1. Copyright © Cengage Learning. All rights reserved.9
Infinite Series
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2. Copyright © Cengage Learning. All rights reserved.
9.8
Power Series
Copyright © Cengage Learning. All rights reserved.

3. Objectives Understand the definition of a power series.
Find the radius and interval of convergence of a power series.
Determine the endpoint convergence of a power series.
Differentiate and integrate a power series.

4. We’ll strip you of any fears,
plug you in,
drill you on the essentials,
make everything plane,
and hopefully electrify you!

5. Power Series

6. We’ve seen power series before…
𝑎 𝑛 𝑥 𝑛
We’ve seen power series before…
Specifically when 𝑎 𝑛 = 𝑓 𝑛 (0) 𝑛!
Does the name “Maclaurin” ring a bell?

7. Power Series
An important function f(x) = ex can be represented exactly by an infinite series called a power series. For example, the Maclaurin series we found is a power series representation for ex
For each real number x, it can be shown that the infinite series converges to the number ex.

8. Power Series

9. Example 1 – Power Series The following power series is centered at 0.
c. The following power series is centered at 1.

10. Power Series A power series can be viewed as a function of x.
The domain of a power series is the set of all x for which
the power series converges. Every power series converges
at its center because
So c always lies in the domain of f.

11. Radius and Interval of Convergence

12. Radius and Interval of Convergence

13. Power Series The domain of a power series can be a single point,
an interval centered at c, or the entire real line.

14. Convergence of Power Series Theorem
The series converges absolutely for all 𝑥
The series converges only for 𝑥=0
The series converges on some symmetric interval,
such as −𝑏, 𝑏 , −𝑏, 𝑏 , −𝑏, 𝑏 or −𝑏,𝑏
(𝑏 is called the 𝐫𝐚𝐝𝐢𝐮𝐬 of convergence.)
Given a Power Series 𝑎 𝑛 𝑥 𝑛 , then either
The only tools we have to find said interval of convergence are the Ratio and Root Tests.
Recall that both tests show convergence when lim 𝑛→∞ |𝑎 𝑛 𝑥 𝑛 |<1

15. Example 2 – Finding the Radius of Convergence
Find the radius of convergence of
Solution: (use the Ratio Test)
Therefore, by the Ratio Test, the series diverges for
|x| > 0 and converges only at its center, 0.
So, the radius of convergence is R = 0.

16. Why is the interval of convergence symmetric?
Well…if a series is convergent then the ratio test yields a value 𝑎 𝑏 <1.
Hence for the power function 𝑎 𝑛 𝑥 𝑛 we would get:
lim 𝑛→∞ 𝑎 𝑛+1 𝑎 𝑛 ∙ |𝑥| 𝑛+1 |𝑥| 𝑛 = 𝑎 𝑏 𝑥 <1
→ 𝑥 < 𝑏 𝑎
→− 𝑏 𝑎 <𝑥< 𝑏 𝑎

17. For example, in this Power Series: a n 𝑥 𝑛
if 𝑎 𝑛 =1, we get
This is a geometric series which converges for 𝑥 𝜖 −1, 1 .
For example, in this Power Series: a n 𝑥 𝑛
𝑥 𝑛 𝑛!
if 𝑎 𝑛 = 1 𝑛! , we get
which converges to 𝑓 𝑥 = 𝑒 𝑥 for all 𝑥.
We have already studied a subset
of all Power Series.

18. Endpoint Convergence

19. Endpoint Convergence
A power series whose radius of convergence is a finite number R, says nothing about the convergence at the endpoints of the interval of convergence. Each endpoint must be tested separately for convergence or divergence.
As a result, the interval of convergence of a power series can take any one of the six forms shown in Figure 9.18.
Figure 9.18

20. Example 3 – Finding the Interval of Convergence
Find the interval of convergence of
Solution (using the Ratio Test):

21. Example 3 – Solution
cont'd
So, by the Ratio Test, the radius of convergence is R = 1.
Moreover, because the series is centered at 0, it converges in the interval (–1, 1).
This interval, however, is not necessarily the interval of convergence.
To determine this, you must test for convergence at each endpoint.
When x = 1, you obtain the divergent harmonic series

22. Example 3 – Solution
cont'd
When x = –1, you obtain the convergent alternating harmonic series
So, the interval of convergence for the series is [–1, 1):

23. The interval of convergence for this series is [−1,1)
𝑛=1 ∞ 𝑥 𝑛 𝑛
Ex: #4 Find the interval of convergence for the series.
Since we are using a test that is only valid for positive series, we must put absolute value around any quantity that might be negative.
Convergence by the Root Test for 𝑥𝜖(−1, 1)
lim 𝑛→∞ 𝑛 |𝑥| 𝑛 𝑛
= lim 𝑛→∞ |𝑥| 𝑛 𝑛
= lim 𝑛→∞ |𝑥|
=|𝑥|<1
Since the Root Test fails if the limit equals 1, we test the endpoints separately.
lim 𝑛→∞ 𝑛 𝑛 =1
If 𝑥=1; 𝑛=1 ∞ 1 𝑛 𝑛 ;divergent p−series
The interval of convergence for this series is [−1,1)
If 𝑥=−1; 𝑛=1 ∞ −1 𝑛 𝑛 ;convergent by the A.S.T.

24. This series is convergent on (−∞,∞) by the Root Test.
𝑛=1 ∞ (−1) 𝑛 𝑥 𝑛 𝑛 𝑛
Ex: #5
lim 𝑛→∞ 𝑛 |−1| 𝑛 |𝑥| 𝑛 𝑛 𝑛
= lim 𝑛→∞ |𝑥| 𝑛
=0<1 for all 𝑥
This series is convergent on (−∞,∞) by the Root Test.
If you get tired, it’s no wonder.
We just did infinitely many problems
twice!

25. Convergence for 𝑥 <10 𝑜𝑟 −10<𝑥<10
𝑛=1 ∞ 𝑛 10 𝑥 𝑛 1 0 𝑛
Ex: #6 1
lim 𝑛→∞ 𝑛 𝑛 10 |𝑥| 𝑛 1 0 𝑛
= lim 𝑛→∞ 𝑛 𝑛 10 |𝑥| 10
= lim 𝑛→∞ |𝑥| 10
= |𝑥| 10 <1
Convergence for 𝑥 <10 𝑜𝑟 −10<𝑥<10
Now check the 2 endpoints.
This series converges for 𝑥 𝜖 (−10, 10)
If 𝑥=10; 𝑛=1 ∞ 𝑛 𝑛 1 0 𝑛
; very divergent!
If 𝑥=−10; 𝑛=1 ∞ 𝑛 10 −10 𝑛 1 0 𝑛 = 𝑛=1 ∞ 𝑛 10 −1 𝑛
; also divergent by nth Term Test

26. 𝑛=1 ∞ (2𝑥−1) 𝑛 𝑛 4 +16
In this case 𝑥 has been replaced with 2𝑥−1, so the center of the symmetric interval would be at 𝑥=1/2.
Ex: #7
lim 𝑛→∞ 𝑛 |2𝑥−1| 𝑛 𝑛 4 +16
= lim 𝑛→∞ 𝑛 |2𝑥−1| 𝑛 𝑛 4
= lim 𝑛→∞ |2𝑥−1| 𝑛 𝑛 4
= 2𝑥−1 <1
−1<2𝑥−1<1
Check the endpoints.
0<2𝑥<2
If 𝑥=0; 𝑛=1 ∞ (−1) 𝑛 𝑛 ;converges by A.S.T.
0<𝑥<1
If 𝑥=1; 𝑛=1 ∞ (1) 𝑛 𝑛 ;compares to convergent p−series
The interval of convergence for this series converges is 0, 1 .
The radius of convergence is 1/2.

27. The interval of convergence is −2, 2 .
𝑛=0 ∞ 𝑥 𝑛
Ex: #28
lim 𝑛→∞ 𝑛 𝑥 𝑛
= lim 𝑛→∞ 𝑥
= 𝑥
<1
−1< 𝑥 <1
−5<𝑥 2 +1<5
Check the endpoints.
−6<𝑥 2 <4
→ 0<𝑥 2 <4
→−2<𝑥<2
𝑥=±2; 𝑛=0 ∞ 𝑛 = 𝑛=0 ∞ 1 𝑛 ;diverges by nth term test.
The interval of convergence is −2, 2 .

28. Ex: Determine the radius and interval of convergence:
𝑛=1 ∞ (−1) 𝑛 𝑛 4 𝑛 𝑥+3 𝑛
Series converges
Series diverges

29. Ex: Determine the radius and interval of convergence:
𝑛=1 ∞ (−1) 𝑛 𝑛 4 𝑛 𝑥+3 𝑛
Radius of Convergence = 4
Test fails (where L=1) Need to test for convergence at endpoints of interval.
Power series diverges at each endpoint. Interval of convergence is

30. Ex: Determine the radius and interval of convergence:
𝑛=1 ∞ 2 𝑛 𝑛 4𝑥−8 𝑛

31. Ex: Determine the radius and interval of convergence:
𝑛=1 ∞ (𝑥−6) 𝑛 𝑛 𝑛

32. Ex: Determine the radius and interval of convergence:
𝑛=1 ∞ 2𝑛 (−3) 𝑛 2𝑛

33. Ex: Determine the radius and interval of convergence:

34. Just 1 more example!

35. Rats! We have to use the Ratio Test when there is a factorial!
𝑛=1 ∞ 1∙3∙5∙⋯∙(2𝑛+1) 𝑛!
Ex: Find the interval
of convergence
Rats! We have to use the Ratio Test when there is a factorial!

36. The interval of convergence is − 1 2 , 1 2
𝑛=1 ∞ 1∙3∙5∙⋯∙(2𝑛+1) 𝑛! 𝑥 𝑛
Ex: #20 (read note in book)
lim 𝑛→∞ 1∙3∙5∙⋯∙(2𝑛+1)(2 𝑛+1 +1) |𝑥| 𝑛+1 (𝑛+1)! 1∙3∙5∙⋯∙(2𝑛+1)|𝑥 | 𝑛 𝑛!
= lim 𝑛→∞ 2𝑛+3 𝑛!|𝑥| 𝑛+1 𝑛!
→ 𝑥 < 1 2
=2|𝑥|<1
Evidently, checking the endpoints is beyond the scope of this course as the text said: “don’t bother checking…the series diverges at the endpoints.”
The interval of convergence is − 1 2 , 1 2

37. Homework
Day 1: Page 666: odd
Day 2: MMM

38. Differentiation and Integration of Power Series

39. Differentiation and Integration of Power Series

40. Example 8 – Intervals of Convergence for f(x), f'(x), and ∫f(x)dx
Consider the function given by
Find the interval of convergence for each of the following.
∫f(x)dx
f(x)
f'(x)

41. Example 8 – Solution By Theorem 9.21, you have and
cont'd
By Theorem 9.21, you have
and
By the Ratio Test, you can show that each series has a
radius of convergence of R = 1.
Considering the interval (–1, 1) you have the following.

42. Example 8(a) – Solution For ∫f(x)dx, the series
cont'd
For ∫f(x)dx, the series
converges for x = ±1, and its interval of convergence is
[–1, 1 ]. See Figure 9.21(a).
Figure 9.21(a)

43. Example 8(b) – Solution For f(x), the series
cont'd
For f(x), the series
converges for x = –1, and diverges for x = 1.
So, its interval of convergence is [–1, 1).
See Figure 9.21(b).
Figure 9.21(b)

44. Example 8(c) – Solution For f'(x), the series
cont'd
For f'(x), the series
diverges for x = ±1, and its interval of convergence is (–1, 1). See Figure 9.21(c).
Figure 9.21(c)

45. EXAMPLE 15.2. Determine the radius and interval of convergence for the power series