Calculus II
Chapter 3 Infinite Sequences and Series
3.1 Sequences
Representing Sequences
A sequence is a list of numbers
in a given order. Each of 𝑎1 , 𝑎2 , 𝑎3 and so on represents a number. These are the terms of the
sequence. The integer 𝑛 is called the index of 𝑎𝑛 and indicates where 𝑎𝑛 occurs in the list.
Example
The sequence
has first term 𝑎1 = 2, second term 𝑎2 = 4, and 𝑛th term 𝑎𝑛 = 2𝑛.
We can think of the sequence
as a function that sends 1 to 𝑎1 , 2 to 𝑎2 , 3 to 𝑎3 , and in general sends the positive integer 𝑛 to
the 𝑛th term 𝑎𝑛 . More precisely, an infinite sequence of numbers is a function whose domain
is the set of positive integers.
Sequences can be described by writing rules that specify their terms, such as
or by listing terms:
Sequences can be represented as points on the real line or as points in the plane where the
horizontal axis 𝑛 is the index number of the term and the vertical axis 𝑎𝑛 is its value.
Example [Fibonacci Sequence]
𝑎1 = 1,
𝑎2 = 1,
𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2 for 𝑛 > 2.
1, 1, 2, 3, 5, 8, 13, 21 …
Convergence and Divergence
A sequence {𝑎𝑛 } converges to the number 𝐿 if we can make the terms 𝑎𝑛 as close to 𝐿 as
we like by taking 𝑛 sufficiently large, and we write
lim 𝑎𝑛 = 𝐿 or 𝑎𝑛 → ∞.
𝑛→∞
If no such number 𝐿 exists, then we say that {𝑎𝑛 } diverges.
Intuitively, {𝑎𝑛 } converges to 𝐿 if we can make the term 𝑎𝑛 as close to 𝐿 as we like by taking
𝑛 sufficiently large.
Example
Show that
Example
Example
A sequence may diverge without diverging to infinity or negative infinity.
Calculating Limits of Sequences
Example
Exercise
Every nonzero multiple of a divergent sequence diverges.
Proof:
Let {𝑎𝑛 } be a divergent sequence and 𝑘 ≠ 0.
Suppose that {𝑘𝑎𝑛 } is convergent.
1
Then {𝑎𝑛 } = {𝑘 (𝑘𝑎𝑛 )} is also convergent. This is impossible.
We conclude that {𝑘𝑎𝑛 } must be divergent.
∎
If {𝑎𝑛 } converges and {𝑏𝑛 } diverges, then {𝑎𝑛 + 𝑏𝑛 } and {𝑎𝑛 − 𝑏𝑛 } both diverge.
Proof:
Suppose that {𝑎𝑛 + 𝑏𝑛 } converges.
Then {𝑏𝑛 } = {(𝑎𝑛 + 𝑏𝑛 ) − 𝑎𝑛 } converges. This is impossible.
We conclude that {𝑎𝑛 + 𝑏𝑛 } must diverge.
Remark:
If {𝑎𝑛 } and {𝑏𝑛 } both diverge, then {𝑎𝑛 + 𝑏𝑛 } may be convergent or divergent.
∎
An immediate consequence of Theorem 2 is that if |𝑏𝑛 | ≤ 𝑐𝑛 and 𝑐𝑛 → 0, then 𝑏𝑛 → 0.
Example
Exercise
Exercise
Example
Example
Exercise
Show that for 𝑥 > 0,
lim 𝑥1/𝑛 = 1.
𝑛→∞
Exercise
Show that if |𝑟| < 1, then
lim 𝑟 𝑛 = 0.
𝑛→∞
∞
𝑟 > 1,
1
𝑟 = 1,
lim 𝑟 𝑛 = {
|𝑟| < 1,
𝑛→∞
0
does not exist 𝑟 ≤ −1.
Remark:
If lim 𝑓(𝑛) = 𝐿, it needs not be true that lim 𝑓(𝑥) = 𝐿.
𝑛→∞
𝑥→∞
For example, lim cos(2𝜋𝑛) = 1 but lim cos(2𝜋𝑥) does not exist.
𝑛→∞
𝑥→∞
Using L’Hôpital’s Rule
Example
Example
Does the sequence whose 𝑛th term is
converge? If so, find
Exercise
Show that
lim 𝑛1/𝑛 = 1.
𝑛→∞
Exercise
Show that for any 𝑟 ∈ ℝ,
𝑟 𝑛
lim (1 + ) = 𝑒 𝑟 .
𝑛→∞
𝑛
Commonly Occurring Limits
Proof:
(6) Choose an integer 𝑀 > |𝑥|. Then for 𝑛 > 𝑀,
𝑛
|𝑥|𝑛
|𝑥|𝑛
|𝑥|𝑛
𝑥𝑛
𝑀𝑚 |𝑥|
0≤| |=
=
≤
=
( ) →0
𝑛!
𝑛!
𝑀! (𝑀 + 1) … (𝑛) 𝑀! 𝑀𝑛−𝑚
𝑀! 𝑀
by (4) since |𝑥|/𝑀 < 1.
The result follows by the Sandwich Theorem.
∎
Example
Bounded Monotonic Sequences
Definition
A sequence {𝑎𝑛 } is bounded above if there is a number 𝑀 such that
𝑎𝑛 ≤ 𝑀 for all 𝑛 ≥ 1.
A sequence {𝑎𝑛 } is bounded below if there is a number 𝑚 such that
𝑎𝑛 ≥ 𝑚 for all 𝑛 ≥ 1.
If a sequence is bounded above and below, then it is called a bounded sequence.
Example
Theorem
Convergent sequences are bounded.
Proof:
Suppose {𝑎𝑛 } converges to 𝐿.
There exists a positive integer 𝑁 such that for 𝑛 > 𝑁,
|𝑎𝑛 − 𝐿| < 1
⇒ |𝑎𝑛 | = |𝑎𝑛 − 𝐿 + 𝐿| ≤ |𝑎𝑛 − 𝐿| + |𝐿| < 1 + |𝐿|
Let 𝑀 = max{|𝑎1 |, … , |𝑎𝑁 |, 1 + |𝐿|}.
Then |𝑎𝑛 | ≤ 𝑀 for all 𝑛 ∈ ℕ.
Therefore, {𝑎𝑛 } is bounded.
∎
Definition
Let {𝑎𝑛 } be a sequence.
i. {𝑎𝑛 } is increasing if 𝑎𝑛 ≤ 𝑎𝑛+1 for all 𝑛.
𝑎1 ≤ 𝑎2 ≤ 𝑎3 ≤ ⋯
ii. {𝑎𝑛 } is decreasing if 𝑎𝑛 ≥ 𝑎𝑛+1 for all 𝑛.
𝑎1 ≥ 𝑎2 ≥ 𝑎3 ≥ ⋯
iii. {𝑎𝑛 } is monotonic if it is increasing or decreasing.
Example
A way to show that {𝑎𝑛 } is monotonic is to define a differentiable function 𝑓(𝑥) satisfying
𝑓(𝑛) = 𝑎𝑛 .
If 𝑓 ′ (𝑥) ≤ 0 for 𝑥 ≥ 𝑁, then {𝑎𝑛 } is decreasing for 𝑛 ≥ 𝑁.
If 𝑓 ′ (𝑥) ≥ 0 for 𝑥 ≥ 𝑁, then {𝑎𝑛 } is increasing for 𝑛 ≥ 𝑁.
Example
Show that the sequence {𝑎𝑛 } defined by
𝑎𝑛 =
10𝑛
𝑛2 + 16
is decreasing for 𝑛 ≥ 4.
Exercise
Show that the sequence {𝑎𝑛 } defined by
𝑎𝑛 =
is decreasing.
𝑛
𝑛+1
Proof:
Suppose that {𝑎𝑛 } is increasing.
Since {𝑎𝑛 } is bounded above, {𝑎𝑛 } has a least upper bound 𝐿. (Completeness Axiom)
Therefore, 𝑎𝑛 ≤ 𝐿 for all 𝑛 ∈ ℕ.
Let 𝜖 > 0.
Then 𝐿 − 𝜖 is not a least upper bound of {𝑎𝑛 }.
There exists 𝑁 ∈ ℕ such that 𝑎𝑁 > 𝐿 − 𝜖.
For 𝑛 ≥ 𝑁, we get
𝐿 − 𝜖 < 𝑎𝑁 ≤ 𝑎𝑛 ≤ 𝐿 < 𝐿 + 𝜖,
that is,
|𝑎𝑛 − 𝐿| < 𝜖.
We conclude that 𝑎𝑛 → 𝐿.
The proof for decreasing sequences bounded below is similar.
∎
3.2 Infinite Series
When we begin to study a given series
we might not know whether it converges or diverges. In either case, it is convenient to use
sigma notation to write the series as
Notice that
Example
Consider
the partial sums form a sequence whose 𝑛th term is
This sequence of partial sums converges to 2 because
We say
Zeno’s Paradox: Achilles and the Tortoise
Assumption:
•
•
•
The speed of Achilles is 1 m/s.
The speed of tortoise is 0.5 m/s.
The initial distance between
Achilles and tortoise is 1 m.
Geometric Series
Geometric series are series of the form
in which first term 𝑎 and the common ratio 𝑟 are fixed real numbers and 𝑎 ≠ 0.
Proof:
If 𝑟 = 1, then
𝑠𝑛 = 𝑎 + 𝑎 + ⋯ + 𝑎 = 𝑛𝑎 → {
∞ if 𝑎 > 0,
−∞ if 𝑎 < 0.
If 𝑟 = −1, then
𝑎
𝑠𝑛 = 𝑎 − 𝑎 + ⋯ + (−1)𝑛 𝑎 = {
0
In this case, {𝑠𝑛 } is divergent.
if 𝑛 is odd,
if 𝑛 is even.
Suppose 𝑟 ≠ 1.
𝑎
If |𝑟| < 1, then 𝑟 𝑛 → 0, and hence 𝑠𝑛 → 1−𝑟.
If |𝑟| > 1, then {𝑟 𝑛 } is divergent, and hence {𝑠𝑛 } is divergent.
∎
Example
Example
Exercise
Telescoping Series
Telescoping sums are finite sums in which pairs of consecutive terms partly cancel each
other, leaving only parts of the initial and final terms.
Let 𝑎𝑛 be the elements of a sequence of numbers. Then
If 𝑎𝑛 → 𝐿, then the telescoping series gives
If {𝑎𝑛 } diverges, then the series diverges.
Example
Exercise
∞
∑ ln
𝑛=1
∞
∑(
𝑛=1
𝑛
𝑛+1
1
1
− )
𝑛+2 𝑛
The 𝒏th-Term Test for a Divergent Series
Proof:
Let 𝑠𝑛 = 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 .
Since ∑𝑎𝑛 is convergent, the sequence {𝑠𝑛 } = {𝑠1 , 𝑠2 , 𝑠3 , … } converges to a number 𝐿.
It follows that {𝑠𝑛+1 } = {𝑠2 , 𝑠3 , 𝑠4 , … } also converges to 𝐿.
Therefore,
lim 𝑎𝑛 = lim (𝑠𝑛+1 − 𝑠𝑛 ) = lim 𝑠𝑛+1 − lim 𝑠𝑛 = 𝐿 − 𝐿 = 0.
𝑛→∞
𝑛→∞
𝑛→∞
𝑛→∞
∎
Example
If 𝑎𝑛 → 0, then the 𝑛th term test on ∑𝑎𝑛 is inconclusive.
Example
∞
1 𝑛
∑( )
2
𝑛=0
∞
𝑛
∑ ln (
)
𝑛+1
𝑛=1
Example
Combining Series
The three rules for series follow from the analogous rules for sequences
Proof:
Let 𝐴𝑛 = 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 and 𝐵𝑛 = 𝑏1 + 𝑏2 + ⋯ + 𝑏𝑛 . Then 𝐴𝑛 → 𝐴 and 𝐵𝑛 → 𝐵.
(1) The partial sum of ∑(𝑎𝑛 + 𝑏𝑛 ) is
𝑠𝑛 = (𝑎1 + 𝑏1 ) + (𝑎2 + 𝑏2 ) + ⋯ + (𝑎𝑛 + 𝑏𝑛 )
= (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 ) + (𝑏1 + 𝑏2 + ⋯ + 𝑏𝑛 )
= 𝐴𝑛 + 𝐵𝑛
By the Sum Rule for sequences, 𝑠𝑛 → 𝐴 + 𝐵.
(2) The partial sum of ∑(𝑎𝑛 − 𝑏𝑛 ) is
𝑠𝑛 = (𝑎1 − 𝑏1 ) + (𝑎2 − 𝑏2 ) + ⋯ + (𝑎𝑛 − 𝑏𝑛 )
= (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 ) − (𝑏1 + 𝑏2 + ⋯ + 𝑏𝑛 )
= 𝐴𝑛 − 𝐵𝑛
By the Difference Rule for sequences, 𝑠𝑛 → 𝐴 − 𝐵.
(3) The partial sum of ∑𝑘𝑎𝑛 is
𝑠𝑛 = 𝑘𝑎1 + 𝑘𝑎2 + ⋯ + 𝑘𝑎𝑛
= 𝑘(𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 )
= 𝑘𝐴𝑛
By the Constant Multiple Rule for sequences, 𝑠𝑛 → 𝐴 + 𝐵.
Remark:
If ∑𝑎𝑛 and ∑𝑏𝑛 both diverge, then ∑(𝑎𝑛 + 𝑏𝑛 ) may be convergent or divergent.
∎
Example
Exercise
Adding or Deleting Terms
We can add a finite number of terms to a series or delete a finite number of terms without
altering the series’ convergence or divergence, although in the case of convergence, this will
usually change the sum.
∞
If ∑∞
𝑛=1 𝑎𝑛 converges, then ∑𝑛=𝑘 𝑎𝑛 converges for any 𝑘 > 1 and
∞
Conversely, if ∑∞
𝑛=𝑘 𝑎𝑛 converges for any 𝑘 > 1, then ∑𝑛=1 𝑎𝑛 converges.
Reindexing
As long as we preserve the order of its terms, we can reindex any series without altering its
convergence.
To raise the starting value of the index ℎ units, replace the 𝑛 in the formula for 𝑎𝑛 by 𝑛 − ℎ:
To lower the starting value of the index ℎ units, replace the 𝑛 in the formula for 𝑎𝑛 by 𝑛 + ℎ:
Example
3.3 The Integral Test
The most basic question we can ask about a series is whether it converges. In this section, we
begin to study this question, starting with series that have nonnegative terms.
Proof:
Each partial sum is greater than or equal to its predecessor because
𝑠𝑛+1 = 𝑠𝑛 + 𝑎𝑛 ≥ 𝑠𝑛 ,
so
𝑠1 ≤ 𝑠2 ≤ 𝑠3 ≤ ⋯ ≤ 𝑠𝑛 ≤ 𝑠𝑛+1 ≤ ⋯.
Since the partial sums form an increasing sequence.
(⇒) Suppose that ∑𝑎𝑛 converges.
Then {𝑠𝑛 } converges.
Since convergent sequence is bounded, we conclude that {𝑠𝑛 } is bounded.
(⇐) Suppose that {𝑠𝑛 } is bounded from above.
By the Monotone Sequence Theorem, {𝑠𝑛 } converges.
Therefore, ∑𝑎𝑛 converges.
∎
Example
Example
Proof:
Without loss of generality, we assume that 𝑁 = 1.
Let 𝑠𝑛 = 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 . We have
𝑛+1
∫
𝑛
𝑓(𝑥) 𝑑𝑥 ≤ 𝑠𝑛 ≤ 𝑎1 + ∫ 𝑓(𝑥) 𝑑𝑥
1
1
∞
(1) Suppose that ∫1 𝑓(𝑥) 𝑑𝑥 converges.
Then
𝑛
∞
𝑠𝑛 ≤ 𝑎1 + ∫ 𝑓(𝑥) 𝑑𝑥 ≤ 𝑎1 + ∫ 𝑓(𝑥) 𝑑𝑥.
1
1
Therefore, {𝑠𝑛 } is bounded.
We conclude that {𝑠𝑛 } is convergent.
∞
(2) Suppose that ∫1 𝑓(𝑥) 𝑑𝑥 diverges to infinity.
Since
𝑛+1
∫
𝑓(𝑥) 𝑑𝑥 ≤ 𝑠𝑛 ,
1
{𝑠𝑛 } is unbounded, and therefore {𝑠𝑛 } diverges.
∎
𝒑-Series Test
The 𝑝-series
(𝑝 a real constant) converges if 𝑝 > 1 and diverges if 𝑝 ≤ 1.
Exercise
Example
Exercise
Example
Determine the convergence or divergence of the series.
Error Estimation
Suppose that a series ∑𝑎𝑛 with positive terms is shown to be convergent by the Integral Test,
and we want to estimate the size of the remainder 𝑅𝑛 , measuring the difference between the
total sum 𝑆 of the series and its 𝑛th partial sum 𝑠𝑛 . That is, we wish to estimate
Example
Approximate the sum of the series
by using the sum of the first 10 terms.
Exercise
3.4 Comparison Tests
Proof:
Let 𝑠𝑛 = 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 and 𝑡𝑛 = 𝑏1 + 𝑏2 + ⋯ + 𝑏𝑛 .
(1) Suppose that ∑𝑏𝑛 converges to 𝐵. Then 𝑡𝑛 → 𝑡. We get
𝑠𝑛 ≤ 𝑡𝑛 ≤ ∑𝑏𝑛 = 𝐵.
This implies that {𝑠𝑛 } is bounded from above.
Therefore, {𝑠𝑛 } converges.
(2) It follows by contrapositive of (1).
∎
Example
(a)
(b)
(c)
Exercise
Exercise
Exercise
The Limit Comparison Test
Proof:
(1) There exists positive integer 𝑁 such that for 𝑛 > 𝑁,
𝑎𝑛
𝑐
𝑐 𝑎𝑛 3𝑐
| − 𝑐| < ⟹ <
<
𝑏𝑛
2
2 𝑏𝑛
2
𝑐
3𝑐
⟹ 𝑏𝑛 < 𝑎𝑛 < 𝑏𝑛 .
2
2
3𝑐
If ∑𝑏𝑛 converges, then ∑ 2 𝑏𝑛 converges.
By the Direct Comparison Test, ∑𝑎𝑛 converges.
𝑐
If ∑𝑏𝑛 diverges, then ∑ 2 𝑏𝑛 diverges.
By the Direct Comparison Test, ∑𝑎𝑛 diverges.
(2) There exists positive integer 𝑁 such that for 𝑛 > 𝑁,
𝑎𝑛
| | < 1 ⟹ 𝑎𝑛 < 𝑏𝑛 .
𝑏𝑛
If ∑𝑏𝑛 converges, then ∑𝑎𝑛 converges by the Direct Comparison Test.
(3) There exists positive integer 𝑁 such that for 𝑛 > 𝑁,
𝑎𝑛
> 1 ⟹ 𝑎𝑛 > 𝑏𝑛 .
𝑏𝑛
If ∑𝑎𝑛 diverges, then ∑𝑏𝑛 diverges by the Direct Comparison Test.
∎
Example
Exercise
Exercise
Example
3.5 Alternating Series
Definition
A series in which the terms are alternately positive and negative is an alternating series.
The 𝑛th term of an alternating series is of the form
where 𝑢𝑛 = |𝑎𝑛 | is a positive number.
Example
Proof:
Without loss of generality, we may assume 𝑁 = 1.
𝑠2𝑚 = (𝑢1 − 𝑢2 ) + (𝑢3 − 𝑢4 ) + ⋯ + (𝑢2𝑚−1 − 𝑢2𝑚 )
𝑠2𝑚+2 = (𝑢1 − 𝑢2 ) + (𝑢3 − 𝑢4 ) + ⋯ + (𝑢2𝑚−1 − 𝑢2𝑚 ) + (𝑢2𝑚+1 − 𝑢2𝑚−2 )
= 𝑠2𝑚 + (𝑢2𝑚+1 − 𝑢2𝑚−2 ) ≥ 𝑠2𝑚
Therefore, the sequence {𝑠2𝑚 } is decreasing.
𝑠2𝑚 = 𝑢1 − (𝑢2 − 𝑢3 ) − ⋯ − (𝑢2𝑚−2 − 𝑢2𝑚−1 ) − 𝑢2𝑚 ≤ 𝑢1
The sequence {𝑠2𝑚 } is bounded above.
By the Monotone Sequence Theorem, {𝑠2𝑚 } converges to a number 𝐿.
Note that
lim 𝑠2𝑚+1 = lim (𝑠2𝑚 + 𝑢2𝑚+1 ) = lim 𝑠2𝑚 + lim 𝑢2𝑚+1 = 𝐿 + 0 = 𝐿.
𝑛→∞
𝑛→∞
𝑛→∞
𝑛→∞
Since both the even and odd partial sums converge to 𝐿, {𝑠𝑛 } also converges to 𝐿.
∎
Example
Exercise
Exercise
Example
Exercise
3.6 Absolute and Conditional Convergence
When some of the terms of a series are positive and others are negative, the series may or
may not converge.
In this series, there is some cancelation in the partial sums, which may be assisting the
convergence property of the series. However, if we make all of the terms positive in the series
to form the new series
we see that it still converges.
For a general series with both positive and negative terms, we can apply the tests for
convergence that we studied before to the series of absolute values of its terms.
Absolute Convergence
Exercise
Proof:
From
−|𝑎𝑛 | ≤ 𝑎𝑛 ≤ |𝑎𝑛 |,
we get
0 ≤ 𝑎𝑛 + |𝑎𝑛 | ≤ 2|𝑎𝑛 |.
Since ∑|𝑎𝑛 | converges, ∑2|𝑎𝑛 | also converges.
By the Direct Comparison Test, ∑(𝑎𝑛 + |𝑎𝑛 |) converges.
Therefore,
∑ 𝑎𝑛 = ∑(𝑎𝑛 + |𝑎𝑛 | − |𝑎𝑛 |) = ∑(𝑎𝑛 + |𝑎𝑛 |) − ∑|𝑎𝑛 |
converges.
Remark:
If ∑|𝑎𝑛 | diverges, then the test is inconclusive.
∎
Example
(a)
(b)
Exercise
Conditional Convergence
For an absolutely convergent series, changing infinitely many of the negative terms in the
series to positive values does not change its property of still being a convergent series.
Other convergent series may behave differently. The convergent alternating harmonic series
has infinitely many negative terms, but if we change its negative terms to positive values, the
resulting series is the divergent harmonic series. So the presence of infinitely many negative
terms is essential to the convergence of this alternating harmonic series.
Example
Exercise
Determine whether the series is absolutely convergent, conditionally convergent, or
divergent.
Rearranging Series
The question of whether a given convergent series is absolutely convergent or conditionally
convergent has a bearing on the question of whether infinite sums behave like
finite sums.
If we rearrange the order of the terms in a finite sum, then of course the value of the
sum remains unchanged. But this is not always the case for an infinite series.
For any real number 𝑟, a given conditionally convergent series can be rearranged so that its
sum is equal to 𝑟.
Example
𝑛+1
The alternating harmonic series ∑∞
/𝑛 converges to some number 𝐿.
𝑛=1(−1)
By the Alternating Series Estimation Theorem,
|𝐿 − 𝑠1 | < 𝑢2
⇒ −1/2 < 𝐿 − 1 < 1/2
⇒ 1/2 < 𝐿 < 3/2
Therefore, 𝐿 ≠ 0.
Now,
1 1 1 1 1 1 1 1 1
𝐿 = 1− + − + − + − + −
+⋯
2 3 4 5 6 7 8 9 10
2 1 2 1 2 1 2 1
2𝐿 = 2 − 1 + − + − + − + − + ⋯
3 2 5 3 7 4 9 5
1
2 1
1
2 1
= (2 − 1) − + ( − ) − + ( − ) + ⋯
2
3 3
4
5 5
1 1 1 1
= 1− + − + −⋯
2 3 4 5
=𝐿
We get 𝐿 = 0, which is not possible.
3.7 The Ratio and Root Test
The Ratio Test
Proof:
(a) Let 𝑟 ∈ ℝ such that 𝜌 < 𝑟 < 1 and let 𝜖 = 𝑟 − 𝜌.
Then there exists a positive integer 𝑁 such that for 𝑛 ≥ 𝑁,
𝑎𝑛+1
𝑎𝑛+1
||
| − 𝜌| < 𝜖 ⟹ |
|<𝜌+𝜖 =𝑟
𝑎𝑛
𝑎𝑛
⟹ |𝑎𝑛+1 | < 𝑟|𝑎𝑛 |.
Hence, we get
|𝑎𝑁+1 | < 𝑟|𝑎𝑁 |
|𝑎𝑁+2 | < 𝑟|𝑎𝑁+1 | < 𝑟 2 |𝑎𝑁 |
|𝑎𝑁+3 | < 𝑟|𝑎𝑁+2 | < 𝑟 3 |𝑎𝑁 |
…
For 𝑛 = 1,2, …, we have
|𝑎𝑁+𝑛 | < 𝑟 𝑛 |𝑎𝑁 |.
Note that
∞
∞
𝑛
∑ 𝑟 |𝑎𝑁 | = |𝑎𝑁 | ∑ 𝑟 𝑛
𝑛=1
𝑛=1
is a convergent geometric series.
By the Direct Comparison Test,
∞
∞
∑ |𝑎𝑛 | = ∑|𝑎𝑁+𝑛 |
𝑛=𝑁+1
𝑛=1
converges.
Therefore, ∑∞
𝑛=1|𝑎𝑛 | converges. We conclude that the series converges absolutely.
(b) If 𝜌 > 1, there exists a positive integer 𝑁 such that for 𝑛 ≥ 𝑁,
𝑎𝑛+1
𝑎𝑛+1
||
| − 𝜌| < 1 − 𝜌 ⟹ |
| > 1.
𝑎𝑛
𝑎𝑛
If 𝜌 = ∞, there exists a positive integer 𝑁 such that for 𝑛 ≥ 𝑁,
𝑎𝑛+1
|
| > 1.
𝑎𝑛
In both cases, |𝑎𝑛+1 | > |𝑎𝑛 | for 𝑛 ≥ 𝑁, that is,
|𝑎𝑁 | < |𝑎𝑁+1 | < |𝑎𝑁+2 | < ⋯
This implies that
lim 𝑎𝑛 ≠ 0.
𝑛→∞
We conclude that the series diverges by 𝑛th-Term Test.
(c) Example:
∞
∑
𝑛=1
∞
∑
𝑛=1
1
𝑛
1
𝑛2
Example
Investigate the convergence of the following series.
Exercise
Exercise
Exercise
Exercise
The Root Test
Proof:
(a) Let 𝑟 ∈ ℝ such that 𝜌 < 𝑟 < 1 and let 𝜖 = 𝑟 − 𝜌.
Then there exists a positive integer 𝑁 such that for 𝑛 ≥ 𝑁,
𝑛
| √|𝑎𝑛 | − 𝜌| < 𝑟 − 𝜌
𝑛
√|𝑎𝑛 | < 𝑟
|𝑎𝑛 | < 𝑟 𝑛 .
𝑛
Note that ∑∞
𝑛=𝑁 𝑟 is a convergent geometric series.
By the Direct Comparison Test, ∑∞
𝑛=𝑁 |𝑎𝑛 | converges.
∞
Therefore, ∑𝑛=1|𝑎𝑛 | converges. We conclude that the series converges absolutely.
(b) If 𝜌 > 1, there exists a positive integer 𝑁 such that for 𝑛 ≥ 𝑁,
𝑛
| √|𝑎𝑛 | − 𝜌| < 1 − 𝜌 ⟹ |𝑎𝑛 | > 1.
If 𝜌 = ∞, there exists a positive integer 𝑁 such that for 𝑛 ≥ 𝑁,
𝑛
| √|𝑎𝑛 || > 1 ⟹ |𝑎𝑛 | > 1.
In both cases,
lim 𝑎𝑛 ≠ 0.
𝑛→∞
We conclude that the series diverges by 𝑛th-Term Test.
(c) Example:
∞
∑
𝑛=1
∞
∑
𝑛=1
1
𝑛
1
𝑛2
Example
Consider the series with terms
Does ∑𝑎𝑛 converge?
Example
Which of the following series converge, and which diverge?
Exercise
Exercise
3.8 Power Series
Now, we study sums that look like “infinite polynomials”. We call these sums power series
because they are defined as infinite series of powers of some variable, in our case 𝑥.
Power Series and Convergence
Example
We think of the partial sum of the series on the right as polynomials 𝑃𝑛 (𝑥) that approximate
the function on the left.
Example
Example
The Radius of Convergence of a Power Series
𝑅 is called the radius of convergence of the power series, and the interval of radius 𝑅
centered at 𝑥 = 𝑎 is called the interval of convergence. The interval of convergence may be
open, closed, or half-open, depending on the particular series.
Exercise
Find the radius of convergence and interval of convergence of the series
Exercise
Find the radius of convergence and interval of convergence of the series
Operations on Power Series
On the intersection of their intervals of convergence, two power series can be added and
subtracted term by term just like series of constants. They can be multiplied just as we
multiply polynomials, but we often limit the computation of the product to the first few
terms, which are the most important.
Example
We can also substitute a function 𝑓(𝑥) for 𝑥 in a convergent power series.
Example
Express 1/(1 − 4𝑥 2 ) as the sum of a power series and find the interval of convergence.
Exercise
Express 1/(1 + 𝑥 2 ) as the sum of a power series and find the interval of convergence.
Exercise
Find a power series representation for 1/(𝑥 + 2).
Exercise
Find a power series representation for 𝑥 3 /(𝑥 + 2).
A power series can be differentiated term by term at each interior point of its interval of
convergence.
Example
Example
Identify the function
A power series can be integrated term by term throughout its interval of convergence.
Example
3.9 Taylor and Maclaurin Series
Series Representations
Within its interval of convergence 𝐼, the sum of a power series is a continuous function with
derivatives of all orders. But what about the other way around? If a function 𝑓(𝑥) has
derivatives of all orders on an interval, can it be expressed as a power series on at least part of
that interval? And if it can, what are its coefficients?
Assume that
∞
𝑓(𝑥) = ∑ 𝑎𝑛 (𝑥 − 𝑎)𝑛
𝑛=0
= 𝑎0 + 𝑎1 (𝑥 − 𝑎) + 𝑎2 (𝑥 − 𝑎)2 + ⋯
By repeated term-by-term differentiation within the interval of convergence 𝐼, we get
𝑓 ′ (𝑥) = 𝑎1 + 2𝑎2 (𝑥 − 𝑎) + 3𝑎3 (𝑥 − 𝑎)2 + ⋯
𝑓 ′′ (𝑥) = 1 ⋅ 2𝑎2 + 2 ⋅ 3(𝑥 − 𝑎) + 3 ⋅ 4(𝑥 − 𝑎)2 + ⋯
𝑓 ′′′ (𝑥) = 1 ⋅ 2 ⋅ 3𝑎3 + 2 ⋅ 3 ⋅ 4(𝑥 − 𝑎) + 3 ⋅ 4 ⋅ 5(𝑥 − 𝑎)2 + ⋯,
with 𝑛th derivative being
𝑓 (𝑛) (𝑥) = 𝑛! 𝑎𝑛 + a sum of terms with (𝑥 − 𝑎) as a factor
𝑓 (𝑛) (𝑎) = 𝑛! 𝑎𝑛
𝑓 (𝑛) (𝑎)
𝑎𝑛 =
𝑛!
If an infinitely differentiable function 𝑓 has a series representation with the center at 𝑎,
then the series must be
∞
𝑓(𝑥) = ∑
𝑛=0
𝑓 (𝑛) (𝑎)
(𝑥 − 𝑎)𝑛
𝑛!
= 𝑓(𝑎) + 𝑓
′ (𝑎)(𝑥
𝑓 ′′ (𝑎)
𝑓 (𝑛) (𝑎)
2
(𝑥 − 𝑎) + ⋯ +
(𝑥 − 𝑎)𝑛 + ⋯
− 𝑎) +
2!
𝑛!
(⋆)
If we start with an arbitrary function 𝑓 that is infinitely differentiable on an interval
containing 𝑥 = 𝑎 and use it to generate the series in (⋆), does the series converge to 𝑓(𝑥) at
each 𝑥 in the interval of convergence?
The answer is maybe—for some functions it will, but for other functions it will not.
Taylor and Maclaurin Series
Example
Find the Taylor series generated by 𝑓(𝑥) = 1/𝑥 at 𝑎 = 2.
Where, if anywhere, does the series converge to 1/𝑥?
Taylor Polynomials
The linearization of a differentiable function 𝑓 at a point 𝑎 is the polynomial of degree at
most 1 given by
We used this linearization to approximate 𝑓(𝑥) at values of 𝑥 near 𝑎.
If 𝑓 has derivatives of higher order at 𝑎, then it has higher-order polynomial approximations
as well, one for each available derivative. These polynomials are called the Taylor
polynomials of 𝑓.
We speak of a Taylor polynomial of order 𝑛 rather than degree 𝑛 because 𝑓 (𝑛) (𝑎) may be
zero.
Example
Find the Taylor series and the Taylor polynomials generated by 𝑓(𝑥) = 𝑒 𝑥 at 𝑥 = 0.
Exercise
Find the Taylor series for 𝑓(𝑥) = 𝑒 𝑥 at 𝑎 = 2.
Example
Find the Taylor series and Taylor polynomials generated by 𝑓(𝑥) = cos 𝑥 at 𝑥 = 0.
Exercise
Find the Taylor series generated by 𝑓(𝑥) = sin 𝑥 at 0.
Exercise
Find the Taylor series generated by 𝑓(𝑥) = sin 𝑥 at 𝜋/3.
Example
By induction, it can be shown that for all 𝑛,
𝑓 (𝑛) (0) = 0.
Therefore, the Taylor series generated by 𝑓 at 𝑥 = 0 is
𝑓 ′′ (0) 2
𝑓(0) + 𝑓 ′ (0)𝑥 +
𝑥 + ⋯ = 0 + 0 + 0 + ⋯.
2!
The series converges for every 𝑥 but converges to 𝑓(𝑥) only at 𝑥 = 0.
That is, the Taylor series generated by 𝑓 at 𝑥 = 0 is not equal to 𝑓(𝑥) over the entire interval
of convergence.
Two questions still remain.
1. For what values of 𝑥 can we normally expect a Taylor series to converge to its
generating function?
2. How accurately do a function’s Taylor polynomials approximate the function on a
given interval?
3.10 Convergence of Taylor Series
When we state Taylor’s theorem in this way, it says that for each 𝑥 ∈ 𝐼,
Equation (1) is called the Taylor’s formula.
The function 𝑅𝑛 (𝑥) is called the remainder of order 𝒏 or the error term for the
approximation of 𝑓 by 𝑃𝑛 (𝑥) over 𝐼.
Example
For what values of 𝑥 can we replace sin 𝑥 by 𝑥 − (𝑥 3 /3!) and obtain an error whose
magnitude is no greater than 3 × 10−4 ?
Exercise
Exercise
Example
Show that the Taylor series generated by 𝑓(𝑥) = 𝑒 𝑥 at 𝑥 = 0 converges to 𝑓(𝑥) for every
real value of 𝑥.
Example
Show that the Taylor series for sin 𝑥 at 𝑥 = 0 converges for all 𝑥.
Example
Show that the Taylor series for cos 𝑥 at 𝑥 = 0 converges for all 𝑥.
Using Taylor Series
Since every Taylor series is a power series, the operations of adding, subtracting, and
multiplying Taylor series are all valid on the intersection of their intervals of convergence.
If the Taylor series generated by 𝑓 at 𝑥 = 0,
converges absolutely for |𝑥| < 𝑅, and if 𝑢 is a continuous function, then the series
converges absolutely on the set of points 𝑥 where |𝑢(𝑥)| < 𝑅.
Example
Using known series, find the first few terms of the Maclaurin series for the given function by
using power series operations.
Example
Find the Maclaurin series for cos 2𝑥.
Exercise
Find the Maclaurin series for 𝑓(𝑥) = ln(1 + 3𝑥 2 )