de Bordeaux 19(2007), 311–323
A new lower bound for
k
(3/2)
kk
parWadim ZUDILIN
R´esum´e. Nous d´emontrons que pour tout entier k sup´erieur `a une constanteKeffectivement calculable, la distance de (3/2)k
` a l’entier le plus proche est minor´ee par 0,5803k
.
Abstract. We prove that, for all integersk exceeding some ef-fectively computable numberK, the distance from (3/2)k
to the nearest integer is greater than 0.5803k
.
1. Historical overview of the problem
Let ⌊ · ⌋ and { · } denote the integer and fractional parts of a number, respectively. It is known [12] that the inequality{(3/2)k} ≤1−(3/4)k for k ≥ 6 implies the explicit formula g(k) = 2k+⌊(3/2)k⌋ −2 for the least integer g = g(k) such that every positive integer can be expressed as a sum of at mostg positivekth powers (Waring’s problem). K. Mahler [10] used Ridout’s extension of Roth’s theorem to show that the inequality
k(3/2)kk ≤Ck, wherekxk= min({x},1− {x}) is the distance from x∈R
to the nearest integer, has finitely many solutions in integers k for any C <1. The particular case C = 3/4 gives one the above value of g(k) for allk≥K, whereK is a certain absolute but ineffective constant. The first non-trivial (i.e., C > 1/2) and effective (in terms of K) estimate of the form
(1)
3 2
k
> Ck for k≥K,
withC= 2−(1−10−64
), was proved by A. Baker and J. Coates [1] by applying
effective estimates of linear forms in logarithms in thep-adic case. F. Beuk-ers [4] improved on this result by showing that inequality (1) is valid with C = 2−0.9 = 0.5358. . . for k≥ K = 5000 (although his proof yielded the better choiceC= 0.5637. . . if one did not require an explicit evaluation of the effective bound forK). Beukers’ proof relied on explicit Pad´e approxi-mations to a tail of the binomial series (1−z)m=Pmn=0 mn(−z)nand was later used by A. Dubickas [7] and L. Habsieger [8] to derive inequality (1) with C = 0.5769 and 0.5770, respectively. The latter work also includes
the estimate k(3/2)kk > 0.57434k for k ≥ 5 using computations from [6] and [9].
By modifying Beukers’ construction, namely, considering Pad´e approxi-mations to a tail of the series
(2) 1
(1−z)m+1 =
∞ X
n=0
m+n
m
zn
and studying the explicitp-adic order of the binomial coefficients involved, we are able to prove
Theorem 1. The following estimate is valid:
3 2
k
>0.5803k= 2−k·0.78512916... for k≥K,
where K is a certain effective constant.
2. Hypergeometric background
The binomial series on the left-hand side of (2) is a special case of the generalized hypergeometric series
(3) q+1Fq
A0, A1, . . . , Aq
B1, . . . , Bq z
=
∞ X
k=0
(A0)k(A1)k· · ·(Aq)k k!(B1)k· · ·(Bq)k
zk,
where
(A)k =
Γ(A+k) Γ(A) =
(
A(A+ 1)· · ·(A+k−1) ifk≥1,
1 ifk= 0,
denotes the Pochhammer symbol (or shifted factorial). The series in (3) converges in the disc |z|< 1, and if one of the parameters A0, A1, . . . , Aq is a non-positive integer (i.e., the series terminates) the definition of the hypergeometric series is valid for allz∈C.
In what follows we will often use the q+1Fq-notation. We will require two classical facts from the theory of generalized hypergeometric series: the Pfaff–Saalsch¨utz summation formula
(4) 3F2
−n, A, B C,1 +A+B−C−n
1
= (C−A)n(C−B)n (C)n(C−A−B)n
(see, e.g., [11], p. 49, (2.3.1.3)) and the Euler–Pochhammer integral for the Gauss2F1-series
(5) 2F1
A, B
C z
= Γ(C)
Γ(B)Γ(C−B) Z 1
0
tB−1(1−t)C−B−1(1−zt)−Adt,
3. Pad´e approximations of the shifted binomial series
Fix two positive integersaand bsatisfying 2a≤b. Formula (2) yields
This motivates (cf. [4]) constructing Pad´e approximations to the function
(7) F(z) =F(a, b;z) =
Remark 1. The connection of F(a, b;z) with Beukers’ auxiliary series H(ea,eb;z) from [4] is as follows:
Taking an arbitrary integer n satisfying n ≤ b, we follow the general recipe of [5], [13]. Consider the polynomial
of degreen. Then
Here the polynomial (10)
has degree at mostn, while the coefficients of the remainder
Rn(z) =
∞ X
l=n rlzl−n
are of the following form:
The assumed condition n ≤ b guarantees that the coefficients rl do not vanish identically (otherwise (−b)n = 0). Moreover, (n−l)n = 0 for l ranging over the set n≤l≤2n−1, therefore rl= 0 for those l, while
rl=
(a+ 2n−1)!(a+b+l)! (a+n−1)!(a+l)!n!b!
×(a+b+l)!/(ba!/+(bb−+nl)!·(l−n)!/(l−2n)!
−n)!·(a+ 2n−1)!/(a+n−1)!
= (a+b+l−n)!(l−n)!
n!(b−n)!(a+l)!(l−2n)! for l≥2n. Finally,
Rn(z) =
∞ X
l=2n
rlzl−n=zn
∞ X
ν=0
rν+2nzν (11)
=zn 1
n!(b−n)! ∞ X
ν=0
(a+b+n+ν)!(n+ν)! ν!(a+ 2n+ν)! z
ν
=zn
a+b+n b−n
·2F1
a+b+n+ 1, n+ 1 a+ 2n+ 1
z
.
Using the integral (5) for the polynomial (8) and remainder (11) we arrive at
Lemma 1. The following representations are valid:
Qn(z−1) =
(a+b+n)! (a+n−1)!n!(b−n)!
Z 1
0
ta+n−1(1−t)b−n(1−z−1t)ndt
and
Rn(z) =
(a+b+n)! (a+n−1)!n!(b−n)!z
nZ 1
0
tn(1−t)a+n−1(1−zt)−(a+b+n+1)dt.
We will also require linear independence of a pair of neighbouring Pad´e approximants, which is the subject of
Lemma 2. We have
(12) Qn+1(x)Pn(x)−Qn(x)Pn+1(x) = (−1)n
a+ 2n+ 1 a+n
a+b+n b−n
x.
Proof. Clearly, the left-hand side in (12) is a polynomial; its constant term is 0 sincePn(0) =Pn+1(0) = 0 by (10). On the other hand,
Qn+1(z−1)Pn(z−1)−Qn(z−1)Pn+1(z−1)
=Qn+1(z−1) Qn(z−1)F(z)−Rn(z)
−Qn(z−1) Qn+1(z−1)F(z)−Rn+1(z)
and from (8), (11) we conclude that the only negative power ofzoriginates with the last summand:
−Qn+1(z−1)Rn(z)
4. Arithmetic constituents
We begin this section by noting that, for any primep >√N,
= min µ∈Z
a+n+µ p
−
a+n
p
−
µ p
+
a+b+n p
−
a+b+µ p
−
n−µ
p
≤ min
0≤µ≤nordp
a+n+µ µ
a+b+n n−µ
.
Set (15)
Φ = Φ(a, b, n) = Y p>√a+b+n
pep and Φ′= Φ′(a, b, n) = Y
p>√a+b+n pe′p.
From (8), (10) and (13), (15) we deduce
Lemma 3. The following inclusions are valid:
Φ−1·
a+n−1 +µ µ
a+b+n n−µ
∈Z for µ= 0,1, . . . , n,
hence
Φ−1Qn(x)∈Z[x] and Φ−1Pn(x)∈Z[x].
Supplementary arithmetic information for the case nreplaced by n+ 1 is given in
Lemma 4. The following inclusions are valid:
(16)
(n+ 1)Φ′−1·
a+n+µ µ
a+b+n+ 1 n+ 1−µ
∈Z for µ= 0,1, . . . , n+ 1,
hence
(n+ 1)Φ′−1Qn+1(x)∈Z[x] and (n+ 1)Φ′−1Pn+1(x)∈Z[x].
Proof. Write
a+n+µ µ
a+b+n+ 1 n+ 1−µ
=
a+n+µ µ
a+b+n n−µ
·a+n+ 1b+n+ 1 −µ .
Therefore, ifp∤n+ 1−µ then (17)
ordp
a+n+µ µ
a+b+n+ 1 n+ 1−µ
≥ordp
a+n+µ µ
a+b+n n−µ
otherwise µ≡n+ 1 (modp), henceµ/p−(n+ 1)/p∈Zyielding
ordp
a+n+µ µ
a+b+n+ 1 n+ 1−µ
(18)
=−
a+n+µ p
+
a+n
p
+
µ p
−
a+b+n+ 1 p
+
a+b+µ p
+
n+ 1−µ p
=−
a+ 2n+ 1 p
+
a+n
p
+
n+ 1 p
= ordp
a+ 2n+ 1 n+ 1
= ordp
a+ 2n n
+ ordp
a+ 2n+ 1 n+ 1
= ordp
a+n+µ µ
a+b+n n−µ
µ=n
+ ordp
a+ 2n+ 1 n+ 1
≥e′p−ordp(n+ 1).
Combination of (17) and (18) gives us the required inclusions (16).
5. Proof of theorem 1
The parametersa,band nwill now depend on an increasing parameter m∈Nin the following way:
a=αm, b=βm, n=γm or n=γm+ 1,
where the choice of the positive integersα,β andγ, satisfying 2α≤β and γ < β, is discussed later. Then Lemma 1 and Laplace’s method give us
C0(z) = lim
m→∞
log|Rn(z)| m (19)
= (α+β+γ) log(α+β+γ)−(α+γ) log(α+γ)
−γlogγ−(β−γ) log(β−γ) +γlog|z| + max
0≤t≤1Re γlogt+ (α+γ) log(1−t)−(α+β+γ) log(1−zt)
and
C1(z) = lim
m→∞
log|Qn(z−1)| m (20)
= (α+β+γ) log(α+β+γ)−(α+γ) log(α+γ)
−γlogγ−(β−γ) log(β−γ) + max
0≤t≤1Re (α+γ) logt+ (β−γ) log(1−t) +γlog(1−z
In addition, from (13)–(15) and the prime number theorem we deduce that
(21)
C2 = lim
m→∞
log Φ(αm, βm, γm)
m =
Z 1
0
ϕ(x) dψ(x),
C2′ = lim m→∞
log Φ′(αm, βm, γm)
m =
Z 1
0
ϕ′(x) dψ(x),
where ψ(x) is the logarithmic derivative of the gamma function and the 1-periodic functionsϕ(x) and ϕ′(x) are defined as follows:
ϕ(x) = min
0≤y<1ϕb(x, y), ϕ
′(x) = min
0≤y<1ϕb
′(x, y),
b
ϕ(x, y) =−{−(α+γ)x}+{−(α+γ)x−y}+{y}
− {(α+β+γ)x}+{(α+β)x+y}+{γx−y}, b
ϕ′(x, y) =−{(α+γ)x+y}+{(α+γ)x}+{y}
− {(α+β+γ)x}+{(α+β)x+y}+{γx−y}.
Lemma 5. The functions ϕ(x) and ϕ′(x) differ on a set of measure 0, hence
(22) C2 =C2′.
Remark 2. The following proof is due to the anonymous referee. In an earlier version of this article, we verify directly assumption (22) for our specific choice of the integer parametersα,β and γ.
Proof. First note that ϕb(x,0) = ϕb′(x,0) ∈ {0,1}, which implies ϕ(x) ∈
{0,1}and ϕ′(x)∈ {0,1}.
From the definition we see that
∆(x, y) =ϕb(x, y)−ϕb′(x, y) =λ((α+γ)x)−λ((α+γ)x+y). Assume that (α+γ)x /∈Z; we plainly obtain ∆(x, y) =−λ((α+γ)x+y)≤0, hence ∆(x, y) = 0 unless y =−{(α+γ)x}. Therefore the only possibility to getϕ(x)6=ϕ′(x) is the following one:
(23) ϕb(x, y)≥1 fory6=−{(α+γ)x} and b
ϕ(x, y) = 0 fory=−{(α+γ)x}.
Sinceϕb′(x,0)≥1, we have{(α+β)x}+{γx}= 1; furthermore,
{(α+β)x+y}+{γx−y} − {(α+β+γ)x}
=
and
{−(α+γ)x−y}+{y} − {−(α+γ)x}= (
0 if 0≤y≤ {−(α+γ)x}, 1 if{−(α+γ)x}< y <1. The above conditions (23) onϕb(x, y) imply 1− {(α+β)x}={−(α+γ)x} or, equivalently, (β−γ)x∈Z.
Finally, the sets{x∈R: (α+γ)x∈Z}and{x∈R: (β−γ)x∈Z}have
measure 0, thus proving the required assertion.
Our final aim is estimating the absolute value ofεk from below, where
3 2
k
=Mk+εk, Mk∈Z, 0<|εk|< 1 2.
Write k ≥ 3 in the form k = 3(βm+ 1) +j with non-negative integers m and j < 3β. Multiply both sides of (9) by Φe−13b−2a+j+1, where Φ =e
Φ(αm, βm, γm) ifn=γmandΦ = Φe ′(αm, βm, γm)/(γm+1) ifn=γm+1, and substitutez= 1/9:
Qn(9)eΦ−12j·
3 2
j
3b−2a+1F
a, b;1 9
(24)
=Pn(9)Φe−13b−2a+j+1+Rn
1 9
e
Φ−13b−2a+j+1.
From (6), (7) we see that
3 2
j
3b−2a+1F
a, b;1 9
≡
3 2
3(b+1)+j
(modZ) =
3 2
k ,
hence the left-hand side equalsM′
k+εkfor someMk′ ∈Zand we may write equality (24) in the form
(25) Qn(9)Φe−12j·εk=Mk′′+Rn
1 9
e
Φ−13b−2a+j+1,
where
Mk′′ =Pn(9)Φe−13b−2a+j+1−Qn(9)Φe−12jMk′ ∈Z
by Lemmas 3 and 4. Lemma 2 guarantees that, for at least one ofn=γmor γm+1, we haveMk′′6= 0; we make the corresponding choice ofn. Assuming furthermore that
(26) C0
1 9
−C2+ (β−2α) log 3<0,
from (19) and (21) we obtain
Rn
1 9
e
Φ−13b−2a+j+1 <
1
where N1 > 0 is an effective absolute constant. Therefore, by (25) and hence from (19), (20) we conclude that
|εk|> parametersα, β, γ, at least under the restrictionβ ≤100) we find that
C0 0 otherwise,
hence
C2=C2′ = 4.46695926. . . .
Using these computations we verify (26),
C0
6. Related results
The above construction allows us to prove similar results for the se-quencesk(4/3)kk and k(5/4)kk using the representations
4 3
2(b+1)
= 2b+1
1 +1 8
−(b+1)
(28)
≡(−1)a2b−3a+1F
a, b;−1 8
(modZ),
where 3a≤b, and
5 4
7b+3
= 2·5b
1 + 3 125
−(2b+1)
(29)
≡(−1)a2·3a·5b−3aF
a,2b;− 3 125
(modZ),
where 3a≤b.
Namely, taking a= 5m, b= 15m, n= 6m(+1) in case (28) and a= 3m, b= 9m,n= 7m(+1) in case (29) and repeating the arguments of Section 5, we arrive at
Theorem 2. The following estimates are valid:
4 3
k
>0.4914k= 3−k·0.64672207... for k≥K1,
5 4
k
>0.5152k= 4−k·0.47839775... for k≥K2,
where K1, K2 are certain effective constants.
The general case of the sequencek(1 + 1/N)kkfor an integerN ≥5 may be treated as in [4] and [2] by using the representation
N + 1
N
b+1
=
1− 1
N + 1
−(b+1) ≡F
0, b; 1 N+ 1
(modZ).
The best result in this direction belongs to M. Bennett [2]: k(1 + 1/N)kk> 3−k for 4≤N ≤k3k.
We would like to conclude this note by mentioning that a stronger argu-ment is required to obtain the effective estimatek(3/2)kk>(3/4)k and its relatives.
Acknowledgements. I am very grateful to the anonymous referee for his valuable comments (incorporated into the current presentation as Re-marks 1–3), providing the proof of Lemma 5 and indicating misprints. My profound gratitude goes to Anne de Roton, for supporting my talk at the
24th Journe´es Arithmetiques as a muse. I thank kindly J. Sondow for several suggestions.
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WadimZudilin
Department of Mechanics and Mathematics Moscow Lomonosov State University Vorobiovy Gory, GSP-2
