Cauchy Condensation Test & Sterlings Approximation by: javier
Calculus II Sequence & Series: Cauchy Condensation Test & Sterlings Approximation ∞ ! n=1
a(n) ≈
n! ≈
√
∞ !
2n a(2n )
n=1
2πn
" n #n e
by: javier
Understanding the Cauchy Condensation Test
The Key Idea
Understanding the Cauchy Condensation Test
The Key Idea ∞
∞
∞
n=1
n=1
n=1
! ! 1! n n 2 a(2 ) ≤ a(n) ≤ 2n a(2n ) 2
Understanding the Cauchy Condensation Test
The Key Idea ∞
∞
∞
n=1
n=1
n=1
! ! 1! n n 2 a(2 ) ≤ a(n) ≤ 2n a(2n ) 2
requires a little bit of work to prove, but well worth it..
Understanding the Cauchy Condensation Test
The Key Idea ∞
∞
∞
n=1
n=1
n=1
! ! 1! n n 2 a(2 ) ≤ a(n) ≤ 2n a(2n ) 2
requires a little bit of work to prove, but well worth it.. from it we get
Understanding the Cauchy Condensation Test
The Key Idea ∞
∞
∞
n=1
n=1
n=1
! ! 1! n n 2 a(2 ) ≤ a(n) ≤ 2n a(2n ) 2
requires a little bit of work to prove, but well worth it.. from it we get ∞ ! n=1
a(n) ≈
∞ ! n=1
2n a(2n )
Examples using CCT !
If a(n) is decreasing
!
and a(n) > 0 for all large n′ s
then
!
a(n) ≈
!
2n · a(2n )
Examples using CCT !
If a(n) is decreasing and a(n) > 0 for all large n′ s
getai
!
then
!
! a(n) ≈ 2n · a(2n ) !1
example:
n
=
[
a
"
II
Bondiverge
Examples using CCT !
If a(n) is decreasing
!
and a(n) > 0 for all large n′ s
then
!
! a(n) ≈ 2n · a(2n ) ! 1
example: a
n2
[ sit
=22¥z¥ .
{ #
"
Examples using CCT !
If a(n) is decreasing
!
and a(n) > 0 for all large n′ s
then
!
a(n) ≈ !
!
2n · a(2n )
1 n(ln n)3
example:
:[
=
=
[
KITH ¥3
# [
nts
Examples using CCT !
If a(n) is decreasing
!
and a(n) > 0 for all large n′ s
then
!
example:
!
2n · a(2n ) 1 n(ln n)(ln(ln n))
a(n) ≈ !
Sterlings Approximation
Sterlings Approximation n! ≈ * !
H
⇐
!
=
√
2πn ·
" n #n e
Ft ( * *
Ftse
(Ey§h
Sterlings Approximation
Sterlings Approximation n! ≈
example:
!
n2 (2n − 1)!
√
2πn ·
" n #n e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
! n! h 5n
√
2πn ·
!
sterling
#
⇐[ Fa
-
=
Ft
[
(
"
mgnehn ,
rn
¢ h
>#e#= £
.
use
root -
test
" n #n e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
! 2n n! nn
Eiden
√
2πn ·
" n #n e
't
.
'
inert =nT#
.µgtY Enter =€z%¥
⇐
=Fu{
"
Sterlings Approximation
Sterlings Approximation n! ≈
example:
! 5n n! nn
√
2πn ·
" n #n e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
1
(n!) n lim n→∞ n
:
den
⇐
"
#¥#sY#n=n÷ =
£
n
√
2πn ·
" n #n e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
! (n!) 1n n
Tete
√
2πn ·
" n #n e
a(n) ≈
n! ≈
√
∞ !
2n a(2n )
n=1
2πn
" n #n e
by: javier
Understanding the Cauchy Condensation Test
The Key Idea
Understanding the Cauchy Condensation Test
The Key Idea ∞
∞
∞
n=1
n=1
n=1
! ! 1! n n 2 a(2 ) ≤ a(n) ≤ 2n a(2n ) 2
Understanding the Cauchy Condensation Test
The Key Idea ∞
∞
∞
n=1
n=1
n=1
! ! 1! n n 2 a(2 ) ≤ a(n) ≤ 2n a(2n ) 2
requires a little bit of work to prove, but well worth it..
Understanding the Cauchy Condensation Test
The Key Idea ∞
∞
∞
n=1
n=1
n=1
! ! 1! n n 2 a(2 ) ≤ a(n) ≤ 2n a(2n ) 2
requires a little bit of work to prove, but well worth it.. from it we get
Understanding the Cauchy Condensation Test
The Key Idea ∞
∞
∞
n=1
n=1
n=1
! ! 1! n n 2 a(2 ) ≤ a(n) ≤ 2n a(2n ) 2
requires a little bit of work to prove, but well worth it.. from it we get ∞ ! n=1
a(n) ≈
∞ ! n=1
2n a(2n )
Examples using CCT !
If a(n) is decreasing
!
and a(n) > 0 for all large n′ s
then
!
a(n) ≈
!
2n · a(2n )
Examples using CCT !
If a(n) is decreasing and a(n) > 0 for all large n′ s
getai
!
then
!
! a(n) ≈ 2n · a(2n ) !1
example:
n
=
[
a
"
II
Bondiverge
Examples using CCT !
If a(n) is decreasing
!
and a(n) > 0 for all large n′ s
then
!
! a(n) ≈ 2n · a(2n ) ! 1
example: a
n2
[ sit
=22¥z¥ .
{ #
"
Examples using CCT !
If a(n) is decreasing
!
and a(n) > 0 for all large n′ s
then
!
a(n) ≈ !
!
2n · a(2n )
1 n(ln n)3
example:
:[
=
=
[
KITH ¥3
# [
nts
Examples using CCT !
If a(n) is decreasing
!
and a(n) > 0 for all large n′ s
then
!
example:
!
2n · a(2n ) 1 n(ln n)(ln(ln n))
a(n) ≈ !
Sterlings Approximation
Sterlings Approximation n! ≈ * !
H
⇐
!
=
√
2πn ·
" n #n e
Ft ( * *
Ftse
(Ey§h
Sterlings Approximation
Sterlings Approximation n! ≈
example:
!
n2 (2n − 1)!
√
2πn ·
" n #n e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
! n! h 5n
√
2πn ·
!
sterling
#
⇐[ Fa
-
=
Ft
[
(
"
mgnehn ,
rn
¢ h
>#e#= £
.
use
root -
test
" n #n e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
! 2n n! nn
Eiden
√
2πn ·
" n #n e
't
.
'
inert =nT#
.µgtY Enter =€z%¥
⇐
=Fu{
"
Sterlings Approximation
Sterlings Approximation n! ≈
example:
! 5n n! nn
√
2πn ·
" n #n e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
1
(n!) n lim n→∞ n
:
den
⇐
"
#¥#sY#n=n÷ =
£
n
√
2πn ·
" n #n e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
! (n!) 1n n
Tete
√
2πn ·
" n #n e
