Calculus II Sequence & Series: Cauchy Condensation Test by: javier
Calculus II Sequence & Series: Cauchy Condensation Test ∞ ∑ n=1
a(n) ≈
∞ ∑
2n a(2n )
n=1
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
then
∑
∑ a(n) ≈ 2n · a(2n ) ∑1
example:
n
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:
n2
Examples using CCT □
If a(n) is decreasing
□
and a(n) > 0 for all large n′ s
then
∑
a(n) ≈ ∑
example:
∑
2n · a(2n )
1 n(ln n)3
Examples using CCT □
If a(n) is decreasing
□
and a(n) > 0 for all large n′ s
then
∑
a(n) ≈ ∑
example:
∑
2n · a(2n ) 1 n(ln n)(ln(ln n))
Sterlings Approximation
Sterlings Approximation n! ≈
( n )n √ 2πn · e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
∑
n2 (2n − 1)!
( n )n √ 2πn · e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
∑ n! 5n
( n )n √ 2πn · e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
∑ 2n n! nn
( n )n √ 2πn · e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
∑ 5n n! nn
( n )n √ 2πn · e
Sterlings Approximation
Sterlings Approximation n! ≈ 1
example:
(n!) n lim n→∞ n
( n )n √ 2πn · e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
∑ (n!) 1n n
( n )n √ 2πn · e
a(n) ≈
∞ ∑
2n a(2n )
n=1
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
then
∑
∑ a(n) ≈ 2n · a(2n ) ∑1
example:
n
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:
n2
Examples using CCT □
If a(n) is decreasing
□
and a(n) > 0 for all large n′ s
then
∑
a(n) ≈ ∑
example:
∑
2n · a(2n )
1 n(ln n)3
Examples using CCT □
If a(n) is decreasing
□
and a(n) > 0 for all large n′ s
then
∑
a(n) ≈ ∑
example:
∑
2n · a(2n ) 1 n(ln n)(ln(ln n))
Sterlings Approximation
Sterlings Approximation n! ≈
( n )n √ 2πn · e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
∑
n2 (2n − 1)!
( n )n √ 2πn · e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
∑ n! 5n
( n )n √ 2πn · e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
∑ 2n n! nn
( n )n √ 2πn · e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
∑ 5n n! nn
( n )n √ 2πn · e
Sterlings Approximation
Sterlings Approximation n! ≈ 1
example:
(n!) n lim n→∞ n
( n )n √ 2πn · e
Sterlings Approximation
Sterlings Approximation n! ≈
example:
∑ (n!) 1n n
( n )n √ 2πn · e
