat. Argentina, Vol. 57, No. 1 (2016)


13

and apply the formula (2) to this equality (Lemma 2 is an auxiliary result). That is, Z 1 Z −z τ pn (τ ) 1 Fn (z)dzdτ 2πi 0 β (−z)n Z 1 Z −z 1 τ pn (τ ) = A (z) + B (z) + C (z)dz dτ. n n n 2πi 0 β (−z)n Integrals along Fn (z) and Cn (z) are estimated in Lemmas 5 and 6 and they give the O-term of Theorem A; these integrals are small in magnitude. The term Bn (z) is a polynomial so the integral corresponding to it is zero. The integral corresponding to An (z) gives the linear form −γan + an dn + bn + en + cn ln 2. This is proved in Lemma 4 with the help of Lemma 3. This is basically the proof of Theorem A. We explain some ideas our theorem. A particular case of [6] gives a proof P∞behind 1 of the irrationality of h=1 (z+2 h ) for rational fixed z using an integral of the form (see [6, formula (1)]) Z n−1 ∞ Y (1 + 2k /zt) (−1/t) X 1 1 dt, Fñ (z) := k n 2πi |t|=1 (1 − 2 t) (1 − 2 t) (z + 2h /t) k=1

h=1

which is slightly similar to our Fn (z). The integral Fñ (z), which can be shown to be small for fixed z and large n, is of the form Fñ (z) = Rac1 (z)

∞ X h=1

1 + Rac2 (z), (z + 2h )

where Raci (z), i = 1, 2, are rational functions in z. On the other hand, Z 1 P∞ 2h Z 1Z ∞ 1 1 τ −z X h=1 τ 1−γ = dτ = dzdτ (1 + τ ) 2πi 1 + τ (z + 2h ) 0 0 β h=1 Z 1Z ∞ X 1 1 = τ −z p0 (τ ) dzdτ ; 2πi 0 β (z + 2h ) h=1

the first equality is well-known (we prove it below) and the second one follows from shifting the curve β to the left picking the residues at the poles z = −2h . So this suggests that the integral Z 1Z 1 τ −z p0 (τ )Fñ (z) dzdτ 2πi 0 β is small. But this is not a linear form in γ and ln 2. Indeed other constants of h R 1 P∞ τ22h apparently different type appear (for example 0 h=1 (1+τ ) dτ ). But the integral Z 1Z 1 τ −z p0 (τ )Fn (z) dzdτ 2πi 0 β Rev. Un. Mat. Argentina, Vol. 57, No. 1 (2016)

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PABLO A. PANZONE

is a linear form in γ and ln 2 only. On the other hand, estimates for this last integral are not good enough because now z is a variable and one needs roughly a factor 1/z n (when z is large) which is not present. This can be corrected with Definition 2 introducing a polynomial pn (τ ) which is not too large (and, of course, without losing the fact that this is a linear form in γ and ln 2). Lemma 1. Let Fn be defined by (1). Then if n ≥ 2, Fn (z) = An (z) + Bn (z) + Cn (z), where An (z) =

∞ X h=1

−

n h n − 1 i hn + m − 1i X 1 (−1)m−1 2m(m−1)/2 h (z + 2 ) m=1 m−1 2 m 2 n X

(−1)

m−1 m(m−1)/2

2

m=1

h n − 1 i h n + m − 1 i θ(n)+m−1 X 1 , m−1 2 m (z + 2h ) 2 h=1

and Bn (z) is a polynomial in z. Also Cn (z) = 0 if 2θ(n) > |z| and if z belongs to the ring shaped region {2θ(n) ≤ m0 2 < |z| < 2m0 +1 }, m0 ∈ N , then Cn (z) = −

m0 n−1 X Y (1 + 2k−h z) h=θ(n) k=1

z n−1 . (z + 2k+h ) (z + 2n+h )

Proof. The formula follows using the theory of residues: the product in (1) has simple poles at t = 21 , . . . , 21n , which gives ∞ X h=1

n k+m X Πn−1 ) 1 k=1 (1 − 2 n h k−m (z + 2 ) m=1 Πk=1,k6=m (1 − 2 )

−

n X

k+m Πn−1 ) k=1 (1 − 2 n k−m Π (1 − 2 ) m=1 k=1,k6=m

θ(n)+m−1

X h=1

1 , (z + 2h )

where we have used that ∞ ∞ m X X X 1 1 1 = − . (z + 2h 2m ) (z + 2h ) (z + 2h ) h=1

h=1

h=1

The term An (z) is obtained if one notices that h i hn + m − 1i k+m Πn−1 ) m−1 m(m−1)/2 n − 1 k=1 (1 − 2 = (−1) 2 . Πnk=1,k6=m (1 − 2k−m ) m−1 2 m 2

(3)

The integrand in formula (1) has a pole of order n − 1 at t = 0, which gives the polynomial Bn (z) whose precise form we shall not need. The term Cn (z) is present due to the possible poles (in the variable t) of P∞ 1 θ(n) > |z| there are no poles in {|t| < 1} and therefore h=θ(n) (z+2h /t) : if 2 Cn ≡ 0. If 2θ(n) ≤ 2m0 < |z| < 2m0 +1 then in the last sum the only poles inside Rev. Un. Mat. Argentina, Vol. 57, No. 1 (2016)


15

{|t| < 1} are those corresponding to h = θ(n), . . . , m0 . Using residues the formula follows. Also we will need an estimate of Z n−1 Y (1 − 2k /t) (−1/t) 1 1 dt, Im (z) := 2πi |t|=1 (1 − 2k t) (1 − 2n t) (z + 2m /t)

(4)

k=1

for one has ∞ X

Fn (z) =

Im (z).

(5)

m=θ(n)

Lemma 2. Let Im be defined as above. Then Im (z) = 0 if 2m < |z| and Im (z) =

n−1 Y k=1

m

if 2

z n−1 (1 + 2k−m z) , (z + 2k+m ) (z + 2n+m )

> |z|.

Proof. Let δ = {|t| = 1} and δm = {|t| = |2m /z| + 1}. Then Z n−1 Y (1 − 2k /t) (−1/t) 1 1 dt, Im (z) = 2πi δ∪δm (1 − 2k t) (1 − 2n t) (z + 2m /t) k=1

for we can change the curve δm to an arbitrarily large circle and therefore the integral over δm is zero. If 2m > |z| then one uses residues for the pole at t = −2m /z in the open annulus formed by the two curves δ, δm and the result follows (if 2m < |z| there is no pole inside this open annulus). Lemma 3. Set p0 (τ ) :=

1 1+τ

and define pr (τ ) recursively by (τ pr−1 (τ ))0 = pr (τ ).

Then: a) If n ≥ 2 then pn−1 (τ ) =

e1 (1+τ )2

+ ··· +

max |pn (τ )| ≤

en (1+τ )n

with ei ∈ Z. Also

0.792(n + 1) n+1

τ ∈[0,1]

ln(n + 2)

n! ,

and one has the closed formula pn (1) =

1 2n+1

n D E X n i=0

i

(−1)i ,

n

for n ≥ 1, where i are the Eulerian numbers. b) If n ≥ 2 then Z 1 Z −z ∞ n X 1 τ pn (τ ) X pk−1 (1) 1 dzdτ = (−1)n−k k + (−1)n (1 − γ). n h 2πi 0 β (−z) (z + 2 ) 2 −1 h=1

k=1


16

PABLO A. PANZONE

c) If n ≥ 2 then Z 1 h pn (τ )τ 2 dτ 0

=

n−1 X

h

n−1−i h(n−1−i)

(−1)

2

n nh

pi (1) + (−1) 2

ln 2 + (−1)

2 X

n+1 nh

2

i=0

j=2h−1 +1

1 . j

Proof. a) The first part follows by induction and is left as an easy exercise to the reader. We prove the inequality stated. First observe that (τ (τ (τ . . . (τ p0 (τ ))0 . . .)0 )0 )0 = pn (τ ), where the number of derivatives is n. But (τ (τ (τ . . . (τ p0 (τ ))0 . . .)0 )0 )0 =

n+1 X

(k−1)

S(n + 1, k)τ k−1 p0

(τ ),

k=1

where S(n, k) are the Stirling numbers of the second kind (recall that these numbers have the property S(1, n) = S(n, n) = 1 and S(n, k) = kS(n − 1, k) + S(n − 1, k − 1), from which the above formula follows by iteration of this last formula). Using these identities we have if τ ∈ [0, 1] |pn (τ )| ≤

n+1 X

(i)

S(n + 1, k) max |p0 (τ )| ≤ i=0,...,n

k=1

n+1 X

S(n + 1, k)n! = B(n + 1)n! ,

k=1

where B(n) is the n-th Bell number, see [14]. The inequality in the lemma follows from a result in [3], namely: if n ∈ N then 0.792n n B(n) < . ln(n + 1) Finally, the closed form formula is proved as follows: using Cauchy’s formula in the above formulae gives Z n+1 X 1 1 pn (1) = S(n + 1, k)(k − 1)! dz. 2πi |z−1|=1 (z − 1)k (1 + z) k=1

But if n ≥ 1 then (−1)n+1 Li−n (z) =

n+1 X

S(n + 1, k)(k − 1)!

k=1

where Lin (z) := pn (1) =

P∞

k

z k=1 kn

(−1)n+1 2πi

1 , (z − 1)k

is the polylogarithm function, which gives Z Li−n (z) (−1)n Li−n (z) dz = dz. 1 + z 2πi |z−1|=1 |z+1|=1 1 + z

Z



17

The proof of the last equality (and the closed form formula) follow from here and the known identity n D E X n n−i 1 z . Li−n (z) = n+1 (1 − z) i i=0 R −n (z) −n (z) Hint: |z|=R Li1+z dz = 0, if R > 2, (say) for | Li1+z | = O(1/|z|2 ) as |z| → ∞. Now use residues. b) From the well-known formula Z 1 P∞ 2h h=1 τ dτ = 1 − γ, (1 + τ ) 0 and, for 0 < τ < 1, 1 2πi

Z β

h τ −z dz = τ 2 h (z + 2 )

(this last formula follows by deforming the curve β, moving it arbitrarily to a left half plane and taking into account the pole at z = −2h ), one has Z 1Z ∞ X 1 1 τ −z p0 (τ ) dzdτ 2πi 0 β (z + 2h ) h=1 Z 1Z Z 1 P∞ 2h ∞ 1 1 τ −z X h=1 τ = dzdτ = dτ = 1 − γ. (6) h 2πi 0 β 1 + τ (z + 2 ) (1 + τ ) 0 h=1

Also Z 0

1

h

h 2 X τ2 (−1)j+1 dτ = ln 2 − . 1+τ j j=1

(7)

(Remark: the above formula for 1 − γ follows from (7) and the identity h

2 X j=2h−1 +1

2h

1 X (−1)j+1 = . j j j=1

(8)

2h R 1 PN h=1 τ Indeed adding (7) for h = 1, . . . , N and using (8) we get 0 (1+τ ) dτ = P2N N ln 2 − j=2 1/j, and the formula follows.) Now observe that integrating by parts and using (τ pr−1 (τ ))0 = pr (τ ) yield Z 1 Z 1 h h ∞ ∞ X X τ2 1 τ2 pr (τ ) dτ = r pr−1 (1) − pr−1 (τ ) dτ. 2hr 2 −1 2h(r−1) 0 0

h=1

h=1

The lemma follows iterating this last formula, using (6) and noting that Z 1 Z −z Z 1 h ∞ ∞ X 1 τ pr (τ ) X 1 τ2 pr (τ ) dzdτ = dτ, 2πi 0 β (−z)r (z + 2h ) 2rh 0 h=1

h=1

(same proof as (6)). This proves part b). Rev. Un. Mat. Argentina, Vol. 57, No. 1 (2016)

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PABLO A. PANZONE

c) Integrating by parts and using (τ pr−1 (τ ))0 = pr (τ ) yield: Z 1 Z 1 h 2h h pn (τ )τ dτ = pn−1 (1) − 2 pn−1 (τ )τ 2 dτ. 0

0

Iterating this one gets Z 1 Z n−1 X h pn (τ )τ 2 dτ = (−1)n−1−i 2h(n−1−i) pi (1) + (−1)n 2nh 0

0

i=0

1

h

τ2 dτ. 1+τ

Now use (7) and (8) to obtain (c).

The following lemma is a kind of linear form for γ. Lemma 4. The following formula holds: if 2 ≤ n then Z 1 Z −z 1 τ pn (τ ) Fn (z)dzdτ = −an γ + an dn + bn + en + cn ln 2 2πi 0 β (−z)n Z 1 Z −z 1 τ pn (τ ) + Cn (z)dzdτ, 2πi 0 β (−z)n where Cn (z) is defined in Lemma 1, γ is Euler’s constant, and an , bn , cn , dn , en are as in Theorem A. Proof. Recall that from Lemma 1 one has Fn (z) = An (z) + Bn (z) + Cn (z). First note that β τ −z z m dz = 0 if m ∈ Z (again deform arbitrarily the curve to a left half-plane), and therefore Z 1 Z −z 1 τ pn (τ ) Bn (z) dzdτ = 0. 2πi 0 β (−z)n R

So we are left with the integral over An . From the definition of An (z) in Lemma 1 and Lemma 3 (b) one gets: Z 1 Z −z τ pn (τ ) 1 An (z)dzdτ 2πi 0 β (−z)n Z 1 Z −z θ(n)+m−1 n X X 1 τ pn (τ ) 1 = an (1 − γ) + an dn − qm,n dzdτ, n (z + 2h ) 2πi (−z) 0 β m=1 h=1

where an , dn , qm,n are defined in Theorem A. Now notice that Z 1 Z −z Z 1 h 1 τ pn (τ ) 1 1 dzdτ = pn (τ )τ 2 dτ, n h hn 2πi 0 β (−z) (z + 2 ) 2 0 P2θ(n)+m−1 1 P n and use Lemma 3 (c). This gives (here e0n := (−1)n m=1 qm,n j=2 j) Z 1 Z −z 1 τ pn (τ ) An (z)dzdτ = an (1 − γ) + an dn + bn + e0n + cn ln 2, 2πi 0 β (−z)n Rev. Un. Mat. Argentina, Vol. 57, No. 1 (2016)


19

where bn , cn are defined in Theorem A. The lemma follows by noticing that an + e0n = en , where en is defined in Theorem A. In the following lemmas (5 and 6) we estimate the integrals that appear in Lemma 4; we remark that the constant implied in the O-symbol in these lemmas is absolute. Lemma 5. If 2 ≤ n then 1 Z 1 Z τ −z p (τ ) 1 n = O . F (z) dzdτ max |p (τ )| n n 2πi 0 β (−z)n 2θ(n)n+n τ ∈[0,1] Proof. We will use formula (5); so we first estimate Im (z) with m ≥ θ(n) ≥ n. Recall that by Lemma 2 one has Im (z) = 0 if |z| > 2m . We chop the curve β in pieces: let βr0 be the part of the curve β in {2r−1 ≤ |z| < 2r }, r = 1, 2, . . . , m. Then using Lemma 2, taking absolute values in the product expression and writing (1 + 2a ) = 2a (1 + 2−a ) whenever a ≥ 1, one gets I (z) 2(n−1−m+r)(n−m+r)/2 1 m max0 ≤ O , ≤ O z∈βr (−z)n 2mn+n+r 2mn+n(n+1)/2 2r if 1 ≤ n − 1 − m + r, where the constant implied in the O-symbol is absolute. The last inequality follows because 2(n−1−m+r)(n−m+r)/2 ≤ 2(n−1)n/2 if r = 1, . . . , m. If n − 1 − m + r ≤ 0 then I (z) 1 1 m ≤ O ≤ O . max0 z∈βr (−z)n 2mn+n+r 2mn+n(n+1)/2 2r Now if Lβr0 (≤ 2r ) is the length of any of the two segments which compose βr0 then Z 1 Z −z Z 1 I (z) τ pn (τ ) m 2r−1 cos(π−α) 0 I (z) ≤ 2L τ |p (τ )| max dzdτ dτ m β n r z∈βr0 (−z)n (−z)n 0 βr0 0 1 ≤ O mn+n+r max |pn (τ )| , 2 τ ∈[0,1] where α is the angle defined at the beginning in the definition of the curves β1 , β2 . Finally let β00 be the part of the curve β in |z| < 1. It is easy to see that I (z) 1 m max0 ≤ O , z∈β0 (−z)n 2mn+n(n+1)/2 and

1

τ −z p (τ ) 1 n I (z) ≤ O max |p (τ )| . dzdτ m n (−z)n 2mn+n(n+1)/2 τ ∈[0,1] 0 β00 All this gathers to give Z 1 Z τ −z p (τ ) Z 1Z τ −z pn (τ ) n I (z)dzdτ = I (z)dzdτ m m Sm n n 0 (−z) (−z) 0 β 0 i=0 βi Z 1Z τ −z p (τ ) 1 n ≤ Im (z) dzdτ = O mn+n max |pn (τ )| , Sm n 0 (−z) 2 τ ∈[0,1] 0 i=0 βi Z

Z


20

PABLO A. PANZONE

and the lemma follows using this last formula and formula (5).

Lemma 6. If 2 ≤ n then 1 Z 1 Z τ −z p (τ ) θ(n)2n(n+1)/2 n = O C (z)dzdτ max |p (τ )| . n n 2πi 0 β (−z)n 2nθ(n)+θ(n) τ ∈[0,1] k−h Qn−1 z) z −1 Proof. Set Dh (z) := k=1 (1+2 . Use the definition of Cn (z) as given (z+2k+h ) (z+2n+h ) Pm0 n in Lemma 1, to notice that Cn (z)/(−z) = ± h=θ(n) Dh (z) for 2m0 < |z| < m0 +1 2 , θ(n) ≤ m0 ∈ N . Using the same notation as in the proof of Lemma 5, ! Pn−1 (n−1)(m0 +1−h)+n(n−1)/2 2 2 k=1 (k−h+m0 +1) P Pn =O . max |Dh (z)| = O n 0 k=1 max{m0 ,k+h} 2m0 z∈βm 2 2 k=1 max{m0 ,k+h} 2m0 0 +1

Now we have two cases. One case is when h ≤ m0 − n and therefore n X max{m0 , k + h} = m0 n. k=1

In this case max |Dh (z)| = O 0

z∈βm

0 +1

2−(h−1)(n−1)+n(n−1)/2 22m0

=O

2−(θ(n)−1)(n−1)+n(n−1)/2 22m0

,

where the last inequality follows because θ(n) ≤ h. The other case is when h satisfies m0 − n + 1 ≤ h ≤ m0 . In this case n X max{m0 , k + h} = m0 n + (h − m0 + n)(h − m0 + n + 1)/2. k=1

Thus max |Dh (z)| = O 0

z∈βm

0 +1

2−(h−1)(n−1)+n(n−1)/2−(h−m0 +n)(h−m0 +n+1)/2

22m0 2−(h−1)(n−1)+n(n−1)/2 2−(θ(n)−1)(n−1)+n(n−1)/2 =O = O , 22m0 22m0

This gives C (z) m 2−(θ(n)−1)(n−1)+n(n−1)/2 n 0 max = O . n 0 z∈βm (−z) 22m0 +1 0 Therefore one has Z 1Z τ −z p (τ ) n C (z) dzdτ n n 0 (−z) 0 βm 0 +1 Z 1 C (z) n 2m0 cos(π−α) 0 ≤ 2Lβm τ |p (τ )| max dτ n 0 0 +1 z∈βm (−z)n 0 +1 0 m 2−(θ(n)−1)(n−1)+n(n−1)/2 0 =O max |p (τ )| . n 22m0 τ ∈[0,1]



21

Now using this last formula and (recall Cn (z) = 0 if |z| < 2θ(n) ) Z 1 Z −z Z 1Z ∞ X τ −z pn (τ ) τ pn (τ ) C (z)dzdτ = Cn (z)dzdτ, n 0 (−z)n (−z)n 0 β 0 βm +1 m0 =θ(n)

0

one gets 1 Z 1 Z τ −z p (τ ) n C (z)dzdτ n 2πi 0 β (−z)n θ(n)2−(θ(n)−1)(n−1)+n(n−1)/2 =O max |p (τ )| . n τ ∈[0,1] 22θ(n) The lemma follows simplifying the O-symbol.

Theorem A is a direct consequence of Lemmas 3 (a), 4, 5 and 6. 3. Proofs of Theorem B and of Corollary We need first the following lemma. Lemma 7. Let an , cn , dn be defined as in Theorem A. Then i) an , cn ∈ Z. ii) dn 2n Πnk=[n/2] (2k − 1) ∈ Z. 3

iii) For some constant 0 < c, Πnk=[n/2] (2k − 1) = O(2 8 n

2

+cn

).

Proof. The proof of (i-ii) is immediate using Lemma 3 (a), i.e. 2n pn−1 (1) ∈ Z. Part (iii) is easy and is left to the reader. Proof of Theorem B. We apply Theorem A in its simplest form, i.e. we take θ(n) = n. In this case the O-term of Theorem A is 1 O n2 , 2 2 −cn ln n where c > 0 is some positive constant. Next one has b0n = an dn + bn . Thus Theorem A yields −γan + b0n + en + cn ln 2 = O Multiplying this by gn = 2[n notices that by Lemma 7

2

ρ]

1

2

n2 2

−cn ln n

.

Πnk=[n/2] (2k − 1) gives the desired result if one 3

2

gn = O(2( 8 +ρ)n

+cn

),

and that an gn , cn gn ∈ Z. This finishes the proof of Theorem B.


22

PABLO A. PANZONE

Proof of Corollary. We prove the remark for the second sequence. Note that if 00 γ were a rational number then γn!an gn ∈ Z for n0 < n. Also n! = O(2c n ln n ) (0 < c00 ) and therefore by Theorem B n!(−γan gn + b0n gn + en gn + cn gn ln 2) → 0, and the assertion follows from this. For the first sequence the proof is similar and is left to the reader.

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Pablo A. Panzone Departamento e Instituto de Matem´ atica, Universidad Nacional del Sur, Av. Alem 1253, 8000 Bah´ıa Blanca, Argentina [email protected]

Received: March 28, 2014 Accepted: September 7, 2015
