CHAPTER V
SOME HYPERGEOMETRIC IDENTITIES ARE ALMOST TRIVIAL
Zeilberger [Z1] once claimed, All binomial identities are verifiable. His reasoning went
as follows. Let F (n, k) be a hypergeometric term. Then
P
k
with polynomial-in-n coefficients. If we want to prove that
F (n, k) satisfies a recurrence
P
k
F (n, k) = f (n) for some
hypergeometric term f (n), then we need to check that f (n) satisfies the same recurrence
as
P
k
F (n, k), and f (n) agrees with
P
k
F (n, k) for all relevant n’s less than or equal to
the sum of the order of the recurrence and the highest integer zero of the leading coefficient
of the recurrence. Since there are only a finite number of cases to check, it is sufficient to
verify
P
k
F (n, k) = f (n) with a pocket calculator.
However, no a priori bounds for the recurrence or the highest integer root of the leading
coefficient were known at the time of the paper. In order to obtain an effective algorithm
for proving a hypergeometric identity, we make use of the mathematical tools developed
in [WZ3]. Using the terminology of [WZ3], we consider only admissible proper-hypergeometric terms to obtain a recurrence of the sum from that of the summand. In this chapter,
we consider the case with one summation index, and calculate an a priori bound for the
number of n’s for which the hypergeometric identity
P
k
F (n, k) = f (n) should be checked
to establish the truth of the identity. These a priori bounds are quite astronomical in size,
but they are finite, and pre-computable. (See the end of this chapter for examples of the
sizes of the bounds.)
42
V. SOME HYPERGEOMETRIC IDENTITIES ARE ALMOST TRIVIAL
43
Main Theorem. Let
Qp
(as n + bs k + cs )! k
ξ
F (n, k) = P (n, k) Qqs=1
(u
s=1 s n + vs k + ws )!
be an admissible proper-hypergeometric term, and P (n, k) be a polynomial with coefficients
in Z. Let
x := max{|as |, |bs |, |cs |, |us |, |vs |, |ws |},
s
y := max{p, q},
¯
¯
z := max ¯[nj k i ]P (n, k)¯ ,
0≤i,j
d := 1 + max{degk P (n, k), degn P (n, k)},
and let n0 be a given integer. If
P
k
F (n, k) = 1 for
2
n0 ≤ n ≤ (3xy)3(d+1)
then
P
k
(2xy)6 5(d+1)(2xy)3 (d+1)(2xy)3
d
z
,
F (n, k) = 1 for all n ≥ n0 .
Note that if we would like to prove that
P
k
F (n, k) = f (n) for some hypergeometric
term f (n), then dividing both sides by f (n), we get
P
k
F (n, k)/f (n) = 1. What remains
is to check whether F (n, k)/f (n) is an admissible proper-hypergeometric term before we
can apply the theorem.
We first state and prove the following theorem which contains a much better bound,
but the bound is in an even more complicated form (5.13).
Theorem 5.1. Let
(5.1)
Qp
(as n + bs k + cs )! k
F (n, k) = P (n, k) Qqs=1
ξ
s=1 (us n + vs k + ws )!
5.1 EXAMPLES
44
be an admissible proper-hypergeometric term, and P (n, k) be a polynomial with coefficients
in Q. Then, given n0 , there exists an effectively computable positive integer n1 such that
if
P
k
F (n, k) = 1 for all n0 ≤ n < n1 , then
P
k
F (n, k) = 1 for all n ≥ n0 . (See (5.13)
for n1 .)
First we claim that it suffices to prove Theorem 5.1 for those polynomials P (n, k) with
integer coefficients. For if P (n, k) is a polynomial with coefficients in Q, then P (n, k) =
P̃ (n, k)/dp where P̃ (n, k) is a polynomial with integer coefficients and dp is the least
common multiple of the denominators of the coefficients of P (n, k). But in order to prove
that
X
F (n, k) =
k
X
k
Qp
(as n + bs k + cs )! k
P (n, k) Qqs=1
ξ = 1,
s=1 (us n + vs k + ws )!
it is equivalent to prove that
X
k
Qp
(as n + bs k + cs )! k
Q
ξ = dp .
P̃ (n, k) qs=1
s=1 (us n + vs k + ws )!
5.1 Examples
In this section, we show some examples of hypergeometric identities whose leading
coefficients, a0 (n), in the recurrence equations vanish at embarrassing places of n, namely
those n where the hypergeometric identities hold.
Why does a vanishing leading coefficient, a0 (n), in the recurrence relation present a
problem? Given a proper-hypergeometric term F (n, k) and an integer n0 , suppose we
want to prove
P
k
F (n, k) = 1 for all n ≥ n0 . We know that
P
k
F (n, k) satisfies some
k-free recurrence of the form
a0 (n)
X
k
F (n, k) + a1 (n)
X
k
F (n − 1, k) + · · · + aJ (n)
X
k
F (n − J, k) = 0,
n ≥ n0 ,
5.1 EXAMPLES
for some positive integer J ≤
P
s
|bs | +
P
s
45
|vs |. (See [WZ3] Theorem 3.1, or a sharper
bound in Chapter 1.) It is easy to check if 1 (the RHS) also satisfies the same recurrence.
That is, we need to check if
(5.2)
a0 (n) + a1 (n) + · · · + aJ (n) = 0
for all n. To do so, we use the fact that if a polynomial, P , of degree d has d + 1 zeros,
then P = 0. Thus it suffices to check that (5.2) is true for 1 + maxj∈[J]0 deg aj (n) different
values of n. In addition, if a0 (n) 6= 0 for all n ≥ n0 , then we can divide by a0 (n) to get
(5.3)
X
F (n, k)
k
=−
a1 (n)
P
k
F (n − 1, k) + a2 (n)
P
k
Now, (5.3) is a Jth order recurrence in n for
F (n − 2, k) + · · · + aJ (n)
a0 (n)
P
k
F (n, k), so if
P
k
P
k
F (n − J, k)
.
F (n, k) = 1 for n0 ≤
©
ª
n ≤ max n0 + J − 1, n0 + maxj∈[J]0 deg aj (n) , then it follows (by induction and using
(5.2)) that
P
k
F (n, k) = 1 for all n ≥ n0 .
However, if a0 (n) vanishes for some n ≥ n0 , then (5.2) and (5.3) does not hold at
that particular n. In order to use the recurrence relation to establish the identity, we
need to know an integer na ≥ J such that a0 (n) does not vanish for all n ≥ na , and
check individually that
ª
maxj∈[J]0 deg aj (n)} .
P
k
©
F (n, k) = 1 for all n ∈ n0 , n0 + 1, . . . , max{na − 1 + J, n0 +
¤
The first example shows that n1 in Theorem 5.1 depends on the coefficients of P (n, k).
Example 1. Suppose we wanted to evaluate the sums
sn =
X
k
µ ¶
n
(k − 9k + 4)
,
k
2
n ≥ 0.
5.1 EXAMPLES
46
We show that sn satisfies the following recurrence relation,
(n − 16)(n − 1)sn+1 − 2n(n − 15)sn = 0.
Let
µ ¶
n
F (n, k) = (k − 9k + 4)
,
k
2
and
fn (x) =
X
F (n, k)xk .
k
Then
µ ¶
n k
(k 2 − 9k + 4)
x
k
k
¡
¢
= (1 + x)n−2 n(n − 1)x2 + nx(1 + x) − 9nx(1 + x) + 4(1 + x)2 .
fn (x) =
X
Therefore sn = fn (1) = 2n−2 (n2 − 17n + 16), and sn+1 = fn+1 (1) = 2n−1 ((n + 1)2 −
17(n + 1) + 16). From the expressions of sn and sn+1 , we conclude that our desired sums
sn satisfy the recurrence
(n − 16)(n − 1)sn+1 − 2n(n − 15)sn = 0,
or equivalently,
sn+1 =
2n(n − 15)
sn
(n − 16)(n − 1)
for
n > 16.
Notice that the recurrence relation can be used to calculate sn successively only if n > 16,
because the leading coefficient vanishes at n = 16 and n = 1. Thus, if we check individually
that fn (1) =
P
k
F (n, k) for n = 0, 1, . . . , 16, 17, then we can use the recurrence relation
to calculate sn for n ≥ 18. In this example, n1 of Theorem 5.1 is 18.
¤
Example 2. In this example, we show that n1 depends on the coefficients of P (n, k) and
might be arbitrarily large.
5.1 EXAMPLES
47
Fix a large n1 ∈ N. We consider more generally,
µ ¶
X
n
and fn (x) =
F (n, k)xk .
F (n, k) = (ak 2 + bk + c)
k
k
We take the given n1 and find a, b, c in Q such that the sum
sn =
X
F (n, k) = fn (1)
k
satisfies a recurrence relation whose leading coefficient, a0 (n) vanishes at n1 − 2.
After a similar calculation as the one in Example 1, we obtain the recurrence
(an2 + n(a + 2b) + 4c)sn+1 − 2(a(n + 1)2 + (n + 1)(a + 2b) + 4c)sn = 0.
The coefficient of sn+1 vanishes at
n=
−(a + 2b) ±
p
(a + 2b)2 − 16ac
.
2a
If the discriminant is greater than 0, then the larger of the two roots is
p
−(a + 2b) + (a + 2b)2 − 16ac
∗
n =
.
2a
If we find some positive integers α and β such that gcd(α, β) = 1 and α > β satisfying
(a + 2b)2 = (α + β)2
16ac = 4αβ
simultaneously, then
p
p
(a + 2b)2 − 16ac = (α − β)2 = α − β.
In this case,
n∗ =
(α + β) + (α − β)
α
= .
2a
a
Since we want n∗ = n1 − 2, we can take a = 1, α = n1 − 2, β = 1. It follows that
b=
and n∗ = α = n1 − 2. ¤
−(α + β) − 1
,
2
c=
αβ
,
4
5.2 TWO APPROXIMATION LEMMAS
48
Example 3. This example shows that even if the summand F (n, k) consists only of
factorial parts, and does not have a polynomial part, then it may happen that the leading
coefficient of the recurrence satisfied by the sum vanishes at some positive integers n where
P
k
F (n, k) is summable for those integers.
Fix a large n1 ∈ N, and take the summand in Saalschütz’ identity,
F (n, k) =
(a + k − 1)! (b + k − 1)! n! (−a − b + c + n − 1 − k)!
.
k! (n − k)! (c + k − 1)!
Then Saalschütz says that
X
F (n, k) =
k
(c − a)n (c − b)n
=: fn .
cn (c − a − b)n
It is easy to check that F (n, k)/fn satisfies the hypothesis of Theorem 5.1. The recurrence
for fn is
(n + c)(n + c − a − b)fn+1 − (n + c − a)(n + c − b)fn = 0.
It suffices to pick a, b in Z− and c := −(n1 − 2). ¤
5.2 Two approximation lemmas
Notation. We use [n] to denote {1, 2, . . . , n}, [n]0 to denote {0} ∪ [n], and [xm y n ]P (x, y)
to denote the coefficient of xm y n in P (x, y). We use P (n, k) 4 Q(n, k) to mean that for
all pairs of integers, (m, l), |[nm k l ]P (n, k)| ≤ |[nm k l ]Q(n, k)|.
We need the following lemmas for the proof of Theorem 5.1.
Lemma 5.2. Let P (n, k) be a polynomial in n and k with integer coefficients, and let
µ=
max
l∈[E]0 ,m∈[D]0
|[nm k l ]P (n, k)|,
D = degn P (n, k),
and
E = degk P (n, k).
5.2 TWO APPROXIMATION LEMMAS
49
Then for every positive integer J,
max
l∈[E]0 ,m∈[D]0 ,j∈[J]0
Proof. Suppose P (n, k) =
|[nm k l ]P (n − j, k)| ≤ (1 + J)D µ.
PE PD
m=0 tlm k
l=0
l m
n . Then, |[k l nm ]P (n − j, k)| is
¯D−m
µ
¶ ¯¯ X
D µ ¶
¯X
m
+
i
D i
¯
i
i¯
(−1) tl,m+i
j ¯≤
J
max
|[nm k l ]P (n, k)|
¯
¯
¯
m
i
l∈[E]0 ,m∈[D]0
i=0
i=0
= (1 + J)D µ
for all j ∈ [J] ∪ {0}, l ∈ [E]0 and m ∈ [D]0 .
Lemma 5.3. Let Q(n, k) =
Qq
s=1 (as n
¤
+ bs k + cs ), where as , bs , cs are integers. Then
max |[nm k l ]Q(n, k)| < 3q
m,l∈[q]0
Y
max{|as |, |bs |, |cs |}.
s∈[q]
Proof. We know that
q
Y
Q(n, k) =
(as n + bs k + cs )
s=1
4 (n + k + 1)q
Y
max{|as |, |bs |, |cs |}.
s∈[q]
Since the absolute value of the largest coefficient of (n + k + 1)q is the trinomial coefficient
¡
¢
q
b q3 c,d q3 e,q−b q3 c−d q3 e
, we get that
µ
¶ Y
q
max{|as |, |bs |, |cs |}
max |[n k ]Q(n, k)| ≤
b 3q c, d 3q e, q − b 3q c − d 3q e
m,l∈[q]0
s∈[q]
Y
< 3q
max{|as |, |bs |, |cs |}. ¤
m l
s∈[q]
5.3 SOLVING A HOMOGENEOUS SYMBOLIC LINEAR SYSTEM
50
5.3 Solving a homogeneous symbolic linear system
Definition. Let M be an l × m matrix over the field of rational functions over Q. Define
the generic rank of M to be the number of non-zero rows in the reduced row-echelon form
of M . Since row rank (M ) = column rank (M ) = rank (M ) from [H, p. 337], the generic
rank is the classical definition of the rank of a matrix over a division ring. Henceforth we
use rank to mean the generic rank.
In this section, we consider a special class of matrices M , l × m such that l < m, and
Mij , the entries of M , are polynomials in x with integer coefficients. Since M is a subset
of the matrices over the field of rational functions, the rank of M is well-defined.
Let x be an m × 1 vector with indeterminate polynomial entries, {an (x)}m
1 , with integer
coefficients. The problem is to solve for x in M x = 0, for some M of non-zero rank. After
obtaining a solution for x, we estimate the degree and the largest coefficient of an (x), for
n ∈ [m].
The following is a procedure for finding x. Let M x = 0 be a system of homogeneous
linear equations such that
(1) M is l × m, l < m,
(2) entries of M are polynomials in x with integer coefficients,
(3) M has rank ρ > 0,
(4) xt = (a1 (x), . . . , am (x)),
(5) wlog, assume a1 (x) is not identically zero.
Then x can be obtained from the following procedure.
Step A. By renumbering the unknowns, if necessary, and permuting the columns of M ,
5.4 SUFFICIENT CONDITIONS FOR A POLYNOMIAL NOT TO VANISH
51
we can arrange that the first ρ columns of M have rank ρ.
Step B. Interchange the rows of the resulting matrix from (1) above to make the ρ × ρ
upper left hand corner of M , called M 0 , a square matrix of rank ρ.
Step C. Set all but the first ρ variables in the new x to 1.
Step D. What remains is a system of ρ inhomogeneous linear equations in x1 , x2 , . . . , xρ ,
say M 0 x0 = y0 . We note that y0 6= 0. For if y0 = 0, and M 0 is of full rank, then
the only solution to M 0 x0 = 0 is the zero solution. But x0 has a1 (x) as its first
member, and a1 (x) is assumed to be non-zero.
Step E. Use Cramer’s rule to find the unknowns x0 , namely:
the nth entry of x0 =
det Mn0
det M 0
(n = 1, . . . , ρ).
In particular,
a1 (x) =
det M10
.
det M 0
Step F. To make the solution for x a polynomial solution, we multiply x by det M 0 . Since
M 0 is obtained from M by interchanging rows and columns, the entries of M 0
are still polynomials with integer coefficients. Therefore det M 0 is a polynomial
over Z. Similarly, each Mn0 has entries over Z[x], for entries of y0 are sums of
some entries in M . Thus the complete solution vector is
the new x = (det M10 , . . . , det Mρ0 , det M 0 , . . . , det M 0 )t .
5.4 Sufficient conditions for a polynomial not to vanish
Given {an (x)}m
1 from the end of §5.3, we find in this section upper bounds for the
degrees and the largest coefficient of the {an (x)}m
1 . With these bounds, we apply
5.4 SUFFICIENT CONDITIONS FOR A POLYNOMIAL NOT TO VANISH
52
Proposition 5.4. Let a(x) ∈ Z[x], let d = deg a, and let m be maxi∈[d]0 |[xi ]a(x)|. Then
a(x) 6= 0 for all x > md.
Proof. Let
a(x) = a0 + a1 x + a2 x2 + · · · + ad xd ,
ad 6= 0,
(all ai ∈ Z).
Then for sufficiently large x,
|a(x)| ≥ xd |ad | − |xd−1 ad−1 + · · · + a0 |
≥ xd |ad | − xd−1 d · max{|a0 |, |a1 |, . . . , |ad−1 |}
= xd−1 (x|ad | − d · max {|aj |}) > 0,
j∈[d−1]0
if
x>
d maxj∈[d−1]0 |aj |
.
|ad |
Since ad 6= 0 and ad ∈ Z, the latter surely holds if x > md. ¤
We estimate the degree of an (x), n ∈ [m], from det M 0 and det Mn0 , the polynomial
solutions for x.
Let
µ := max {deg Mij (x) | i ∈ [l], j ∈ [m] } ,
then deg det M 0 (x) ≤ ρµ for the following reason. Since M 0 is obtained from M by inter0
≤ µ, for i, j ∈ [ρ]. We conclude that deg(det M 0 (x)) ≤
changing rows and columns, deg Mij
ρµ because rank(M 0 ) = ρ. Similarly, deg(det Mn0 (x)) ≤ ρµ because Mn0 is obtained from
M by interchanging rows and columns of M , and by adding some columns of M together
to make y0 , then replacing the nth column of M 0 by y0 to get Mn0 .
5.5 THE LEADING COEFFICIENT, a0 (n), OF THE RECURRENCE
53
Next we estimate the heights of the an (x)’s, i.e., maxk∈[ρµ] |[xk ]an (x)| for n ∈ [m]. Let
¯
©¯
ª
c := max ¯[xk ]Mij (x)¯ | i ∈ [l], j ∈ [m], k ∈ [µ] ,
then
0
max |[xk ]Mij
(x)| ≤ c and
i,j,k
max |[xk ](Mn0 )ij (x)| ≤ (m − ρ)c,
i,j,k
again from the ways M 0 and Mn0 are obtained from M . With an upper bound for the
maximum coefficient of M 0 , we estimate maxk |[xk ] det M 0 (x)| using the definition of the
determinant. By the definition of the determinant, we have
det M 0 =
X
sgn(σ)e1σ(1) e2σ(2) . . . eρσ(ρ)
σ∈Sρ
4
X
|e1σ(1) e2σ(2) . . . eρσ(ρ) |
σ∈Sρ
4 ρ! cρ (xµ + xµ−1 + · · · + x + 1)ρ .
Thus
¯
¯
max ¯[xk ] det M 0 ¯ ≤ ρ! cρ (µ + 1)ρ .
k
Similarly,
¯
¯
max ¯[xk ] det Mn0 ¯ ≤ ρ! ((m − ρ)c)ρ (µ + 1)ρ .
k
5.5 The leading coefficient, a0 (n), of the recurrence
We estimate the degree of the leading coefficient, a0 (n), in the recurrence of F (n, k)
as a polynomial in n, and na , the positive integer with the property that for all n ≥ na ,
a0 (n) 6= 0. The plan for achieving this goal consists of the following four stages:
Stage 1. Take a given admissible proper-hypergeometric term F (n, k), and use Theorem
3.2A of [WZ3] to say that F (n, k) satisfies a recurrence of the form:
(5.4) a0 (n)F (n, k) + a1 (n)F (n − 1, k) + · · · + aJ (n)F (n − J, k) = G(n, k) − G(n, k − 1),
5.5 THE LEADING COEFFICIENT, a0 (n), OF THE RECURRENCE
54
where the aj (n)’s are unknown polynomials in n. Divide (5.4) by F (n, k) and
put the resulting sum of rational functions over a common denominator.
Stage 2. Equate the coefficient of each power of k in the common numerator to 0, and
solve the resulting homogeneous linear equations for the unknowns aj (n)’s and
ci (n)’s (see (5.5) for ci (n)’s) by Cramer’s rule for a0 (n) only, in the form
a0 (n) =
det M10
.
det M 0
(See §5.3 Step E.)
Stage 3. Observe that a0 (n) = 0 exactly when det M10 = 0. Therefore, we express
det M10 as a polynomial in n, and obtain an upper bound for the degree of
det M10 (see §5.6 formula (5.11)) and the largest coefficient of det M10 (see §5.6
formula (5.12)).
Stage 4. Use the simple fact that if a(x) is a polynomial over Z, d is the degree of a(x)
and m is maxi∈[d]0 |[xi ]a(x)|, then a(x) 6= 0 for all x > md. (See Proposition
5.4 in §5.4.) Thus we use the estimates in Stage 3 to obtain an na such that
for all n > na , a0 (n) 6= 0.
We now proceed to do Stage 1 of the plan in detail. Let an admissible proper-hypergeometric term F (n, k) be given such that P (n, k) in F (n, k) has integer coefficients. Recall
that
Qp
(as n + bs k + cs )! k
F (n, k) = P (n, k) Qqs=1
ξ .
(u
s=1 s n + vs k + ws )!
Then Theorem 3.2A of [WZ3] guarantees us the existence of polynomials a0 (n), a1 (n), . . . ,
aJ (n), not all zero, an integer, J ≤
P
s
|bs | +
P
s
|vs |, and a function G(n, k) such that
5.5 THE LEADING COEFFICIENT, a0 (n), OF THE RECURRENCE
55
G(n, k) = R(n, k)F (n, k) for some rational function R and such that
(5.4) a0 (n)F (n, k) + a1 (n)F (n − 1, k) + · · · + aJ (n)F (n − J, k) = G(n, k) − G(n, k − 1).
Without loss of generality, assume a0 (n) is not identically zero. From Chapter 2, we may
assume that R(n, k) has the form
PN
i=0 ci (n)k
i
DR (n, k)
for some polynomials, ci (n) (i ∈ [N]0 ), where
N = degk P (n, k) + J(A + (U − A)+ ) + (I − 1)(B + (V − B)+ ).
(See Theorem 1.4 for the definitions of A, B, A, B, U and V .) Dividing both sides of (5.4)
by F (n, k), we get
(5.5) a0 (n) + a1 (n)
F (n − 1, k)
F (n − J, k)
+ · · · + aJ (n)
F (n, k)
F (n, k)
PN
PN
i
ci (n)(k − 1)i F (n, k − 1)
i=0 ci (n)k
=
− i=0
.
DR (n, k)
DR (n, k − 1)
F (n, k)
Putting (5.5) over a common denominator D(n, k), we find that we can take
D(n, k) := P (n, k)
p
Y
(as )+ J+(bs )+ I
(as n + bs k + cs )
s=1
q
Y
+ J+(−v
(us n + vs k + ws + 1)(−us )
s=1
×
p
Y
(as n + bs k + cs − bs + 1)
(bs )+
s=1
q
Y
(−vs )+
(us n + vs k + ws − vs )
s=1
Next we collect all terms of (5.5) to the left side to get
(5.6)
+
s) I
L0 + L1 + · · · + LJ − R1 + R2
= 0,
D(n, k)
.
5.5 THE LEADING COEFFICIENT, a0 (n), OF THE RECURRENCE
where for 0 ≤ j ≤ J,
Lj (n, k) := aj (n)P (n − j, k)
×
Y
Y
+
(as n + bs k + cs + 1)j(−as )
s
(as n + bs k + cs − (as )+ J + 1)(J−j)(as )
+
s
×
Y
(as n + bs k + cs − bs + 1)(bs )
+
s
×
Y
I(bs )+
(as n + bs k + cs − (as )+ J)
s
×
Y
(J−j)(−us )+
(us n + vs k + ws + (−us )+ J)
s
×
Y
(us n + vs k + ws )
j(us )+
s
×
Y
+
(us n + vs k + ws + (−us )+ J + 1)I(−vs )
s
×
Y
(−vs )+
(us n + vs k + ws − vs )
,
s
and
Ã
R1 (n, k) :=
N
X
!
ci (n)k
i
Y
+
(as n + bs k + cs − (as )+ J − (bs )+ I + 1)(bs )
s
i=0
×
Y
+
(as n + bs k + cs − bs + 1)(bs )
s
×
Y
(−vs )+
(us n + vs k + ws + (−us )+ J + (−vs )+ I)
s
×
Y
(us n + vs k + ws − vs )
(−vs )+
,
s
and
!
µ ¶
i
R2 (n, k) :=
(−1)j
ci (n)
k i−j
j
j=0
i=j
Y
Y
(bs )+
+
(as n + bs k + cs + 1)(−bs )
×
(as n + bs k + cs )
ÃN
X
N
X
s
×
Y
s
s
+
(vs )
(us n + vs k + ws )
Y
+
(us n + vs k + ws + 1)(−vs ) .
s
56
5.5 THE LEADING COEFFICIENT, a0 (n), OF THE RECURRENCE
57
We now do Stage 2 of the plan. Our goal is to find an expression for a0 (n). To solve
for the unknown polynomials in n,
a0 , a1 , . . . aJ , c0 , c1 , . . . , cN ,
we expand the terms of (5.6), and collect like powers of k. Since LHS of (5.6) is zero,
the coefficients of k l must be identically zero. This yields a system of linear homogeneous
equations. We then express the system as M x = 0, xt = (a0 , a1 , . . . , aJ , c0 , c1 , . . . , cN ),
and the ith row of M corresponds to the coefficients of k i−1 in the common numerator of
(5.6). We are now set to apply the procedure in §5.3 for finding a polynomial a0 (n) using
Cramer’s rule.
Stage 3 of the plan consists of three steps. First, we find an upper bound for the
maximum degree of the entries of M . (See Lemma 5.5.) Second, we find an upper bound
for the largest coefficient of the entries of M . (See Lemma 5.6.) With these estimates for
the entries of M , we find upper bounds for the degree and the largest coefficient of det M10
(= a0 (n)) using §5.4.
Step 1. An upper bound for the maximum degree of all entries of M regarded as a
polynomial in n.
Lemma 5.5. Let
µ1 := degn P (n, k) + (I + 1)B̃ + J(Ã + (Ũ − Ã)+ ),
where
B̃ =
X
b+
s +
s:as 6=0
Ũ =
X
s∈[q]
us ,
X
(−vs )+ ,
à =
s:us 6=0
X
a+
s +
s∈[p]
à =
X
s∈[p]
X
s∈[q]
as ,
(−us )+ ,
5.5 THE LEADING COEFFICIENT, a0 (n), OF THE RECURRENCE
and let
µ2 := max

X

|bs | +
X
|vs |,
2
us 6=0
as 6=0
X
b+
s +2
as 6=0
X
(−vs )+
us 6=0
58



.
Then
max deg Mij (n) ≤ max{µ1 , µ2 }.
i,j
Proof. Let µ be the maximum degree of the entries of M . By the setup, M1j is the
coefficient of k 0 , that is, P (n − j, k) times the first elementary symmetric function when
the product
Lj
aj (n)P (n − j, k)
is viewed as a product of terms of the type (bs k + ds (n)), where ds (n) is a polynomial in
n of degree 1 at most. Therefore, deg Mij ≤ deg M1j for 1 ≤ j ≤ J + 1. But
Ã
!
X
X
deg M1j (n) = degn P (n, k) + (J + 1 − j)
(as )+ +
(−us )+
Ã
+ (I + 1)
s
X
+
(bs ) +
as 6=0
X
+
(−vs )
!
s
Ã
X
X
+ (j − 1)
(−as )+ +
(us )+
s
us 6=0
!
,
s
for 1 ≤ j ≤ J + 1. Thus
max deg M1j (n) = degn P (n, k) + (I + 1)B̃ + J(Ã + (Ũ − Ã)+ ) = µ1 ,
j
where
B̃ =
X
b+
s +
as 6=0
Ũ =
X
X
(−vs )+ ,
à =
us 6=0
us ,
à =
X
X
a+
s +
X
(−us )+ ,
as .
For 1 ≤ i ≤ ν, J + 2 ≤ j ≤ J + 2 + N, Mij is the polynomial (in n) multiplied by
k i−1 cj−(J+2) (n). To find deg Mij , we compute R1 and R2 to conclude that


X

X
X
X
+
+
deg Mij ≤ max
= µ2 .
|bs | +
|vs |, 2
bs + 2
(−vs )


as 6=0
us 6=0
as 6=0
us 6=0
5.5 THE LEADING COEFFICIENT, a0 (n), OF THE RECURRENCE
Thus µ ≤ max{µ1 , µ2 }.
59
¤
Step 2. An upper bound for maxi,j,l |[nl ]Mi,j (n)|.
We note that Mij is the polynomial in n multiplied either by aj−1 (n)k i−1 for all i ∈ [ν]
and j ∈ [J + 1], or by cj−(J+2) k i−1 for i ∈ [ν] and j ≥ J + 2. First, we compute an upper
bound for maxi,j,l |[nl ]Mi,j (n)| for all i and j ∈ [J + 1]. The parameters in cs and ws are
fixed arbitrarily in this step.
Let
P (n, k) :=
E X
D
X
tlm nm k l ,
l=0 m=0
µ3 := (1 + J)D max{|tlm | : l ∈ [E]0 and m ∈ [D]0 },
Y
Y
s:bs >0
i∈[b+
s ]
µ4 :=
×
Y
max{|as |, |bs |, |cs − bs + i|}
Y
max{|us |, |vs |, |ws − vs + 1 − i|}
s:vs <0 i∈[(−vs )+ ]
×
Y
Y
max{|as |, |bs |, |cs − J(a+
s ) + 1 − i|}
s:bs >0 i∈[I(b+
s )]
×
Y
Y
max{|us |, |vs |, |ws + (−us )+ J + i|},
s:vs <0 i∈[I(−vs )+ ]
µ5 := max
0≤j≤J
×
Y
µ Y
Y
max{|as |, |bs |, |cs + i|}
s:as <0 i∈[j(−as )+ ]
Y
max{|as |, |bs |, |cs − Ja+
s + i|}
s:as >0 i∈[(J−j)a+
s ]
×
Y
Y
max{|us |, |vs |, |ws + J(−us )+ + 1 − i|}
s:us <0 i∈[(J−j)(−us )+ ]
×
Y
Y
¶
max{|us |, |vs |, |ws + 1 − i|} ,
s:us >0 i∈[j(us )+ ]
and
e1 := J(Ã + (Ũ − Ã)+ ) + (I + 1)(
X
s
b+
s +
X
s
(−vs )+ ).
5.5 THE LEADING COEFFICIENT, a0 (n), OF THE RECURRENCE
60
We know from Lemma 5.3 that
Lj
4 P (n − j, k)µ4 µ5 (n + k + 1)e1 .
aj
Moreover, Lemma 5.2 states that
max |[nm k l ]P (n − j, k)| ≤ µ3
m,l
for all j ∈ [J]0 . We conclude that the largest coefficient of Mij for i ∈ [ν] and j ∈ [J + 1]
is bounded above by
(D + 1)(E + 1)µ3 µ4 µ5 3e1 .
Now we find an upper bound for the largest coefficient of Mi,j+J+2 for i ∈ [ν] and
j ≥ 0. As we observed before, Mi,j+J+2 is the polynomial in n multiplied by cj k i−1 . By
expanding R1 and R2 , we get
Mi,j+J+2 = −[k
i−1−j
µ ¶
j
X
R2
l j
(−1)
[k i−1−j+l ] P
] PN
+
.
¡l¢
P
j
l−j
l
l
j (−1)
l cl j k
l=1 cl k
l=0
R1
Let
Y µ Y
µ6 :=
s:bs >0
×
max{|as |, |bs |, |cs −
−
Ib+
s
+ i|} × max{|as |, |bs |, |cs −
Y
Y
i∈[(−vs
max{|us |, |vs |, |ws − vs + 1 − i|}
¶
× max{|us |, |vs |, |ws + J(−us ) + I(−vs ) + 1 − i|} ,
Y
max{|as |, |bs |, |cs + i|}
s:bs <0 i∈[(−bs )+ ]
×
Y
+ i|}
)+ ]
+
µ7 :=
b+
s
i∈[b+
s ]
Y µ
s:vs <0
¶
Ja+
s
Y
s:vs <0 i∈[(−vs
Y
Y
max{|as |, |bs |, |cs + 1 − i|}
s:bs >0 i∈[b+
s ]
max{|us |, |vs |, |ws + i|}
)+ ]
+
Y
Y
s:vs >0
i∈[vs+ ]
max{|us |, |vs |, |ws + 1 − i|},
5.5 THE LEADING COEFFICIENT, a0 (n), OF THE RECURRENCE
61
and
e2 := 2
e3 :=
Ã
X
X
b+
s +
s
|bs | +
s
X
X
!
(−vs )+
,
s
|vs |.
s
By Lemma 5.3,
¯
Ã
!¯
¯
¯
R2
¯ m l
¯
¡i¢
max ¯[n k ] P
P
¯
j
i−j ¯
m,l ¯
(−1)
c
k
j≥0
i≥j i j
µ
resp.
¯
µ
¶¯¶
¯ m l
¯
R
1
¯
max ¯¯[n k ] P
i
m,l
ci k ¯
i
is bounded above by
3e3 µ7
(resp. 3e2 µ6 ).
Thus, from the expression of Mij , we conclude that the largest coefficient is bounded above
by
2N 3e3 µ7 + 3e2 µ6 .
Let
©
µ8 := max (D + 1)(E + 1)3e1 µ3 µ4 µ5 ,
ª
2N 3e3 µ7 + 3e2 µ6 .
Then the largest coefficient of Mij for j ∈ [N + J + 2] and i ∈ [ν] is bounded above by µ8 .
Thus we have
Lemma 5.6. The absolute value of the largest coefficient of the entries of M is bounded
by µ8 .
Step 3. Upper bounds for deg det M10 and maxi |[ni ] det M10 |.
We take the M 0 and M10 obtained from §5.3, and perform the computation of §5.4 to get
deg det M10 ≤ rank(M 0 ) × max{µ1 , µ2 },
5.6 PROOF OF THEOREM 5.1
62
where µ1 and µ2 are from Lemma 5.5, and
¡
¢ρ
max |[ni ] det M10 | ≤ ρ! (J + N + 2 − ρ)µ8 (1 + max{µ1 , µ2 }) ,
i
where ρ :=rank(M 0 ).
Stage 4 is the application of Proposition 5.4 using the bounds we just calculated in Step
3 above. We have now completed the computation needed for the proof of Theorem 5.1.
5.6 Proof of Theorem 5.1
Proof of Theorem 5.1. Let (n, k) ∈ Z2 be a point at which F (n, k) 6= 0, and such that
F (n − j, k − i) is well-defined for all i ∈ [I]0 and j ∈ [J]0 , where I and J are some integers
bounded above by the expressions found in [WZ3, Theorem 3.1] or a sharper bound from
Theorem 1.4. By Theorem 3.2A of [WZ3], there exist polynomials a0 (n), a1 (n), . . . , aJ (n)
not all identically zero, and a function G(n, k) such that G(n, k) = R(n, k)F (n, k) for some
rational function R and such that a0 (n) is not identically 0 and
(5.7) a0 (n)F (n, k) + a1 (n)F (n − 1, k) + · · · + aJ (n)F (n − J, k) = G(n, k) − G(n, k − 1).
From Step 3 of Chapter 2, we know that for
N := degk P (n, k) + J(A + (U − A)+ ) + (I − 1)(B + (V − B)+ ),
where A, B, U, V, A, B are defined in Theorem 1.4, the rational function R(n, k) in (5.7)
assumes the form
PN
i=0 ci (n)k
DR (n, k)
i
.
Substituting the expression for R(n, k) into (5.7), then dividing both sides by F (n, k), we
get an equation of rational functions
(5.8)
a0 (n) + a1 (n)
F (n − J, k)
R(n, k − 1)F (n, k − 1)
F (n − 1, k)
+ · · · + aJ (n)
= R(n, k) −
.
F (n, k)
F (n, k)
F (n, k)
5.6 PROOF OF THEOREM 5.1
63
A common denominator for (5.8) is
D(n, k) := P (n, k)
Y
(as )+ J+(bs )+ I
(as n+bs k +cs )
s∈[p]
×
Y
+ J+(−v
(us n+vs k +ws +1)(−us )
+
s) I
s∈[q]
Y
+
(as n + bs k + cs − bs + 1)(bs )
s∈[p]
Y
(−vs )+
(us n + vs k + ws − vs )
.
s∈[q]
Thus, (5.8) is equivalent to
L0 + L1 + · · · + LJ − R1 + R2
= 0,
D(n, k)
(5.9)
where Li ’s and Ri ’s are defined following (5.6) in §5.5.
Expanding (5.9), collecting the coefficients of like powers of k, and setting them to
zero, we get a system of homogeneous linear equations with unknown polynomials in n,
namely a0 , a1 , . . . , aJ , c0 , c1 , . . . , cN . Let us use M x = 0 to represent the system. Let ν be
1 + deg(common numerator of (5.9)), then
ν ≤ 1 + degk P (n, k) + J(A + (U − A)+ ) + I(B + (V − B)+ ) + B.
The matrix M is ν by 2 + J + N, of rank ρ > 0, and the ith row of M corresponds to
the coefficient of k i−1 in the numerator of (5.9). Furthermore, a0 (n) is assumed not to be
identically zero. The stage is now set to apply the procedure for solving for a0 (n) in §5.3.
In the solution set thus obtained, all of the aj ’s and ci ’s are either equal to 1 or are
certain rational functions. To get a polynomial solution (that may have common polynomial-in-n factors), we multiply x by det M 0 . Henceforth, we take det M10 as a polynomial
solution for a0 (n).
Our goal is to bound real zeros of a0 (n) from above for the following reasons. If |a0 (n)| >
0 for all n ≥ na , then summing (5.7) over k yields a recurrence for
(5.10)
a0 (n)
X
k
F (n, k) + a1 (n)
X
k
F (n − 1, k) + · · · + aJ (n)
P
X
k
k
F (n, k), i.e.,
F (n − J, k) = 0.
5.6 PROOF OF THEOREM 5.1
64
To show that 1 also satisfies the recurrence (5.10), i. e. that
a0 (n) + a1 (n) + · · · + aJ (n) = 0
∀n,
we use the fact that if a polynomial P of degree d has d + 1 zeros, then P = 0. Therefore
it suffices to show that 1 satisfies (5.10) for
n0 ≤ n ≤ max{na + J − 1, n0 + max deg aj (n)}.
j∈[J]0
If it does, then we can use (5.10) to calculate 1 and
1=−
F (n, k) = −
k
k
F (n, k) in the following way:
a1 (n)f (n − 1) + · · · + aJ (n)f (n − J)
,
a0 (n)
and
X
P
a1 (n)
P
k
F (n − 1, k) + · · · + aJ (n)
a0 (n)
P
k
F (n − J, k)
,
where a0 (n) 6= 0 (n ≥ na ). We thus have
(a)
P
k
F (n, k) = 1 for n0 ≤ n ≤ max{na + J − 1, n0 + maxj∈[J]0 deg aj (n)};
(b) both
(c) both
P
k
P
k
F (n, k) and 1 are uniquely defined for all n ≥ na ;
F (n, k) and 1 satisfy the same recurrence relation.
By induction, (a), (b), and (c) imply that
P
k
F (n, k) = 1 for all n ≥ n0 .
We devote the rest of the proof to estimating deg det M10 , maxj∈[J]0 deg aj (n), and n1 .
Observations.
(1) The entries of M are polynomials in n with integer coefficients because P (n, k) is
assumed to have integer coefficients, and ws and cs are fixed integer parameters.
(2) The maximum degrees of entries of M 0 and Mi0 (i ∈ [ρ]) are bounded by the
maximum degree of the entries of M because of the way we obtain M 0 and Mi0
(i ∈ [ρ]) from M .
5.6 PROOF OF THEOREM 5.1
65
Let a00 + a01 n + a02 n2 + · · · + a0d nd = a0 (n) (= det M10 ). Then by Proposition 5.4,
a0 (n) 6= 0, if
n>d·
max |a0j |.
0≤j≤d−1
To find an upper bound for d, we use Lemma 5.5 which gives us a degree bound µ for
the entries of M . From the second observation above, µ is also a degree bound for Mi0
(i ∈ [ρ]) and M 0 . Thus
(5.11)
deg det Mi0 ≤ ρµ ≤ νµ
i ∈ [ρ],
and
deg det M 0 ≤ ρµ ≤ νµ,
where ν ≤ 1 + degk P (n, k) + J(A + (U − A)+ ) + I(B + (V − B)+ ) + B. From Step F of
§5.3, we know that
max deg aj (n) ≤ max{deg det M10 , deg det M20 , . . . , deg det Mρ0 , deg det M } ≤ νµ.
j∈[J]0
To estimate maxj∈[d]0 {|a0j |}, we use Lemma 5.6 which gives us µ8 , a bound for the
coefficients of the entries of M . From the way we obtained M10 from M , the absolute value
of the largest coefficient of M10 is bounded by (N + J + 1)µ8 .
Finally we compute an upper bound for the absolute value of the largest coefficient of
det M10 . Let ω := (N + J + 1)µ8 and µ := max{µ1 , µ2 }. In other words, ω is the computed
upper bound for the absolute value of the largest coefficient of the entries of M10 ; and µ is
the computed upper bound for the maximum degree of the entries of M10 .
5.6 PROOF OF THEOREM 5.1
66
By the definition of the determinant, we have
det M10 =
X
sgn(σ)e1σ(1) e2σ(2) . . . eρσ(ρ)
σ∈Sρ
4
X
|e1σ(1) e2σ(2) . . . eρσ(ρ) |
σ∈Sρ
4 ν! ω ν (nµ + nµ−1 + · · · + n + 1)ν .
Hence
¯
¯
max ¯[ni ] det M10 ¯ ≤ ν! ω ν (µ + 1)ν .
(5.12)
i
Putting (5.11) and (5.12) together and using Proposition 5.4, we conclude that a0 (n)
does not vanish for all n ≥ µν × ν! ω ν (µ + 1)ν (=: na ). Knowing na , we calculate
(5.13)
n1 = max{na + J − 1, n0 + max deg aj (n)}.
j∈[J]0
(See the discussion following formula (5.10) for the way we arrive at n1 .) Since
max deg aj (n) ≤ νµ ¿ na ,
j∈[J]0
and n0 is usually very small compared to ν, we can take n1 to be na +J −1 for Theorem 5.1.
Otherwise, n1 = max{µνν! ω ν (µ + 1)ν + J − 1, n0 + µν} will do.
¤
We next compute a cruder but simpler n1 . Given F (n, k), an admissible proper-hypergeometric term, let tlm ∈ Z,
P (n, k) =
E X
D
X
l m
tlm k n ,
l=0 m=0
and
Qp
(as n + bs k + cs )! k
F (n, k) = P (n, k) Qqs=1
ξ ,
t=1 (ut n + vt k + wt )!
such that F (n, k) satisfies a non-trivial recurrence relation,
PI,J
i,j
αij (n)F (n − j, k − i) = 0
for some positive integers I, J bounded by the result of Theorem 3.1 in [WZ3], or a sharper
bound in Theorem 1.4. Let
x := max{|tlm |, |as |, |bs |, |cs |, |ut |, |vt |, |wt | : l ∈ [E]0 , m ∈ [D]0 , s ∈ [p], t ∈ [q]}.
5.6 PROOF OF THEOREM 5.1
67
Then
µ ≤ x(p + q)(J + I + 1) + D,
µ3 ≤ (J + 1)D x,
µ4 ≤ ((J + I + 1)x)2(p+q)Ix ,
µ5 ≤ ((J + 1)x)(p+q)Jx ,
e1 ≤ (p + q)x(J + I + 1),
e2 ≤ 2(p + q)x,
e3 ≤ (p + q)x,
µ6 ≤ (x(J + I + 1))2x(p+q) ,
µ7 ≤ (2x)x(p+q) ,
ν ≤ 1 + E + x(p + q)(J + I + 1),
µ8 ≤ (D + 1)(E + 1)(6x(J + I + 1))2x(p+q)(I+J)+D+E ,
N ≤ E + x(p + q)(J + I − 1),
where the estimates are obtained directly from the expressions defining the variables in
Section 5.5. Thus
ω := (N + J + 1)µ8
≤ (E + x(p + q)(J + I − 1) + J + 1)(D + 1)(E + 1)(6x(J + I + 1))2x(p+q)(I+J)+D+E .
Using the estimate obtained in the proof of Theorem 5.1, i.e., n1 := µν ·ν! ω ν (µ+1)ν +J −1,
we get
¢f +h
¡
n1 = (f + h)(f + D)(f + h)! (f + g)(f + h + J)gh(6x(J + I + 1))2f +D+E
,
5.6 PROOF OF THEOREM 5.1
68
where
f := x(p + q)(J + I + 1),
g := D + 1,
h := E + 1.
Before proving the Main Theorem, we prove first the following corollary that is simpler
than the Main Theorem, then we give the proof of the Main Theorem following the method
used in the proof of Corollary 5.7.
Corollary 5.7. Let
Qp
(as n + bs k + cs )! k
ξ
F (n, k) = Qqs=1
s=1 (us n + vs k + ws )!
be an admissible proper-hypergeometric term (free of P (n, k)), let
x := max{|as |, |bs |, |cs |, |us |, |vs |, |ws |},
s
y := max{p, q},
and let n0 be a given integer. If
P
k
F (n, k) = 1 for
6
n0 ≤ n ≤ (3xy)3(2xy) ,
then
P
k
F (n, k) = 1 for all n ≥ n0 .
Proof. From Theorem 1.4, we know that
J ≤ B + (V − B)+
and
I ≤ 1 + δ + (A + (U − A)+ − 1)(B + (V − B)+ ),
5.6 PROOF OF THEOREM 5.1
69
where δ := degk P (n, k),
U :=
X
us ,
s:vs 6=0
A :=
X
X
V :=
+
(as ) +
s:bs 6=0
X
vs ,
s
A :=
X
as ,
B :=
X
s
s:bs 6=0
+
(−us ) ,
bs ,
X
X
B :=
(bs )+ +
(−vs )+ .
s
s:vs 6=0
s
Since P (n, k) = 1 in F (n, k), we have that δ = 0. Let
x := max{|as |, |bs |, |cs |, |us |, |vs |, |ws |}
s
and y := max{p, q}. Then
J ≤ B + (V − B)+
= max
nX
(bs )+ +
X
s
(−vs )+ ,
s
o
X
X
(−bs )+ +
(vs )+
s
s
≤ 2xy.
Similarly,
I ≤ 1 + (2xy − 1)(2xy) = (2xy)2 − 2xy + 1 < (2xy)2 .
We express upper bounds for N, e1 , e2 , e3 , µ, and µi , for all i ∈ [8], in terms of x and
y. (See §5.5 for the definitions of e1 , e2 , e3 , µ, and µi , for all i ∈ [8].) From Step 3 of
Chapter 2, we know that
N := degk P (n, k) + J(A + (U − A)+ ) + (I − 1)(B + (V − B)+ )
≤ 0 + (2xy)2 + ((2xy)2 − 2xy)(2xy) = (2xy)3 .
Next we compute bounds for e1 , e2 , and e3 :
´
³X
X
¡
¢
(−vs )+
e1 := J Ã + (Ũ − Ã)+ + (I + 1)
(bs )+ +
s
2
2
s
3
≤ (2xy) + ((2xy) − 2xy + 2)(2xy) = (2xy) + 4xy,
5.6 PROOF OF THEOREM 5.1
e2 := 2
e3 :=
70
³X
´
X
(bs )+ +
(−vs )+ ≤ 4xy,
s
X
|bs | +
X
s
s
|vs | ≤ 2xy.
s
Now we compute the bounds for µ and µi , for i ∈ [8]:
µ1 = degn P (n, k) + (I + 1)B̃ + J(Ã + (Ũ − Ã)+ )
≤ 0 + ((2xy)2 − 2xy + 2)2xy + (2xy)(2xy)
= (2xy)3 + 4xy,
µ2 = max
n X
|bs | +
s:as 6=0
X
|vs | ,
2
s:us 6=0
X
X
(bs )+ + 2
s:as 6=0
(−vs )+
s:us 6=0
≤ 4xy,
µ3 = 1
because P (n, k) = 1,
2
µ4 ≤ (2x)2xy ((2xy + 1)x)2xy((2xy)
−2xy+1)
,
µ5 ≤ ((2xy + 1)x)2(2xy)xy ,
µ6 ≤ ((J + I + 1)x · 2x)2xy
≤ ((2xy + (2xy)2 − 2xy + 2)2x2 )2xy
¡
¢2xy
= 2((2xy)2 + 2)x2
,
µ7 ≤ (2x)2xy ,
µ8 := max{3e1 µ4 µ5 , 2N 3e3 µ7 + 3e2 µ6 }
n
3
2
≤ max 3(2xy) +4xy (2x)2xy ((2xy + 1)x)2xy((2xy) +1)
,
3
2(2xy) 32xy (2x)2xy + 34xy 22xy (x2 ((2xy)2 + 2))2xy
3
= 22xy 3(2xy)
+4xy 2xy((2xy)2 +2)
x
2
(2xy + 1)2xy((2xy)
+1)
,
o
o
5.6 PROOF OF THEOREM 5.1
71
µ ≤ max{µ1 , µ2 }
≤ (2xy)3 + 4xy.
From Theorem 5.1, we know that n1 = max{n0 + µν, µν · ν! ω ν (µ + 1)ν + J − 1}, where
ν ≤ 1 + degk P (n, k) + J(A + (U − A)+ ) + I(B + (V − B)+ ) + B
and
ω = (N + J + 1)µ8 .
Estimating ν and ω in terms of x and y, we get
ν ≤ 1 + (2xy)2 + ((2xy)2 − 2xy + 1)2xy + 2xy
= 1 + 4xy + (2xy)3 ,
and
3
ω ≤ ((2xy)3 + 2xy + 1)22xy 3(2xy)
+4xy 2xy((2xy)2 +2)
x
2
(2xy + 1)2xy((2xy)
+1)
.
Thus
¡
¢ν
n1 ≤ ((2xy)3 + 4xy)((2xy)3 + 4xy + 1) ((2xy)3 + 4xy + 1)2 ω
6
6
6
< 33(2xy) x3(2xy) y 2(2xy)
6
< (3xy)3(2xy) .
¤
6
In practice n0 and µν are both much smaller than (3xy)3(2xy) ; therefore we take
6
(3xy)3(2xy) as a bound for n1 .
We restate the
5.6 PROOF OF THEOREM 5.1
72
Main Theorem. Let
Qp
(as n + bs k + cs )! k
ξ
F (n, k) = P (n, k) Qqs=1
s=1 (us n + vs k + ws )!
be an admissible proper-hypergeometric term, and P (n, k) be a polynomial with coefficients
in Z. Let
x := max{|as |, |bs |, |cs |, |us |, |vs |, |ws |},
s
y := max{p, q},
¯
¯
z := max ¯[nj k i ]P (n, k)¯ ,
0≤i,j
d := 1 + max{degk P (n, k), degn P (n, k)},
and let n0 be a given integer. If
P
k
F (n, k) = 1 for
2
n0 ≤ n ≤ (3xy)3(d+1)
then
P
k
(2xy)6 5(d+1)(2xy)3 (d+1)(2xy)3
d
z
,
F (n, k) = 1 for all n ≥ n0 .
Proof. From Theorem 1.4, we know that
J ≤ B + (V − B)+
and
I ≤ 1 + δ + (A + (U − A)+ − 1)(B + (V − B)+ ),
where δ := degk P (n, k),
U :=
X
us ,
s:vs 6=0
A :=
X
s:bs 6=0
(as )+ +
X
V :=
X
vs ,
A :=
s
(−us )+ ,
s:vs 6=0
X
as ,
B :=
s
s:bs 6=0
B :=
X
X
X
(−vs )+ .
(bs )+ +
s
s
bs ,
5.6 PROOF OF THEOREM 5.1
73
We see that J ≤ 2xy, and that
I ≤ J + I ≤ d + (2xy)2 .
We express upper bounds for N, e1 , e2 , e3 , µ, and µi for all i ∈ [8], in terms of x, y, z,
and d. (See §5.5 for the definitions of e1 , e2 , e3 , µ, and µi for all i ∈ [8].) From Step 3 of
Chapter 2 we know that
N := degk P (n, k) + J(A + (U − A)+ ) + (I − 1)(B + (V − B)+ )
< d + (2xy)(d − 1 + (2xy)2 )
< (2xy)3 + d(2xy + 1).
Next we compute bounds for e1 , e2 , and e3 :
³X
´
X
e1 ≤ J(Ã + (Ũ − Ã)+ ) + (I + 1)
(bs )+ +
(−vs )+
s
s
≤ (2xy)(d + 1 + (2xy)2 ) = (2xy)3 + (d + 1)2xy,
³X
´
X
e2 := 2
(bs )+ +
(−vs )+ ≤ 4xy,
e3 :=
X
s
|bs | +
X
s
s
|vs | ≤ 2xy.
s
Now we compute the bounds for µi , for i ∈ [8]:
µ1 = degn P (n, k) + (I + 1)B̃ + J(Ã + (Ũ − Ã)+ )
≤ d − 1 + (2xy)(d + 1 + (2xy)2 )
= (2xy)3 + (d + 1)2xy + d − 1,
µ2 = max
n X
s:as 6=0
≤ 4xy,
|bs | +
X
s:us 6=0
|vs | ,
2
X
(bs )+ + 2
s:as 6=0
X
(−vs )+
s:us 6=0
o
5.6 PROOF OF THEOREM 5.1
74
µ3 = (1 + J)d z ≤ (1 + 2xy)d z,
2
µ4 ≤ (2x)2xy ((2xy + 1)x)2xy((2xy)
−2xy+d)
,
2
µ5 ≤ ((2xy + 1)x)(2xy) ,
µ6 ≤ ((J + I + 1)2x2 )2xy ≤ (2x2 )2xy ((2xy)2 + d + 1)2xy ,
µ7 ≤ (2x)2xy .
Let r := 2xy. Then
µ8 ≤ max{d2 3e1 µ3 µ4 µ5 , 2N 3e3 µ7 + 3e2 µ6 }
n
3
2
2
≤ max d2 3r +(d+1)r (1 + r)d z(2x)r ((r + 1)x)r(r +d)+r
2r
3
2r
3
= z2r (3x)r
o
+d(r+1) r
3 (2x)r + 32r (2x2 )r (r2 + d + 1)r
n
3
3
3
= max z2r 3r +(d+1)r d2 xr +(d+1)r (1 + r)r +d(r+1)
+(d+1)r+d
3
(3x)r + 2r (3x)2r (r2 + d + 1)r
+(d+1)r 2
d (1 + r)r
3
+d(r+1)
,
,
o
.
Thus
µ ≤ max{µ1 , µ2 } ≤ r3 + (d + 1)r + d + 1.
From Theorem 5.1, we know that n1 ≤ µν · ν! ω ν (µ + 1)ν + J − 1 where
ν ≤ 1 + degk P (n, k) + J(A + (U − A)+ ) + I(B + (V − B)+ ) + B
and
ω = (N + J + 1)µ8 .
Estimating ν and ω, we get
ν ≤ d + r(d + r2 ) + r = r3 + r(d + 1) + d,
5.7 GENERALIZATIONS OF THEOREM 5.1
75
where, still, r = 2xy, and
ω ≤ (r3 + d(r + 1) + r + 1)z2r (3x)r
3
+(d+1)r 2
d (1 + r)r
3
+d(r+1)
.
Therefore
n1 < (r3 + r(d + 1) + d)2 (ν 2 ω)ν
< (r3 + r(d + 1) + d)2+2(r
³
3
+r(d+1)+d)
3
r 3 +(d+1)r 2
r
× (r + (d + 1)(r + 1))z2 (3x)
3
3
d (1 + r)
4
< ((d + 1)r3 )3(d+1)r z (d+1)r 2(d+1)r 3(d+1)
< (3x)3(d+1)
2 6
r
3
d5(d+1)r y 2(d+1)
2 6
< (3xy)3(d+1)
r
3
2 6
r
r 3 +d(r+1)
z (d+1)r
2 6
r
3
d2(d+1)r x(d+1)
2 6
r
´r3 +r(d+1)+d
(1 + r)(d+1)
2 6
r
3
3
d5(d+1)r z (d+1)r . ¤
The following is a comparison of the ‘sharp’ and crude estimates in two relevant hypergeometric series.
Examples. First we calculate n1 for
P ¡n¢
k
k
= 2n . Sharp n1 = 4 × 3 × 4! (4 × 18)4 + 1 <
1011 , and crude n1 = 9 × 10 × 10! (50 × 1812 )10 < 10177 .
Second we calculate n1 for
P ¡n¢2
k
k
=
¡2n¢
13
<
n . Sharp n1 = 12 × 13 × 13! (10 × 13!/6!)
10115 , and crude n1 = 36 × 37 × 37! (37 × 28 × 3660 )37 < 103613 .
5.7 Generalizations of Theorem 5.1
We consider in the following theorem hypergeometric identities of the type
P
k
F (n, k) =
f (n) where F (n, k) is an admissible proper-hypergeometric term and f (n) is a hypergeometric term. (Instead of f = 1 as in Theorem 5.1.) In Theorem 5.9, the object of interest
will be identities of the form
P
proper-hypergeometric terms.
k
F (n, k) =
P
k
G(n, k) where F and G are both admissible
5.7 GENERALIZATIONS OF THEOREM 5.1
76
Theorem 5.8. Let
Qp
(as n + bs k + cs )! k
Q
F (n, k) = P (n, k) qs=1
ξ
s=1 (us n + vs k + ws )!
be an admissible proper-hypergeometric term where P is a polynomial with coefficients in
Z. Let
Qr
Q(n) s=1 (αs n + βs )! n
f (n) =
ζ
Q
S(n) ts=1 (µs n + νs )!
be a hypergeometric term where Q and S are polynomials with coefficients in Q. Let
x := max{|as |, |bs |, |cs |, |us |, |vs |, |ws |},
s
y := max{p, q},
¯
¯
z := max ¯[nj k i ]P (n, k)¯ ,
0≤i,j
d := 1 + max{degk P (n, k), degn P (n, k)},
and let n0 be a given integer. If
P
k
F (n, k) = f (n) for n0 ≤ n ≤ n1 , then
P
k
F (n, k) =
f (n) for all n ≥ n0 , where
½
n1 := max (3xy)3(d+1)
2
(2xy)6 5(d+1)(2xy)3 (d+1)(2xy)3
d
2
z
,
6
n0 + (d + 1) (2xy) + deg Q + (2xy + 1) deg S + 2xy
³X
|αs | +
s
X
s
Proof. By Theorem 3.1 of [WZ3], we know that there exist a positive integer J ≤
P
s
P
|vs | and polynomials a0 (n), a1 (n), . . . , aJ (n) such that
(5.14)
a0 (n)F (n, k) + a1 (n)F (n − 1, k) + · · · + aJ (n)F (n − J, k) = 0.
Since F (n, k) is admissible, we can sum (5.14) over k to get
a0 (n)
X
k
F (n, k) + a1 (n)
X
k
F (n − 1, k) + · · · + aJ (n)
X
k
´¾
|µs | .
F (n − J, k) = 0.
s
|bs |+
5.7 GENERALIZATIONS OF THEOREM 5.1
77
From Theorem 5.1, we know that a0 (n) 6= 0 for all n ≥ na . (See the line above (5.13) for
the definition of na .) Therefore
X
F (n, k) = −
a1 (n)
P
k
k
for all n ≥ na . From the hypothesis,
f (n) = −
(5.15)
F (n − 1, k) + · · · + aJ (n)
a0 (n)
P
k
P
k
F (n − J, k)
F (n, k) = f (n) for n0 ≤ n ≤ n1 . Hence
a1 (n)f (n − 1) + · · · + aJ (n)f (n − J)
a0 (n)
for na ≤ n ≤ n1 . Dividing both sides of (5.15) by f (n) we get
(5.16)
a0 (n) + a1 (n)
f (n − 1)
f (n − 2)
f (n − J)
+ a2 (n)
+ · · · + aJ (n)
= 0.
f (n)
f (n)
f (n)
Putting (5.16) over a common denominator, we find that the numerator polynomial
(5.17)
a0 (n)f0 (n) + a1 (n)f1 (n) + · · · + aJ (n)fJ (n) = 0
where the fj ’s (j ∈ [J]0 ) are all polynomials of degree at most
deg Q + (J + 1) deg S + J
³X
|αs | +
s
X
´
|µs |
s
and the aj ’s (j ∈ [J]0 ) are polynomials of degree at most νµ. (See (5.11).) Since
½
n1 ≥ max na +J −1
,
n0 +νµ+deg Q+deg S +(J +1) deg S +J
³X
s
|αs |+
X
´¾
|µs | ,
s
we have more zeros than the degree of the polynomial in (5.17). Therefore, the numerator
polynomial of (5.16) is identically zero, or equivalently,
a0 (n)f (n) + a1 (n)f (n − 1) + · · · + aJ (n)f (n − J) = 0.
Thus we have (5.15) for all n ≥ na .
5.7 GENERALIZATIONS OF THEOREM 5.1
The facts that f (n) and
P
k
P
k
78
F (n, k) satisfy the same recurrence relation, and that
F (n, k) = f (n) for n0 ≤ n ≤ n1 imply (by induction on n) that
P
k
F (n, k) = f (n) for
all n ≥ n0 . ¤
The following theorem has a sum of hypergeometric terms on both sides of the equal
sign.
Theorem 5.9. Let
Qp
(as n + bs k + cs )! k
F (n, k) = P (n, k) Qqs=1
ξ
s=1 (us n + vs k + ws )!
and
Qr
(αs n + βs k + γs )! k
ζ
G(n, k) = Q(n, k) Qs=1
t
(δ
n
+
φ
k
+
ψ
)!
s
s
s
s=1
be admissible proper-hypergeometric terms where P and Q are polynomials with coefficients
in Z. Let
x := max{|as |, |bs |, |cs |, |us |, |vs |, |ws |},
f := max{|αs |, |βs |, |γs |, |δs |, |φs |, |ψs |},
y := max{p, q},
g := max{r, t},
s
s
¯
¯
z := max ¯[nj k i ]P (n, k)¯ ,
0≤i,j
d := 1 + max{degk P (n, k), degn P (n, k)},
e := 1 + max{degk Q(n, k), degn Q(n, k)},
let n0 be a given integer, and assume wlog that xy ≤ f g. If
n0 ≤ n ≤ n1 , then
P
k
F (n, k) =
P
k
P
k
F (n, k) =
P
k
G(n, k) for
G(n, k) for all n ≥ n0 , where
n
n1 := max n0 + 2(max{d, e} + 1)2 (2f g)7
,
(3xy)3(d+1)
2
(2xy)6 5(d+1)(2xy)3 (d+1)(2xy)3
d
z
o
5.7 GENERALIZATIONS OF THEOREM 5.1
79
Proof. By Theorem 3.1 of [WZ3], there exist a positive integer J ≤
P
s
|bs | +
P
s
|vs | and
polynomials a0 (n), a1 (n), . . . , aJ (n) such that
(5.18)
a0 (n)F (n, k) + a1 (n)F (n − 1, k) + · · · + aJ (n)F (n − J, k) = 0.
P
Similarly, there exist a positive integer I ≤
s
|βs |+
P
s
|φs | and polynomials b0 (n), b1 (n),
. . . , bI (n) such that
(5.19)
b0 (n)G(n, k) + b1 (n)G(n − 1, k) + · · · + bI (n)G(n − I, k) = 0.
Summing (5.18) and (5.19) over k, we get
(5.20)
a0 (n)
X
F (n, k) + a1 (n)
k
X
F (n − 1, k) + · · · + aJ (n)
X
k
F (n − J, k) = 0
k
and
(5.21)
b0 (n)
X
G(n, k) + b1 (n)
k
X
G(n − 1, k) + · · · + bI (n)
k
X
G(n − I, k) = 0,
k
since both F and G are admissible.
By the hypothesis, xy ≤ f g, so both I and J are bounded by 2f g. Our goal is to show
that
P
k
G(n, k) satisfies the same recurrence relation as
is achieved if
P
k
G(n, k) satisfies (5.20) or
P
k
P
k
F (n, k), or visa versa. This
F (n, k) satisfies (5.21) for
µ
n
o¶
n0 ≤ n ≤ n0 + (2f g + 1) 1 + max max deg aj (n), max deg bi (n)
,
j∈[J]0
i∈[I]0
because there are (I + 1)(1 + maxi∈[I]0 deg bi (n)) indeterminates in
X
i∈[I]0
0≤r≤maxj∈[I]0 deg bj (n)
cr,i nr
X
k
G(n − i, k) = 0,
5.7 GENERALIZATIONS OF THEOREM 5.1
80
and (J + 1)(1 + maxj∈[J]0 deg aj (n)) indeterminates in the corresponding recurrence for
F (n, k).
We know from Theorem 5.1 that
I ≤ 2f g
and
J ≤ 2xy ≤ 2f g
and
max deg bi (n) ≤ (e + 1)2 (2f g)6 ;
i∈[I]0
max deg aj (n) ≤ (d + 1)2 (2xy)6 ≤ (d + 1)2 (2f g)6 .
j∈[J]0
Thus
n
max (I + 1)(1 + max deg bi (n))
i∈[I]0
where
o
(J + 1)(1 + max deg aj (n)) ≤ n2 ,
,
j∈[J]0
µ
n
n2 := (2f g + 1) 1 + max max deg aj (n)
,
j∈[J]0
Since n1 − n0 > n2 , and
P
k
P
k
F (n, k) =
X
P
F (n, k) = −
k
F (n, k) and
a1 (n)
P
k
for n ≥ na . We conclude that if
k
k
F (n, k) =
P
k
k
P
k
i∈[I]0
G(n, k) for n0 ≤ n ≤ n1 , we conclude that
G(n, k) satisfies the same recurrence relation as
n ≥ na . Therefore, both
P
P
o¶
max deg bi (n) .
P
k
F (n, k). Further, a0 (n) 6= 0 for
G(n, k) are determined inductively by
F (n − 1, k) + · · · + aJ (n)
a0 (n)
P
k
F (n, k) =
G(n, k) for all n ≥ n0 . ¤
P
k
P
k
F (n − J, k)
G(n, k) for n0 ≤ n ≤ n1 , then