Section 11.8 Lecture Slides

1. Chapter 11: Series, Sequences, and Power Series
Section 11.8: Power Series
Alea Wittig
SUNY Albany

2. Outline
Power Series
Radius and Interval of Convergence

3. Power Series

4. Power Series
A power series centered at a is a series of the form
∞
X
n=0
cn(x − a)n
= c0 + c1(x − a) + c2(x − a)2
+ c3(x − a)3
+ . . .
where x is a variable and the cn’s are constants called the
coefficients of the series.
eg
∞
X
n=0
5n
(x − 2)n

5. ▶ A power series may converge for some values of x and diverge
for other values of x.
▶ We will call the set of values for which a power series
converges its interval of convergence.
▶ The sum of the series on its interval of convergence is a
function
f (x) = c0 + c1x + c2x2
+ . . .
kind of like an ’infinite polynomial’.

6. eg The power series centered at a = 0 with coefficients cn = 1 for
all n:
f (x) =
∞
X
n=0
cn(x − a)n
=
∞
X
n=0
xn
is the geometric series
∞
X
n=0
xn
=
∞
X
n=1
xn−1
=
∞
X
n=1
arn−1
where a = 1, r = x
which converges to 1
1−x for |x| < 1 and diverges for all other
values of x =⇒ its interval of convergence is (−1, 1).
▶ At x = 1
2 : f
1
2
=
∞
X
n=0
1
2n
= 1
▶ f is not defined at x = 2 where
P∞
n=0 2n
diverges

7. ▶ To determine the values x for which the power series
∞
X
n=0
cn(x − a)n
= c0 + c1(x − a) + c2(x − a)2
+ . . .
converges we usually use the ratio (or root) test.
▶ The ratio/root test is inconclusive when lim
n→∞

10. an+1
an

13. =1 so we
will have to use a different method in those cases.

14. The Ratio Test
(i) If lim
n→∞

17. an+1
an

20. = L 1 then the series
P
an is absolutely
convergent (and therefore convergent.)
(ii) If lim
n→∞

23. an+1
an

26. = L 1 then the series
P
an is divergent.
(iii) If lim
n→∞

29. an+1
an

32. = 1 then the ratio text is inconclusive.

33. The Root Test
(i) If lim
n→∞
n
p
|an| = L 1, then the series
P
an is absolutely
convergent. (and therefore convergent.)
(ii) If lim
n→∞
n
p
|an| = L 1
or lim
n→∞
n
p
|an| = ∞, then the series
P
an is divergent.
(iii) If lim
n→∞
n
p
|an| = 1, the Root Test is inconclusive.

34. Example 1
1/3
For what values of x does the following power series converge?
∞
X
n=1
(x − 3)n
n

35. Example 1
2/3
▶ We proceed by using the ratio test:
lim
n→∞

38. an+1
an

41. = lim
n→∞

46. (x − 3)n+1
n + 1
·
n
(x − 3)n

51. = lim
n→∞

56. (x − 3)
n+1
(x − 3)n

61. n
n + 1
= |x − 3| lim
n→∞
n
n + 1
= |x − 3|
▶ So ratio test tells us:
▶ the power series converges if |x − 3| 1
▶ the power series diverges if |x − 3| 1
▶ nothing when |x − 3| = 1 (test is inconclusive when L = 1).

62. Example 1
3/3
|x − 3| 1 ⇐⇒ −1 x − 3 1 ⇐⇒ 2 x 4
▶ So ratio test tells us the series is convergent on (2, 4) and
divergent on (4, ∞) and (−∞, 2).
▶ We need to check the endpoints x = 2 and x = 4 using a
different test/method:
▶ x = 2 :
X (x − 3)n
n
=
X (2 − 3)n
n
=
X (−1)n
n
this is the alternating harmonic series which we already know
converges by AST.
▶ x = 4 :
X (x − 3)n
n
=
X (4 − 3)n
n
X 1
n
this is the harmonic series which diverges by the p-series test.
▶ Therefore the interval of convergence is [2, 4)

63. Example 2
1/2
For what values of x does the following series converge?
∞
X
n=0
n!xn

64. Example 2
2/2 ▶ We start by using the ratio test:
lim
n→∞

67. an+1
an

70. = lim
n→∞

73. (n + 1)!xn+1
n!xn

76. = lim
n→∞
(n + 1)!
n!

79. x
n+1
xn

82. = |x| lim
n→∞
(n + 1)!
n!
= |x| lim
n→∞
(n + 1)n!
n!
= |x| lim
n→∞
(n + 1)
=
(
∞ if x ̸= 0
0 if x = 0
▶ Now since the series diverges everywhere but x = 0, its
interval of convergence is just the single value {0}.

83. Example 3
1/2
For what values of x does the following series converge?
∞
X
n=0
xn
(2n)!

84. Example 3
2/2
▶ As usual we proceed by applying the ratio test:
lim
n→∞

87. an+1
an

90. = lim
n→∞

93. xn+1
(2(n + 1))!
(2n)!
xn

96. = lim
n→∞

99. x
n+1
xn

102. (2n)!
(2n + 2)!
= |x| lim
n→∞
(2n)!
(2n + 2)(2n + 1)
(2n)!
= |x| lim
n→∞
1
(2n + 2)(2n + 1)
= 0 for all values of x
▶ Since the limit is 0 for all values of x, the series converges for
all values of x by the ratio test.
▶ The interval of convergence is (−∞, ∞)

103. Radius and Interval of Convergence

104. Convergence of a Power Series
For a power series
∞
X
n=0
cn(x − a)n
there are only three possibilities:
(i) The series converges only when x = a.
(ii) The series converges for all x.
(iii) There is a positive number R such that the series converges if
|x − a| R and diverges if |x − a| R

105. Radius and Interval of Convergence
▶ R is called the radius of convergence.
▶ The interval of convergence, I, is the interval that consists
of all values of x for which the series converges.
(i) The series converges only
when x = a.
(ii) The series converges for all
x.
(iii) There is a positive number
R such that the series
converges if |x − a| R
and diverges if |x − a| R
(i) R = 0
I = {a}
(ii) R = ∞
I = (−∞, ∞)
(iii) R = R
▶ I = (a − R, a + R) or
▶ I = (a − R, a + R] or
▶ I = [a − R, a + R) or
▶ I = [a − R, a + R]

106. Example 4
1/3
Find the radius of convergence and interval of convergence of the
series
∞
X
n=0
n(x + 2)n
3n+1

107. Example 4
2/3 ▶ We proceed by applying ratio test
lim
n→∞

110. an+1
an

113. = lim
n→∞

116. (n + 1)(x + 2)n+1
3n+2
3n+1
n(x + 2)n

119. = lim
n→∞

124. (x + 2)
n+1
(x + 2)n

129. n + 1
n
!
3n+1
3n+2
!
= |x + 2| lim
n→∞
n + 1
n
!
3n+1
3 ·
3n+1
!
=
|x + 2|
3
lim
n→∞
n + 1
n
=
|x + 2|
3
▶ Ratio test tells us that the series converges if |x+2|
3 1, ie, if
|x + 2| 3.
▶ Therefore the radius of convergence is R = 3 .

130. Example 4
3/3
▶ Now we need to find the interval of convergence.
|x + 2| 3 ⇐⇒ −3 x + 2 3 ⇐⇒ −5 x 1
▶ We need to check the endpoints x = −5 and x = 1 for
convergence using a different method than ratio/root test
because it is at the points that ratio/root are inconclusive.
▶ x = −5 :
∞
X
n=0
n3n
3n+1
=
∞
X
n=0
n(−3)n
3n+1
=
1
3
∞
X
n=0
(−1)n
n =⇒ diverges by test
for divergence
▶ x = 1 :
∞
X
n=0
n3n
3n+1
=
∞
X
n=0
n
3
=⇒ diverges by test for divergence
▶ So the interval of convergence is I = (−5, 1)