Transcript 9 5 9 6 Convergence Tests 02

9-5/9-6: Convergence Tests, 2
Objectives:
Assignment:
1. To determine whether • P. 637-638: 11-27 eoo,
a series converges or
43, 45, 47-61 eoo, 67diverges by the
70
Alternating Series and • P. 645-647: 13-29 odd,
Ratio Tests
51-65 odd, 87
• Homework Supplement
Warm Up
Find the first 5 terms of
∞
−1
𝑛=1
𝑛+1
∞
𝑛=1
−1
𝑛+1 1
𝑛
1
1
1
1 1
= 1 −
+ − + −⋯
𝑛
5
2
3 4
Alternating Series
.
Objective 1
You will be able to
determine whether a
series converges or
diverges by the
Alternating Series and
Ratio Tests
Alternating Series
An alternating series has terms that
alternate in sign.
∞
−1
𝑛+1
𝑛=1
∞
−1
𝑛
1 2 3 4 5
=− + − + − +⋯
𝑛+1
2 3 4 5 6
𝑛
𝑛=1
∞
𝑛=0
1
−
2
1
1 1 1 1
=1− + − + −⋯
𝑛
2 3 4 5
𝑛
1 1 1 1
=1− + − +
−⋯
2 4 8 16
An alternating
series will
converge if the
terms are
decreasing in
magnitude
toward zero.
Alternating Series Test
Let 𝑎𝑛 > 0. The
alternating series
Terms are decreasing
in magnitude
∞
−1 𝑛 𝑎𝑛
𝑎𝑛+1 ≤ 𝑎𝑛 , for all 𝑛
𝑛=1
∞
and
−1
𝑛+1 𝑎
𝑛
𝑛=1
converge if:
and
lim 𝑎𝑛 = 0
𝑛→∞
Alternating Series Test
Graphically, you can see that if the
alternating terms of a series are decreasing
in magnitude toward zero, their sum is finite.
Exercise 1
Determine whether the series converges or
diverges.
∞
−1
𝑛+1
𝑛=1
1
1 1 1 1
=1− + − + −⋯
𝑛
2 3 4 5
Alternating
Harmonic Series
Alternating Series Test:
1
=0
𝑛→∞ 𝑛
lim
1
1
≤
𝑛+1 𝑛
𝑛≤𝑛+1
0≤1
Converges by
the Alternating
Series Test
Exercise 2
Determine whether the series
converges or diverges.
𝑛 3𝑛
−1
∞
𝑛=1 4𝑛−1
Protip: Alternating Series
If lim 𝑎𝑛 ≠ 0, then the series diverges by
𝑛→∞
the 𝑛th Term Divergence Test.
Perform
the limit
test first
Check the
inequality to
show terms
are
decreasing
In some cases, you can
substitute 𝑓 𝑥 for 𝑎𝑛 and
check the 1st derivative to
see if the terms are
decreasing.
Exercise 3
Determine whether the series
converges or diverges.
∞
𝑛=1
−1
2
𝑛+1 𝑛
𝑛3 +1
Approximating Alternating Series
For convergent geometric series, we have
a handy formula for the infinite sum, but this
is not the case for other sums.
A geometric series with ratio 𝑟
diverges if 𝑟 ≥ 1. If 0 < 𝑟 < 1,
then the series converges to the sum
∞
𝑎𝑟 𝑛 =
𝑛=0
𝑎
1−𝑟
Approximating Alternating Series
However, we can approximate a series by using a
partial sum. With Alternating Series, as with Taylor
polynomials, we have a way to bound the error of
the approximation.
𝑓 𝑛+1 𝑧 is the maximum value of the
𝑛 + 1 th derivative between 𝑥 and 𝑐.
𝑅𝑛 𝑥
≤
𝑓
𝑥≤𝑧≤𝑐
𝑛+1
𝑧
𝑥−𝑐
𝑛+1 !
𝑛+1
𝑐≤𝑧≤𝑥
Remainder
𝑆𝑁 , the 𝑛th partial sum, is only approximately equal
to 𝑆, the actual sum of the series:
𝑆 ≈ 𝑆𝑁
𝑆 = 𝑆𝑁 + 𝑅𝑁
If we added (or subtracted) a little something to (or
from) our approximation, we would arrive at the
exact value. That little something is called the
remainder.
Remainder
𝑆𝑁 , the 𝑛th partial sum, is only approximately equal
to 𝑆, the actual sum of the series:
𝑆 ≈ 𝑆𝑁
𝑆 = 𝑆𝑁 + 𝑅𝑁
𝑆 − 𝑆𝑁 = 𝑅𝑁
𝑆 − 𝑆𝑁 = 𝑅𝑁
Remainder
𝑆 − 𝑆4 = 𝑅4 ≤ 𝑏5
First
Neglected
Term
Remainder
First
Neglected
Term
𝑏6 ≥ 𝑅4 = 𝑆 − 𝑆5
Alternating Series Remainder
If a convergent alternating series satisfies the
condition 𝑎𝑛+1 ≤ 𝑎𝑛 , the absolute value of
the remainder 𝑅𝑁 involved in approximating
the sum 𝑆 by 𝑆𝑁 is less than (or equal to) the
first neglected term.
𝑆 − 𝑆𝑁 = 𝑅𝑁 ≤ 𝑎𝑁+1
First
Neglected
Term
Exercise 4
∞
𝑛=1
Approximate
−1
𝑛+1
1
𝑛!
by its first 6 terms.
Then calculate the remainder of the approximation.
1 1 1
1
1
91
𝑆6 = 1 − + −
+
−
=
≈ 0.63194
2 6 24 120 720
144
First Neglected Term
𝑆 − 𝑆6 ≤ 0.0002
7th
𝑆 − 0.63194 ≤ 0.0002
𝑆 − 𝑆6 = 𝑅6 ≤ 𝑎7
𝑅6
−0.0002 ≤ 𝑆 − 0.63194 ≤ 0.0002
1
≤
≈ 0.0002
5040
0.63174 ≤ 𝑆 ≤ 0.63214
Absolute Convergence
Sometimes a series contains both positive
and negative terms but is not an alternating
series:
∞
cos 𝑛
cos 2 cos 3 cos 4 cos 5
=
cos
1
+
+
+
+
+⋯
𝑛2
4
9
16
25
𝑛=1
+
−
−
−
+
In such cases, it’s easier to check the convergence
of the absolute value of its terms.
Absolutely Convergent
A series Σ𝑎𝑛 is absolutely convergent if
Σ 𝑎𝑛 converges.
Exercise 5
Determine if the series is absolutely convergent.
∞
𝑛=1
cos 𝑛
cos 2 cos 3 cos 4 cos 5
=
cos
1
+
+
+
+
+⋯
𝑛2
4
9
16
25
−1 ≤ cos 𝑛 ≤ 1
cos 𝑛 ≤ 1
cos 𝑛
1
≤ 2
2
𝑛
𝑛
Therefore,
∞ cos 𝑛
𝑛=1 𝑛2
Divide by 𝑛2 , which is positive
Convergent 𝑝-Series
is absolutely convergent by the Direct Comparison Test.
Exercise 5
Determine if the series is absolutely convergent.
∞
𝑛=1
cos 𝑛
cos 2 cos 3 cos 4 cos 5
=
cos
1
+
+
+
+
+⋯
𝑛2
4
9
16
25
−1 ≤ cos 𝑛 ≤ 1
cos 𝑛 ≤ 1
cos 𝑛
1
≤ 2
2
𝑛
𝑛
Therefore,
∞ cos 𝑛
𝑛=1 𝑛2
Divide by 𝑛2 , which is positive
Convergent 𝑝-Series
is absolutely convergent by the Direct Comparison Test.
It can be
shown that if a
series is
absolutely
convergent,
then it is
convergent.
Absolutely Convergent
A series Σ𝑎𝑛 is absolutely convergent if
Σ 𝑎𝑛 converges.
Absolute
Convergence
Theorem
If Σ 𝑎𝑛 converges, then Σ𝑎𝑛 converges.
Exercise 6
Is the converse of the Absolute Convergence
Theorem true? In other words, if Σ𝑎𝑛 converges, is
Σ 𝑎𝑛 convergent?
∞
−1
𝑛+1
𝑛+1
1
𝑛
𝑛=1
−1
1
1 1 1 1
=1− + − + −⋯
𝑛
2 3 4 5
1
=
𝑛
Divergent
Harmonic Series
Alternating
Harmonic Series
Converges by
the Alternating
Series Test
Thus, the
converse is not
true. This is an
example of a
conditionally
convergent
series.
Conditionally Convergent
A series Σ𝑎𝑛 is conditional convergent if
Σ𝑎𝑛 converges, but Σ 𝑎𝑛 diverges.
Conditionally
convergent
series are a bit
wonky. If you
rearrange the
terms, you get
a different sum.
Riemann proved that a
conditionally convergent series
can be rearranged to sum to
any real number.
Conditionally Convergent
A series Σ𝑎𝑛 is conditional convergent if
Σ𝑎𝑛 converges, but Σ 𝑎𝑛 diverges.
∞
Conditionally
convergent
series are a bit
wonky. If you
rearrange the
terms, you get
a different sum.
−1
𝑛=1
𝑛+1
1
1 1 1 1
= 1 − + − + − ⋯ = ln 2
𝑛
2 3 4 5
However…
Exercise 7
Determine whether each series is convergent or
divergent. Classify any convergent series as
absolutely or conditionally convergent.
1.
−1 𝑛 𝑛!
∞
𝑛=0 2𝑛
2.
−1 𝑛
∞
𝑛=1
𝑛
3.
−1 𝑛 𝑛+1 /2
∞
𝑛=1
3𝑛
Common Ratio
Recall that a geometric series converges if its
common ratio is less than 1 and diverges if it is
greater than 1.
A geometric series with ratio 𝑟
diverges if 𝑟 ≥ 1. If 0 < 𝑟 < 1,
then the series converges to the sum
∞
𝑎
𝑎𝑟 =
1−𝑟
𝑛
𝑛=0
This fact can be
used to prove the
Ratio Test, where
if the limit of the
ratio of term is
less than 1, then
the series is
convergent.
Ratio Test
Let Σ𝑎𝑛 be a series of nonzero terms.
lim
𝑛→∞
𝑎𝑛+1
<1
𝑎𝑛
Σ𝑎𝑛
converges
absolutely
lim
Σ𝑎𝑛
diverges
𝑎𝑛+1
lim
>1
𝑛→∞ 𝑎𝑛
𝑛→∞
𝑎𝑛+1
=1
𝑎𝑛
The Ratio
Test is
inconclusive
Protip: The Ratio Test
The Ratio Test plays well with quickly
converging/diverging series, like ones containing
factorial or exponential terms, but it does not play
well with 𝑝-series.
Divergent Harmonic Series:
Convergent 𝑝-Series:
1
Σ
𝑛
1
𝑛
= 1
lim 𝑛 + 1 = lim
1
𝑛→∞ 𝑛 + 1
𝑛→∞
𝑛
1
Σ 2
𝑛
lim
𝑛→∞
1
𝑛+1
1
𝑛2
2
𝑛2
= lim
𝑛→∞ 𝑛 + 1
2
= 1
Exercise 8
Determine the convergence or divergence of
𝑛
∞ 2
𝑛=0 𝑛! .
Ratio Test:
lim
𝑛→∞
2𝑛+1
𝑛+1
2
𝑛!
𝑛+1 !
∙
=
= lim
𝑛→∞ 𝑛 + 1 ! 2𝑛
2𝑛
𝑛!
Therefore,
2𝑛
Σ
𝑛!
2
lim
= 0 <1
𝑛→∞ 𝑛 + 1
converges absolutely by the Ratio Test.
Exercise 9
Determine whether each series is convergent or
divergent.
1.
2 𝑛+1
∞ 𝑛 2
𝑛=0 3𝑛
2.
𝑛
∞ 𝑛
𝑛=1 𝑛!
3.
∞
𝑛=1
−1
𝑛
𝑛
𝑛+1
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
Test
Series
∞
𝑛
Alternating Series
−1 𝑎𝑛
𝑛=1
Convergence
𝑎𝑛
𝑛=1
Comments
𝑛→∞
lim 𝑎𝑛 = 0
Remainder:
𝑎𝑛+1 ≤ 𝑎𝑛
𝑅𝑁 ≤ 𝑎𝑁+1
∞
Ratio
Divergence
lim
𝑛→∞
𝑎𝑛+1
<1
𝑎𝑛
lim
𝑛→∞
𝑎𝑛+1
>1
𝑎𝑛
Test is inconclusive:
lim
𝑛→∞
𝑎𝑛+1
=1
𝑎𝑛
Exercise 10
Determine the convergence or divergence of
each series.
9-5/9-6: Convergence Tests, 2
Objectives:
Assignment:
1. To determine whether • P. 637-638: 11-27 eoo,
a series converges or
43, 45, 47-61 eoo, 67diverges by the
70
Alternating Series and • P. 645-647: 13-29 odd,
Ratio Tests
51-65 odd, 87
• Homework Supplement