Transcript 9 3 9 4 Convergence Tests 01

9-3/9-4: Convergence Tests, I
Objectives:
1. To determine whether
a series of positive
terms converges or
diverges by the
Integral, 𝑝-series,
Direct Comparison, or
Limit Comparison
Tests
Assignment:
• P. 620-621: 1-7 odd, 11,
15, 17, 21, 25, 29-35
odd
• P. 628-630: 3-27 eoo,
29-36, 55-60
• Homework Supplement
Warm Up 1
Determine whether
∞
𝑛=1
∞ 1
𝑛=1 𝑛
1
1 1 1
=1+ + + +⋯
𝑛
2 3 4
converges or diverges.
Harmonic Series
𝑛th Term Test:
1
lim = 0
𝑛→∞ 𝑛
No Conclusion
Warm Up 2
Show that 𝑓 𝑥 = 𝑒 −𝑥 is positive, continuous,
and decreasing for 𝑥 ≥ 1.
These are the necessary conditions
for the Integral Test.
You will be able to determine whether a
series of positive terms converges or
diverges by the Integral, 𝑝-series, Direct
Comparison, or Limit Comparison Tests
Objective 1
The Integral Test
If 𝑓 is positive, continuous, and
decreasing for 𝑥 ≥ 1 and
𝑎𝑛 = 𝑓 𝑛 , then
∞
∞
𝑎𝑛
𝑛=1
𝑓 𝑥 𝑑𝑥
and
1
either both converge* or both diverge.
*not necessarily to the same value
Exercise 1
𝑛
∞
𝑛=1 𝑛2 +1
Determine whether
diverges.
Integral Test:
𝑥
,
𝑥 2 +1
𝑓 𝑥 =
converges or
𝑥≥1
𝑥
Positive?
∞
Continuous?
1
𝑥
𝑑𝑥 = ∞
2
𝑥 +1
Decreasing?
𝑥 2 + 1 − 𝑥 2𝑥
𝑓′ 𝑥 =
𝑥2 + 1 2
So Σ 2 also
𝑥 +1
diverges by the
Integral Test.
[Insert Calculus here]
−𝑥 2 + 1
= 2
𝑥 +1 2
<0
𝑓 is decreasing
Protip: The Integral Test
When applying the Integral Test:
Be sure the
conditions are met:
Note that the test does
not tell you to what the
series converges.
Positive?
Continuous?
Decreasing?
∞
If 1 𝑓 𝑥 𝑑𝑥 converges to 𝐿,
then Σ𝑎𝑛 does not necessarily
converge to 𝐿.
Exercise 2
Determine whether
diverges.
1
∞
𝑛=1 𝑛2 +1
converges or
Exercise 3
Determine whether
diverges.
1
∞
𝑛=2 𝑛ln 𝑛
converges or
𝑝-Series
A simple test for convergence involves the
𝒑-series:
∞
𝑛=1
1
1
1
1
1
= 𝑝+ 𝑝+ 𝑝+ 𝑝+⋯
𝑝
𝑛
1
2
3
4
where 𝑝 is a
positive constant
Harmonic Series
When 𝑝 = 1, the 𝑝-series is known as the
harmonic series, named after the points on
a vibrating string that produce multiples of its
fundamental frequency.
∞
The Harmonic
Series
𝑛=1
1
1 1 1
=1+ + + +⋯
𝑛
2 3 4
𝑝-Series
Integral Test:
Let 𝑓 𝑥 =
1
,
𝑥𝑝
where 𝑝 is a positive integer.
Positive?
∞
Continuous?
1
1
𝑑𝑥
𝑝
𝑥
Decreasing?
𝑓′
𝑥 = −𝑝 ∙
𝑥 −𝑝−1
= −𝑝 ∙ 𝑥
− 𝑝+1
Converges if 𝑝 > 1
Diverges if 0 < 𝑝 ≤ 1
−𝑝
= 𝑝+1 < 0
𝑥
𝑓 is decreasing
Convergence of 𝑝-Series
∞
𝑛=1
The 𝒑-series
1
1
1
1
1
= 𝑝+ 𝑝+ 𝑝+ 𝑝+⋯
𝑝
𝑛
1
2
3
4
Converges* if 𝑝 > 1
Diverges if 0 < 𝑝 ≤ 1
*Again, we only know if
the series converges, not
to what it converges.
Exercise 4
Determine whether
diverges.
∞ 1
𝑛=1 𝑛
converges or
DIVERGES!
∞
The Harmonic
Series
𝑛=1
1
1 1 1
=1+ + + +⋯
𝑛
2 3 4
Exercise 5
Determine whether
diverges.
1
∞
𝑛=1 𝑛2
converges or
Shown by
Euler to
converge to
𝜋2
6
Exercise 6
1. Assume 0 < 𝑎𝑛 ≤ 𝑏𝑛 for all 𝑛. If ∞
𝑛=1 𝑏𝑛
converges, what must be true about
∞
𝑛=1 𝑎𝑛 ?
2. Assume 0 < 𝑏𝑛 ≤ 𝑎𝑛 for all 𝑛. If ∞
𝑛=1 𝑏𝑛
diverges, what must be true about
∞
𝑛=1 𝑎𝑛 ?
The Direct Comparison Test
If 0 < 𝑎𝑛 ≤ 𝑏𝑛 for all 𝑛 and
∞
𝑛=1 𝑏𝑛 converges, then
∞
𝑛=1 𝑎𝑛 also converges.
If a series is bigger than
a divergent series,
then it diverges.
If a series is smaller than
a convergent series,
then it converges.
If 0 < 𝑏𝑛 ≤ 𝑎𝑛 for all 𝑛 and
∞
𝑛=1 𝑏𝑛 diverges, then
∞
𝑛=1 𝑎𝑛 also diverges.
Exercise 7
Determine whether
1
∞
𝑛=1 2𝑛+1 2
converges or
diverges.
Direct Comparison Test:
Compare to Σ
1
𝑛2
Converges by 𝑝-series test
So our series needs to be smaller than
1
2𝑛 + 1
2
1
≤ 2
𝑛
𝑛2 ≤ 4𝑛2 + 4𝑛 + 1
1
𝑛
0 ≤ 3𝑛2 + 4𝑛 + 1
Therefore,
1
Σ
2
2𝑛+1
converges by
the Direct
Comparison
Test.
Protip: The Direct Comparison Test
When applying the Direct Comparison Test:
Choose a comparison series 𝑏𝑛 that
Step 1
closely resembles given series 𝑎𝑛
Step 2 of new series 𝑏𝑛
Discuss convergence
If new series
converges:
𝑎𝑛 ≤ 𝑏𝑛
3
Set upStep
inequality
If the above inequality is true, write
Step 4
conclusion by the Direct Comparison Test
If new series
diverges:
𝑏𝑛 ≤ 𝑎𝑛
Exercise 8
Determine whether
diverges.
1
∞
𝑛=1 2+3𝑛
converges or
Exercise 9
Determine whether
diverges.
𝑛
3
∞
𝑛=1 2𝑛 −1
converges or
Exercise 10
Determine whether
diverges.
1
∞
𝑛=1 3𝑛+2
converges or
If you can’t
establish an
inequality; try
The Limit
Comparison
Test
The Limit Comparison Test
Suppose 𝑎𝑛 > 0, 𝑏𝑛 > 0,
𝑎𝑛
and lim
𝑛→∞ 𝑏𝑛
= 𝐿 > 0,
then the two series Σ𝑎𝑛 and Σ𝑏𝑛 either both
converge* or both diverge.
*Yet again, this
test does not
tell you to what
the series
converges.
Exercise 9
Determine whether
diverges.
1
∞
𝑛=1 3𝑛+2
converges or
Limit Comparison Test:
Compare to Σ
1
lim 3𝑛 + 2
1
𝑛→∞
𝑛
1
𝑛
Divergent harmonic series
1
𝑛
𝑛
1
= lim
∙ = lim
=
𝑛→∞ 3𝑛 + 2 1
𝑛→∞ 3𝑛 + 2
3
Finite and
positive
Therefore,
1
Σ
3𝑛+2
diverges by
the Limit
Comparison
Test
Protip: The Limit Comparison Test
When applying the Limit Comparison Test:
Choose a comparison series 𝑏𝑛 that
Step 1
closely resembles given series 𝑎𝑛
Step 2 of new series 𝑏𝑛
Discuss convergence
3
SetStep
up limit
If the limit above is finite and positive, then
Step 4
both series have the same behavior.
𝑎𝑛
lim
𝑛→∞ 𝑏𝑛
Exercise 10
Determine whether
diverges.
1
∞
𝑛=1 2+ 𝑛
converges or
Exercise 11
Determine whether
diverges.
1
∞
𝑛=1 sin 𝑛
converges or
Exercise 12
Determine whether
diverges.
𝑛
𝑛2
∞
𝑛=1 4𝑛3 +1
converges or
Test
Series
Convergence
Divergence
Comments
lim 𝑎𝑛 ≠ 0
Cannot be used to
show convergence
𝑟 ≥1
𝑎
𝑆=
1−𝑟
∞
𝑛th Term
𝑎𝑛
𝑛→∞
𝑛=1
∞
𝑎𝑟
Geometric
𝑛
𝑟 <1
𝑛=𝟎
∞
∞
𝑎𝑛 , 𝑎𝑛 = 𝑓 𝑛
Integral
∞
𝑝-Series
𝑛=1
Direct
Comparison
Limit
Comparison
The Harmonic
Series Diverges!
1
1
𝑛𝑝
𝑝>1
0<𝑝≤1
𝑎𝑛
0 < 𝑎𝑛 ≤ 𝑏𝑛 , and
Σ𝑏𝑛 converges
0 < 𝑏𝑛 ≤ 𝑎𝑛 , and
Σ𝑏𝑛 diverges
∞
𝑛=1
∞
𝑎𝑛
𝑛=1
𝑓 𝑥 𝑑𝑥 converges
1
𝑛=1
𝑓 𝑥 𝑑𝑥 diverges
𝑓 must positive,
continuous, and
decreasing
∞
𝑎𝑛
𝑛→∞ 𝑏𝑛
lim
= 𝐿 > 0,
and Σ𝑏𝑛 converges
𝑎𝑛
𝑛→∞ 𝑏𝑛
lim
= 𝐿 > 0,
and Σ𝑏𝑛 diverges
9-3/9-4: Convergence Tests, I
Objectives:
1. To determine whether
a series of positive
terms converges or
diverges by the
Integral, 𝑝-series,
Direct Comparison, or
Limit Comparison
Tests
Assignment:
• P. 620-621: 1-7 odd, 11,
15, 17, 21, 25, 29-35
odd
• P. 628-630: 3-27 eoo,
29-36, 55-60
• Homework Supplement