Chapter 1
Hypergeometric Series
1.1
The general hypergeometric series
Definition 1.1. A series
function of k.
P∞
k=0 tk
is called hypergeometric if tk+1 /tk is a rational
Qk−1
Definition 1.2. The rising factorial j=0 (a+j) = a(a+1)(a+2) · · · (a+k −1)
is denoted by the Pochhammer symbol (a)k .
Note that (a)0 = 1 and k! = (1)k . P
∞
As we said hypergeometric series k=0 tk has the property that tk+1 /tk is
a rational function R(k). Without loss of generality, we may write R(k) in the
form
R(k) =
(k + a1 )(k + a2 ) · · · (k + ap )
z,
(k + 1)(k + b1 ) · · · (k + bq )
and use this form to motivate the following standard hypergeometric notation:
Definition 1.3.
p Fq
∞
X
a1 , a2 , . . . , ap
(a1 )k (a2 )k · · · (ap )k k
; z :=
z .
b1 , b 2 , . . . , bq
k!(b1 )k · · · (bq )k
k=0
We will assume that none of the bj is a nonpositive integer, as this would
cause there to be zeros in the denominator.
Example 1.4.
1.
z
e =
∞
X
zk
k=0
2.
cos z =
k!
= 0 F0
∞
X
(−1)k z 2k
k=0
(2k)!
1
−
; z .
−
= 0 F1
−
1
2
; −
z2
.
4
3.
sin z =
∞
X
(−1)k z 2k+1
k=0
(2k + 1)!
4.
sin−1 z = z 2 F1
5.
"
1 1
2, 2
3
2
= z 0 F1
−
#
; z2 ,
1
1
= 1 F0
; z ,
1−z
−
3
2
z2
; −
.
4
|z| < 1
|z| < 1
Notice that if at least one of the aj is a negative integer, say −n, then for
all k > n, the term tk = 0, and thus the series has only finitely many nonzero
terms, so the question of convergence of the series does not arise.
On the other hand, if none of the aj is a negative integer and z 6= 0, then
the series has infinitely many nonzero terms.
h
i
2 ,...,ap
Theorem 1.5. The series p Fq ab11,a
;
z
,b2 ,...,bq
• converges absolutely for all z if p < q + 1,
• diverges for all z 6= 0 if p > q + 1,
• converges absolutely for |z| < 1 if p = q + 1,
• diverges for |z| > 1 if p = q + 1.
Proof. Apply the ratio test from elementary calculus.
The ratio test, however, provides no convergence information when |z| = 1
and p = q + 1.
Let us now turn our attention to the particular case where p = q + 1 = 2.
1.2
Gauß’s Hypergeometric Series
The hypergeometric series
2 F1
a, b
; z
c
(1.1)
was studied extensively by K. F. Gauß, and he delivered a famous lecture on
such series in January 1812. By the results of the previous section, we know
that (1.1) converges absolutely when |z| < 1 and diverges for |z| > 1. In order
to study the case where |z| = 1, we will need some background results.
2
1.2.1
Background material
Definition 1.6 (Big “O” notation). Let ζ1 , ζ2 , ζ3 , . . . and z1 , z2 , . . . be a pair
of sequences where ζk /zk < C for all sufficiently large k and C is a constant
(independent of k). We then say that the sequence {ζk } is of order zk and write
ζk = O(zk ).
Example 1.7.
2
=O
k3
because ( k23 )/( k12 ) =
2
k
1
k2
< 1 for all k > 2.
Proposition 1.8 (The Binomial Series). For p constant and |z| < 1,
(1 + z)p = 1 + pz + O(z 2 )
Proof. The Maclaurin series expansion of f (z) is
f (z) =
∞
X
f (k) (z)
k=0
k!
zk ,
so with f (z) = (1 + z)p , we have f (0) = 1, f 0 (0) = p. The convergence follows
by the ratio test. (Note that if p is a nonnegative integer, then the Maclaurin
series reduces to a finite sum, and thus is valid for all z.)
P∞
Theorem 1.9 (De Morgan). Let k=1 uk be a series whose terms satisfy
lim
k→∞
The series
such that
uk+1
= 1.
uk
P∞
uk converges absolutely if there exists a positive real number C
uk+1
lim k
− 1 = −(1 + C).
k→∞
uk
P∞
Proof. Let vk = Ak −(1+c/2) , where A is a constant. Clearly,
k=1 vk is a
convergent series (by the “p-series test” from elementary calculus). Notice that
vk+1
=
vk
k=1
k
k+1
1+c/2
=
1+
1
k
−(1+c/2)
=1−
1 + c/2
+O
k
1
k2
,
where the last equality follows from Proposition 1.8 with z = 1/k and p =
−(1 + c/2). Thus,
vk+1
c
lim k
−1 =− 1+
.
k→∞
vk
2
A
constant A will ensure
that for all k, |uk | < vk . Since
P suitable choice of the P
P
vk converges, so does
|uk |, and thus
uk is absolutely convergent.
3
Corollary 1.10 (Raabe’s test). If
1
,
k2
P
(where B is a constant independent of k), then
uk is absolutely convergent if
B < −1.
B
uk+1
=1+ +O
uk
k
1.2.2
The convergence of Gauß’s series when |z| = 1.
h
i
P∞
Consider k=0 tk = 2 F1 a,b
;
z
where |z| = 1. The numbers a, b, and c are
c
complex; let us write them as a√= a1 + ia2 , b = b1 + ib2 and c = c1 + ic2 , where
a1 , a2 , b1 , b2 , c1 , c2 ∈ R and i = −1. Then we have
−1
tk+1
(k + a)(k + b)
a
b
1
c
=
= (1 + )(1 + ) (1 + )(1 + )
tk
(k + 1)(k + c)
k
k
k
k
a+b
1
1+c
1
= 1+
+O
1−
+O
k
k2
k
k2
a+b−c−1
1
+O
= 1+
k
k2
a1 + b1 − c1 − 1 + i(a2 + b2 − c2 )
1
= 1+
+O
k
k2
s
2 2
2
a1 + b1 − c1 − 1
a2 + b2 − c2
1
=
1+
+
+O
k
k
k2
a1 + b1 − c1 − 1
1
=1+
+O
.
k
k2
Thus by Corollary 1.10, the series converges absolutely if <(a + b − c) < 0,
where <z denotes the real part of the complex number z. It turns out that even
more is known than just whether or on the series converges in the z = 1 case.
Gauß gave an explicit formula for the sum of the series:
Theorem 1.11 (Gauß’s Hypergeometric Summation Formula). If <(a+
b − c) < 0,
a, b
Γ(c)Γ(c − a − b)
F
;
1
=
.
2 1
c
Γ(c − a)Γ(c − b)
Sketch of proof. First, show that
a, b
(c − a)(c − b)
a, b
F
;
1
=
F
;
1
,
2 1
2 1
c
c(c − a − b)
c+1
and more generally that
a, b
(c − a)n (c − b)n
a, b
; 1
; 1 =
2 F1
2 F1
c
(c)n (c − a − b)n
c+n
4
for n ∈ Z+ . Then show that
lim
n→∞
(c − a)n (c − b)n
Γ(c − a)Γ(c − b)
=
(c)n (c − a − b)n
Γ(c)Γ(c − a − b)
and
lim 2 F1
n→∞
a, b
; 1 = 1.
c+n
If a = −n where n is a positive integer, then Theorem 1.11 simplifies to
Corollary 1.12 (Chu-Vandermonde).
−n, b
(c − b)n
; 1 =
.
2 F1
c
(c)n
Of course, this result follows easily as a consequence of Theorem 1.11. We
choose to present an independent proof, however.
Proof. Let
F (n, k) :=
(
(−n)k (b)k (c)n
k!(c)k (c−b)n
0,
let
R(n, k) :=
if k = 0
,
if k < 0
k(1 − c − k)
,
n(n + c − 1)
let
G(n, k) := F (n, k)R(n, k),
and let
f (n) :=
∞
X
k=0
(c)n
−n, b
F (n, k) =
; 1 .
2 F1
(c − b)n
c
kc + kn − kb + nb − k
kc + kn − kb + nb − k
=
n(n + c − 1)
n(n + c − 1)
(c + n − 1)n − (n − k)(c + n − b − 1)
(b + k)(n − k) + k(k − 1 + c)
⇐⇒
=
n(c + n − 1)
n(n + c − 1)
(n − k)(c + n − b − 1)
(n − k)(b + k)(k + 1)(k + c)
k(k − 1 + c)
⇐⇒ 1 −
=
+
n(c + n − 1)
n(n + c − 1)(k + 1)(k + c)
n(n + c − 1)
1 − F (n − 1, k)
F (n, k + 1)R(n, k + 1)
⇐⇒
=
− R(n, k)
F (n, k)
F (n, k)
⇐⇒ F (n, k) − F (n − 1, k) = F (n, k + 1)R(n, k + 1) − F (n, k)R(n, k)
⇐⇒ F (n, k) − F (n − 1, k) = G(n, k + 1) − G(n, k)
∞
∞
X
X
=⇒
{F (n, k) − F (n − 1, k)} =
{G(n, k + 1) − G(n, k)}
k=−∞
k=−∞
=⇒ f (n) − f (n − 1) = 0
=⇒ f (n) = f (n − 1)
for n ∈ Z+ .
5
Thus f (n) is constant for all n ∈ Z+ , so all we need to do to find f (n) for
general n is to evaluate it at some particular value of n, say n = 0.
∞
X
(0)k (b)k (c)0
f (n) = f (0) =
= 1.
k!(c)k (c − b)0
k=0
Thus,
−n, b
(c)n
; 1 = 1,
2 F1
(c − b)n
c
or, equivalently,
2 F1
−n, b
(c − b)k
; 1 =
.
c
(c)k
Remark 1.13. The novelty of the preceding proof is that it was produced automatically by a computer programmed to carry out the so-called “WZ algorithm”
due to Wilf and Zeilberger. Zeilberger is professor here at Rutgers.
Remark 1.14. Corollary 1.12 was discovered in 1770 by the French mathematician Vandermonde and was called “Vandermonde’s sum” in the literature
for many years. Recently, however, it was noticed that this identity had been
discovered more than four and a half centuries earlier and had appeared in a
book written in 1303 by the Chinese mathematician Chu Shih-Chieh, so we
now call it the “Chu-Vandermonde sum.” This same book from 1303 contains
an illustration of what we call “Pascal’s triangle” and refers to it as an “ancient
method.”
1.3
Exercises
1. Prove that
2. Consider the series
10 + 20k
=O
k3
1
k2
.
∞
X
k!
(α)k
k=0
where α is a positive real constant. Find the values of α for which the
series converges absolutely.
3. Find the Maclaurin series for tan−1 (z), write the series in hypergeometric “p Fq ” notation, and find the set of z for which the series converges
absolutely.
6
4. Find conditions on the complex numbers a, b, and c which cause
a, b
; i
2 F1
c
to be absolutely convergent.
5. Show that
d2
a, b
a(a + 1)b(b + 1)
a + 2, b + 2
F
;
z
=
F
;
z
.
2 1
2 1
dz 2
c
c(c + 1)
c+2
6. Prove that y = 2 F1
h
a,b
c
z(1 − z)
i
; z is a solution to the differential equation
i dy
d2 y h
+ c − (a + b + 1)z
− aby = 0.
2
dz
dz
7