Analysis with an introduction to proof 5th by steven lay ch0

Chapter 8
Infinite Series

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 8.3, Slide 1


Section 8.3
Power Series

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 8.3, Slide 2


Up to this point we have dealt with infinite series whose terms were fixed numbers.
We broaden our perspective now to consider series whose terms are variables.
The simplest kind is known as a power series, and the main question will be determining
the set of values of the variable for which the series is convergent.

Definition 8.3.1
Let

of real numbers. The series
( an be) n∞a=sequence
0

∞

∑ an x n

n=0
is called a power series.

= a0 + a1x + a2 x 2 + a3 x 3 + L
The number an is called the nth coefficient of the series.

Example 8.3.2
n

Consider the power series whose coefficients are all equal to 1: Σ x .
 

This is the geometric series, and it converges iff | x | < 1.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 8.3, Slide 3


Theorem 8.3.3
1/n
n
Let Σ a  x be a power series and let α = lim sup | a  | . Define R by
n
n

 1
 α ,
R =  + ∞,


 0,

if 0 < α < + ∞,
if α = 0,
if α = + ∞.

Then the series converges absolutely whenever | x | < R and diverges whenever | x | > R.
(When R = + ∞, we take this to mean that the series converges absolutely for all real x.
When R = 0, then the series converges only at x = 0.)

Proof:
n
Let b = a  x and apply the root test (Theorem 8.2.8).
n
n

β = lim sup | b  |

n

1/n

n 1/n

= lim sup | a  x  |

n

If lim sup | a  |

n

= | x | lim sup | a  |

n

1/n

1/n

=α∈

, we have

= |  x   |α .

Thus if α = 0, then β = 0 < 1, and the series converges (absolutely) for all real x.
If 0 < α < + ∞, then the series converges when | x | α < 1 and diverges when | x | α > 1.
That is, Σ a  x converges when | x | < 1/α = R and diverges when | x | > 1/α = R.
n n
n
If α = + ∞, then for x ≠ 0 we have β = + ∞, so Σ a  x diverges when | x | > 0 = R.
n
Certainly, the series will converge when x = 0, for then all the terms except the first are zero. ♦

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 8.3, Slide 4

From Theorem 3 we see that the set of values C for which a power series converges
will either be {0},

, or a bounded interval centered at 0.

The R that is obtained in the theorem is referred to as the radius of convergence
and the set C is called the interval of convergence.
We think of {0} as an interval of zero radius and

as an interval of infinite radius.

When R = + ∞, we may denote the interval of convergence

by (− ∞, ∞).

Notice that when R is a positive real number, the theorem says nothing about the
convergence or divergence of the series at the endpoints of the interval of convergence.
It is usually necessary to check the endpoints individually for convergence using
one of the other tests in Section 8.2.
The following Ratio Criterion is also useful in determining the radius of convergence.

Theorem (Ratio
8.3.4 Criterion)
n
The radius of convergence R of a power series Σ a  x is equal to lim     | a /a        |, provided that this limit exists.
n
n →∞ n n + 1

Note that this ratio is the reciprocal of the ratio in the ratio test for sequences (Theorem 8.2.7.)

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 8.3, Slide 5


Example 8.3.5
n

(a) The interval of convergence for the geometric series Σ x is ( –1, 1).
 

(b) For the series

∑ 1n x n

we have

lim

an
n +1
= lim
= 1,
n
an+1

so the radius of convergence is 1.
When x = 1, we get the divergent harmonic series:

∑ 1n .

∑

When x = – 1, we get the convergent alternating harmonic series:

( −1)
n

n

So the interval of convergence is [ – 1, 1).

(c) For the series

∑

1 n
x
n 2we have

an
(n + 1) 2
lim
= lim
= 1,
an+1
n2

so the radius of convergence is 1.

When x = 1, we get the convergent p-series with p = 2:

When x = – 1, the series is also absolutely convergent since

∑ n1

2

.

( −1) n
n2

= 12 .
n

So the interval of convergence is [ – 1, 1].

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 8.3, Slide 6


Example 8.3.7
(a) For the series

∑ n1! x n

lim

we have

an
(n + 1)!
= lim
= lim (n + 1) = + ∞.
an+1
n!

Thus the radius of convergence is + ∞, and the interval of convergence is

∑ nn xn

(b) For the series

.

, it is easier to use the root formula:

α = lim an

1/ n

= lim n n

1/ n

= lim n = + ∞.

Thus R = 0 and the interval of convergence is {0}.

Example 8.3.9
Consider the series

∞

1 2
1 4
1 6
1 8
x +
x
+
x
+
x +L =
2
3
4
1 ×3
2 ×3
3 ×3
4 ×3
2
Letting y = x , we may apply the ratio criterion to the series

n +1

1 2n
∑ n ×3n x .

n =1
∞
Σn =1 (3− n / n) y n
and obtain

an
3 (n + 1)
= lim
= 3.
n
an +1
3 ( n)

lim

Thus the series in y converges when | y | < 3.
 

Since it also converges when y = − 3

 

 

(−

)

but diverges when y = 3, its interval of convergence is [− 3, 3).
2

But y = x , so the original series has
Note that

3, 3

as its interval of convergence.

x is=not−included,
3 because this corresponds to y = +3.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

 

Section 8.3, Slide 7


Sometimes we wish to consider more general power series of the form

∞

∑ an ( x − x0 )n ,

where x0 is a fixed real number.

n=0

By making the substitution y = x – x0, we can apply the familiar techniques to the series

∞

∑ an y n .

n=0

If we find that the series in y converges when | y | < R, we conclude that the original series
 

 

converges when | x – x0 | < R.
 

 

Example 8.3.10
For the series

2

3

1 + ( x − 1) + ( x − 1) + ( x − 1) + L =

∞

∑ ( x − 1)n

n=0
we have R = 1, so it converges when | x – 1 | < 1.

That is, when 0 < x < 2.

Since it diverges at x = 0 and x = 2, the interval of convergence is (0, 2).
In fact, using the formula for the geometric series, we see that for all x in (0, 2)

1
1
=
.
1 − ( x − 1)
2−x

the sum of the series is equal to

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 8.3, Slide 8