Problem Set Twelve: Raabe's Theorem Revisited
Theorem: Let ( a n ) be a positive sequence for which ρ  lim n [ a n /a n  1  1] exists.
(1) If ρ  1 then there are constants c and p with 0 < c, 1 < p and a n  c n  p for all n.
Moreover if ρ  1 then the sequence ( n a n ) is eventually decreasing.
(2) If ρ  1 then ( n a n ) is eventually increasing.
Lemma: If 1 < p then Γ(n) /  (n  p)  n  p .
Proof. Use the log-convexity of the Gamma Function. n  1  α n  β (n  p) for β  1 / p and
n Γ(n)  Γ(n  1)  Γ(n)
α
β
Γ(n  p)  n Γ(n)
1 α
β
 Γ(n  p)  Γ(n) /  (n  p)  n
p
Proof of (1): Fix 1  p  ρ . Since p  lim n [ a n /a n  1  1] there is a m  N so that n  m implies
p  n[
an
 1]
a n 1
. In particular 1  n [
an
 1]
a n 1
makes (n  1) a n 1  n a n for all n  m .
Let c  max { 1 p a 1 , 2 p a 2 , 3 p a 3 , ... , m p a m , a m Γ(m  p) / Γ(m) } . Clearly a n  c n  p for 1  n  m .
For n > m
p  n[
an
 1] 
a n 1
 a
a n   n
 a n 1
Γ(m)
Γ(m  p)
np
1
n
  a n 1

 a
  n2
an 

p
n
  am2
 ... 
 a
  m 1

an

a n 1
  a m 1

 a
 m
a n 1

an
n
np
,

(n  1)(n  2) ... (m  1)(m)
a m 
a m , and

(n  p  1)(n  p  2) ... (m  p  1)(m  p)

(n  1)(n  2) ... (m  1)(m) Γ(m)
(n  p  1)(n  p  2) ... (m  p  1)(m  p) Γ(m  p)
(n  1)(n  2) ... (m  1) Γ(m  1)
(n  p  1)(n  p  2) ... (m  p  1) Γ(m  p  1)
 ... 
(n  1) Γ(n  1)
(n  p  1) Γ(n  p  1)
am 
Γ(n)
Γ(n  p)
am
am
a m  a mn
p
so that a n  cn  p for each n > m.
Proof of (2): Since lim n [ a n /a n 1  1]  1 there is a m  N so that n  m implies n [
that makes n a n  (n  1) a n 1 for all n  m .
an
a n 1
 1]  1
and
Corollary (Raabe’s Test): Let ( a n ) be a positive sequence for which ρ  lim n [ a n /a n  1  1] exists.
(1) If ρ  1 then

(2) If ρ  1 then


k 1
converges by comparison with a convergent p-series
ak

k 1
diverges by comparison with the harmonic series.
ak
PROBLEMS
Problem 12-1: Use a change of variable to check each of these identities.
1
x 1
 0  ln(1/t)  d t
(a, Alternate Definition) For x > 0, Γ(x) 

0
(b, Laplace Transform of t x ) For x   1 and s > 0,
(c, Connection with Gaussian) Γ(1/2) 
Problem 12-2: Γ(1/2) 
G
2

x

0

 0   0

e
2
 
G  d x

and Γ(1/2)  2 G 
π
2
0
t
 1/2
e
. To see why let G 

0
 (x  y )
2
e
π

e
x
2
t
dt  2

0
x
t e
e
x
s t
d t  Γ(x  1)/s

2
 (x )
0
e
x 1
dx
2
d x and work through this calculation.
  y2

e
dydx

 0


π/2   r 2

dydx 
e
r d r d θ  π/4,
0
0



.
1
Problem 12-3: Show that Γ(n  ) 
2
(1)(3)(5). ..(2n  1)
2
n
π

(2n)!
2
2n
π
.
(n! )
Problem 12-4: For positive a and b, show (b)(a  b)(2 a  b)(3 a  b)...( (n  1) a  b) 
a
n
Γ(n  (b/a))
Γ(b/a)
.
Problem 12-5: We proved these consequences of the log-convexity of the Gamma function:
(i)
Γ(n  p)
Γ(n)
 n
p
for 0 < p < 1
(ii) (n  1) p 
Use those inequalities to estimate the size of a n 
Γ(n  p)
Γ(n)
for 0 < p.
(1)(6)(11) (16)...(5n  4)
(2)(7)(12) (17)...(5n  3)
. Does the series
converge or diverge?
Problem 12-6: Prove  c  0  x  0 , Γ(x)  c/x . Any such equality implies lim Γ(x)   .
x 0

 n 1 a n