9 3 9 4 Convergence Tests 01
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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, π-series,
Direct Comparison, or
Limit Comparison
Tests
Assignment:
β’ P. 620-621: 1-7 odd, 11,
15, 17, 21, 25, 29-35
odd
β’ P. 628-630: 3-27 eoo,
29-36, 55-60
β’ Homework Supplement
Warm Up 1
Determine whether
β
π=1
β 1
π=1 π
1
1 1 1
=1+ + + +β―
π
2 3 4
converges or diverges.
Harmonic Series
πth Term Test:
1
lim = 0
πββ π
No Conclusion
Warm Up 2
Show that π π₯ = π βπ₯ is positive, continuous,
and decreasing for π₯ β₯ 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, π-series, Direct
Comparison, or Limit Comparison Tests
Objective 1
The Integral Test
If π is positive, continuous, and
decreasing for π₯ β₯ 1 and
ππ = π π , then
β
β
ππ
π=1
π π₯ ππ₯
and
1
either both converge* or both diverge.
*not necessarily to the same value
Exercise 1
π
β
π=1 π2 +1
Determine whether
diverges.
Integral Test:
π₯
,
π₯ 2 +1
π π₯ =
converges or
π₯β₯1
π₯
Positive?
β
Continuous?
1
π₯
ππ₯ = β
2
π₯ +1
Decreasing?
π₯ 2 + 1 β π₯ 2π₯
πβ² π₯ =
π₯2 + 1 2
So Ξ£ 2 also
π₯ +1
diverges by the
Integral Test.
[Insert Calculus here]
βπ₯ 2 + 1
= 2
π₯ +1 2
<0
π 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?
β
If 1 π π₯ ππ₯ converges to πΏ,
then Ξ£ππ does not necessarily
converge to πΏ.
Exercise 2
Determine whether
diverges.
1
β
π=1 π2 +1
converges or
Exercise 3
Determine whether
diverges.
1
β
π=2 πln π
converges or
π-Series
A simple test for convergence involves the
π-series:
β
π=1
1
1
1
1
1
= π+ π+ π+ π+β―
π
π
1
2
3
4
where π is a
positive constant
Harmonic Series
When π = 1, the π-series is known as the
harmonic series, named after the points on
a vibrating string that produce multiples of its
fundamental frequency.
β
The Harmonic
Series
π=1
1
1 1 1
=1+ + + +β―
π
2 3 4
π-Series
Integral Test:
Let π π₯ =
1
,
π₯π
where π is a positive integer.
Positive?
β
Continuous?
1
1
ππ₯
π
π₯
Decreasing?
πβ²
π₯ = βπ β
π₯ βπβ1
= βπ β π₯
β π+1
Converges if π > 1
Diverges if 0 < π β€ 1
βπ
= π+1 < 0
π₯
π is decreasing
Convergence of π-Series
β
π=1
The π-series
1
1
1
1
1
= π+ π+ π+ π+β―
π
π
1
2
3
4
Converges* if π > 1
Diverges if 0 < π β€ 1
*Again, we only know if
the series converges, not
to what it converges.
Exercise 4
Determine whether
diverges.
β 1
π=1 π
converges or
DIVERGES!
β
The Harmonic
Series
π=1
1
1 1 1
=1+ + + +β―
π
2 3 4
Exercise 5
Determine whether
diverges.
1
β
π=1 π2
converges or
Shown by
Euler to
converge to
π2
6
Exercise 6
1. Assume 0 < ππ β€ ππ for all π. If β
π=1 ππ
converges, what must be true about
β
π=1 ππ ?
2. Assume 0 < ππ β€ ππ for all π. If β
π=1 ππ
diverges, what must be true about
β
π=1 ππ ?
The Direct Comparison Test
If 0 < ππ β€ ππ for all π and
β
π=1 ππ converges, then
β
π=1 ππ 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 < ππ β€ ππ for all π and
β
π=1 ππ diverges, then
β
π=1 ππ also diverges.
Exercise 7
Determine whether
1
β
π=1 2π+1 2
converges or
diverges.
Direct Comparison Test:
Compare to Ξ£
1
π2
Converges by π-series test
So our series needs to be smaller than
1
2π + 1
2
1
β€ 2
π
π2 β€ 4π2 + 4π + 1
1
π
0 β€ 3π2 + 4π + 1
Therefore,
1
Ξ£
2
2π+1
converges by
the Direct
Comparison
Test.
Protip: The Direct Comparison Test
When applying the Direct Comparison Test:
Choose a comparison series ππ that
Step 1
closely resembles given series ππ
Step 2 of new series ππ
Discuss convergence
If new series
converges:
ππ β€ ππ
3
Set upStep
inequality
If the above inequality is true, write
Step 4
conclusion by the Direct Comparison Test
If new series
diverges:
ππ β€ ππ
Exercise 8
Determine whether
diverges.
1
β
π=1 2+3π
converges or
Exercise 9
Determine whether
diverges.
π
3
β
π=1 2π β1
converges or
Exercise 10
Determine whether
diverges.
1
β
π=1 3π+2
converges or
If you canβt
establish an
inequality; try
The Limit
Comparison
Test
The Limit Comparison Test
Suppose ππ > 0, ππ > 0,
ππ
and lim
πββ ππ
= πΏ > 0,
then the two series Ξ£ππ and Ξ£ππ 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
β
π=1 3π+2
converges or
Limit Comparison Test:
Compare to Ξ£
1
lim 3π + 2
1
πββ
π
1
π
Divergent harmonic series
1
π
π
1
= lim
β = lim
=
πββ 3π + 2 1
πββ 3π + 2
3
Finite and
positive
Therefore,
1
Ξ£
3π+2
diverges by
the Limit
Comparison
Test
Protip: The Limit Comparison Test
When applying the Limit Comparison Test:
Choose a comparison series ππ that
Step 1
closely resembles given series ππ
Step 2 of new series ππ
Discuss convergence
3
SetStep
up limit
If the limit above is finite and positive, then
Step 4
both series have the same behavior.
ππ
lim
πββ ππ
Exercise 10
Determine whether
diverges.
1
β
π=1 2+ π
converges or
Exercise 11
Determine whether
diverges.
1
β
π=1 sin π
converges or
Exercise 12
Determine whether
diverges.
π
π2
β
π=1 4π3 +1
converges or
Test
Series
Convergence
Divergence
Comments
lim ππ β 0
Cannot be used to
show convergence
π β₯1
π
π=
1βπ
β
πth Term
ππ
πββ
π=1
β
ππ
Geometric
π
π <1
π=π
β
β
ππ , ππ = π π
Integral
β
π-Series
π=1
Direct
Comparison
Limit
Comparison
The Harmonic
Series Diverges!
1
1
ππ
π>1
0<πβ€1
ππ
0 < ππ β€ ππ , and
Ξ£ππ converges
0 < ππ β€ ππ , and
Ξ£ππ diverges
β
π=1
β
ππ
π=1
π π₯ ππ₯ converges
1
π=1
π π₯ ππ₯ diverges
π must positive,
continuous, and
decreasing
β
ππ
πββ ππ
lim
= πΏ > 0,
and Ξ£ππ converges
ππ
πββ ππ
lim
= πΏ > 0,
and Ξ£ππ diverges
9-3/9-4: Convergence Tests, I
Objectives:
1. To determine whether
a series of positive
terms converges or
diverges by the
Integral, π-series,
Direct Comparison, or
Limit Comparison
Tests
Assignment:
β’ P. 620-621: 1-7 odd, 11,
15, 17, 21, 25, 29-35
odd
β’ P. 628-630: 3-27 eoo,
29-36, 55-60
β’ Homework Supplement