Final Review – Exam 3
Sequences & Series
Improper Integrals
Chapter 10: Sequences and Series
1)
2)
3)
4)
5)
6)
7)
Sequences
Series
Sequence of Partial Sums
Named series: geometric, telescoping, p-series.
Tests for testing series
Absolute and Conditional Convergences
Error bound estimations for integral test and
alternating series test.
Please refer to lecture slides (notes) or textbook for details.
Sequences (10.1 & 10.2)
• Convergence of sequences
• Definitions of bounded and monotonic
sequences
• Theorem: Every bounded and monotonic
sequence is convergent.
Example 1 (exam 3 problem 2)
Given 𝑎𝑘+1 − 𝑎𝑘 ≥ 0 and 𝑎𝑘 < 17 for all 𝑘 ≥ 1.
Which one of the following is most completely
describes the sequence 𝑎𝑘 ?
a) monotonic and convergent
b) monotonic and divergent
c) bounded and convergent
d) bounded and divergent
e) monotonic, bounded and convergent
f) monotonic, bounded and divergent
Sequence of Partial Sums and Series (10.3)
The 𝑛th Partial Sum: The sum of the first n terms of an
infinite series .
∞
Infinite series:
Partial Sums:
𝑎𝑘 = 𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + ⋯
S1  a1
𝑘=1
S 2  a1  a2
The sums 𝑆1 , 𝑆2 , 𝑆3 , . . . form
a sequence 𝑆𝑛 . It is called the
sequence of partial sums.
S3  a1  a2  a3
S 4  a1  a2  a3  a4
n
S n  a1  a2  a3  a4  ...  an   ak
k 1
Sequence of Partial Sums and Series (10.3)
∞
𝑎𝑘 = 𝑎1 + 𝑎2 + 𝑎3 + ⋯
For the infinite series
𝑘=1
If the sequence of partial sums 𝑆𝑛 converges to 𝑆, i.e.,
lim 𝑆𝑛 = 𝑆 exists as a real number, then the series
𝑛→∞
∞
𝑎𝑘 converges to 𝑆. We say the sum of the series is 𝑆,
∞
𝑘=1
and write 𝑎1 + 𝑎2 + 𝑎3 + ⋯ = 𝑆 or
𝑎𝑘 = 𝑆
𝑘=1
If the sequence of partial sums 𝑆𝑛 diverges, the series also diverges.
Remark: if
∞
𝑘=1 𝑎𝑘
converges, then 𝑎𝑘 converges to zero (10.4).
Example 2 (exam 3 problem 1)
12𝑛
Given 𝑆𝑛 =
is the 𝑛th partial
2𝑛+3
series ∞
𝑘=1 𝑎𝑘 . Fill in the blanks.
𝑆𝑛 ______
F
𝑎𝑘 ______
B
∞
𝑘=1 𝑎𝑘
F
___
A.
B.
C.
D.
E.
F.
G.
sum of the
Diverges
Converges to 0
Converges to 1
Converges to 2
Converges to 3
Converges to 6
Converges to 12
Named Series
• Geometric series
∞
𝑎𝑟 𝑘−1 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 + 𝑎𝑟 4 + ⋯
𝑘=1
is convergent if 𝑟 < 1 and its sum is
𝑎
1−𝑟
.
𝒇𝒊𝒓𝒔𝒕 𝒕𝒆𝒓𝒎
The sum of a convergent geometric series is
𝟏 − 𝒓𝒂𝒕𝒊𝒐
If | 𝑟 | ≥ 1, the geometric series is divergent.
• Telescoping series
Example 3 (exam 3 problem 7)
Determine whether the series is convergent or
divergent. If it’s convergent, find its sum if
possible.
4
8
16
3 −2+ − +
−⋯
3
9
27
Named Series
• 𝒑-series
∞
1
𝑘𝑝
𝑘=1
The 𝑝-series converges if 𝑝 > 1 and diverges if 𝑝 ≤ 1.
∞
1
𝑘
When 𝑝 = 1, the series
is called Harmonic series:
𝑘=1
Notice that Harmonic series is divergent since 𝑝 = 1.
Example 4 (exam 3 problems 5 and 6)
Determine whether the series is convergent or
divergent.
∞
−4 𝑘+1 𝑘 8
5𝑘!
𝑘=1
∞
−3
𝑘=1
𝑘
1
𝜋+
𝑘
𝑘
Absolute & Conditional Convergence (10.6)
Definition:
if the series
𝑎𝑘 is called absolutely convergent
|𝑎𝑘 | is convergent.
Definition: 𝑎𝑘 is called conditionally convergent
if 𝑎𝑘 is convergent but |𝑎𝑘 | is divergent.
Remark:
• Any positive term series that is convergent is
absolutely convergent.
• A positive term series cannot be conditionally
convergent.
How to handle the questions ?
Determine whether the series is Conditionally
Convergent, Absolutely Convergent or Divergent.
1) For absolute convergence:
 Use Ratio Test or Root Test, or
 Use the definition.
2) For conditional convergence:
 Use the definition.
Determine whether the series is Convergent or Divergent:
 use any test that applies.
How to use the definition to show the series 𝒂𝒌
is absolutely or conditionally convergent?
1) To show the series
 show the series
𝒂𝒌 is absolutely convergent:
𝒂𝒌 is convergent
2) To show the series 𝒂𝒌 is conditionally
convergent, we need to show two things:
 show the series
and
 show the series
𝒂𝒌 is convergent
𝒂𝒌 is divergent
Example 5 (exam 3 problem 8)
Prove that the series is conditionally convergent.
∞
−1
3
𝑘=1
𝑘−1
𝑘
Example 6 (exam 3 problem 9)
Show that the series is absolutely convergent.
∞
𝑘=1
sin𝑘
𝑘 𝑘
Estimation for Integral Test (10.4)
Suppose that the series ∞
𝑘=1 𝑎𝑘 is convergent (by
the Integral Test).
Let 𝒇 be a continuous, positive, decreasing function
on 1, ∞ , and let 𝑎𝑘 = 𝑓 𝑘 . Then the remainder
𝑅𝑛 = 𝑆 − 𝑆𝑛 satisfies
∞
∞
𝑓 𝑥 𝑑𝑥 ≤ 𝑅𝑛 ≤
𝑛+1
𝑓 𝑥 𝑑𝑥
𝑛
The exact value of the sum is bounded as follows:
∞
𝑆𝑛 +
∞
𝑓 𝑥 𝑑𝑥 ≤ S ≤ 𝑆𝑛 +
𝑛+1
𝑓 𝑥 𝑑𝑥
𝑛
Alternating Series Estimation Theorem (10.6)
∞
𝑘=1
Suppose that the series
−1 𝑘−1 𝑏𝑘 is
convergent by the Alternating Series Test.
𝑛
𝑘=1
Let 𝑆𝑛 =
−1 𝑘−1 𝑏𝑘 be an approximation
to the sum 𝑆 of the series.
Then
𝑅𝑛
=
𝑆 − 𝑆𝑛
≤ 𝑏𝑛+1
the first neglected term
In other words, the remainder (the error) is the less than
or equal to the first neglected term. It is the first term
that is not used in the approximation.
Example 7 (exam 3 problem 3)
Determine the minimum number of terms need
to approximate the sum of the series to within
0.09 = 9 × 10−2 .
∞
−1 𝑘−1
𝑘
𝑘=1
Type I - Improper Integrals (8.8)
1)
∞
𝑓
𝑎
𝑥 𝑑𝑥 =
𝑡
lim 𝑎 𝑓
𝑡→∞
𝑥 𝑑𝑥 if the limit on the
right side exists as a finite number and the integral
𝑡
𝑓 𝑥 𝑑𝑥 exists for every number 𝑡 ≥ 𝑎.
𝑎
2)
𝑎
−∞
𝑓 𝑥 𝑑𝑥 =
𝑎
lim 𝑡 𝑓
𝑡 → −∞
𝑥 𝑑𝑥 if the limit on
the right side exists as a finite number and the
𝑎
integral 𝑡 𝑓 𝑥 𝑑𝑥 exists for every number 𝑡 ≤ 𝑎.
If the limits exist we say the integral converges or is convergent.
Otherwise, we say the integral diverges or is divergent.
3)
∞
𝑓
−∞
𝑎
∞
𝑓
𝑎
𝑥 𝑑𝑥 =
𝑓 𝑥 𝑑𝑥 +
𝑥 𝑑𝑥 if both
−∞
integrals on the right side are convergent.
Type II - Improper Integrals (8.8)
1) If 𝑓 is continuous on [𝑎, 𝑏) and 𝑓 has a vertical
𝑡
asymptote at 𝑏 : 𝑏
a
𝑓(𝑥) 𝑑𝑥 = lim −
𝑡→𝑏
𝑓(𝑥) 𝑑𝑥
a
2) If 𝑓 is continuous on (𝑎, 𝑏] and 𝑓 has a vertical
𝑏
asymptote at 𝑎 : 𝑏
a
𝑓(𝑥) 𝑑𝑥 = lim +
𝑡→𝑎
𝑓(𝑥) 𝑑𝑥
𝑡
(provide that the limits on the right side exist)
3) If 𝑓 is continuous on [𝑎, 𝑏] and 𝑓 has a vertical
asymptote at 𝑐 inside [𝑎, 𝑏].
𝑏
𝑓(𝑥) 𝑑𝑥 =
a
(provide that the
𝑐
𝑏
integrals on the right
𝑓(𝑥) 𝑑𝑥 + 𝑓(𝑥) 𝑑𝑥
side converge)
𝑎
𝑐
Example 8 (exam 3 problem 10)
Evaluate the integral.
𝜋
0
6
cos(3𝑥)
𝑑𝑥
sin 3𝑥 − 1