§11.8 Power Series

§11.8 Power Series

Definition：  

0 1

 

2

 

²

^...

0















^





a x c a x c c a x

c

ⁿ

n

n

is called a power series

about a (centered at a) (or in (x - a)).

Question：For what values of x do the corresponding series converge?

越靠近

a (中心)，越可能收斂。

 收斂半徑(radius of convergence)

 收斂區間(interval of convergence) Example 1：

Find the interval of convergence and radius of convergence of  

3 .

1



^





n

n

n x

Solution：

Using ratio test, we get

 

  ^{1 3 .}

lim 3 3

1 3 lim

1



 

 















 x

n x n n

x n x

n n n

n

If x  3  1 , that is

x

 ( 2 , 4 ) , the power series converges (absolutely).

 Check end points：

 

 

. 1 : e convergenc of

radius

. 4 , 2 : e convergenc of

interval

. divergence

n 4 1

x

e.

convergenc 2 1

1 n

1













 

















r x n

n

n

Example 2：

Find the interval of convergence and radius of convergence of

   



^





1 2 2

2

! 2

1

n n

n n

n x

Solution：

By ratio test

 

 

 

 

 

 

















 



 













r R n x

x n x

n x

n

n n n

n

n

: e convergenc of

radius

, - : e convergenc of

interval 1 0 lim 4

! 2

! 1 lim 2

2 2

2 2 2 ) 2 1 ( 2

1 2

＊註：重量級在分母   n !

²

 x

²ⁿ

. Example 3：

Find the values of x for which the series 

^

   







1

1 3 1 1

n

n n

n

x

n converges

Solution：

Example 4：

Find the interval of convergence and radius of convergence of 

^

1

!

n

x

n

n

Solution：

By ratio test

 

 

0.

r : e convergenc of

radius

. 0 : e convergenc of

interval

) exist

(not 1 lim















 n x R

n

＊註：重量級 

ⁿ

! 

^xn

 在分子。

Example 5：

Find the interval of convergence and radius of convergence of

 

1 . 2

1



^





n n n

n x

Solution：

By ratio test

 

2 . , 3 2 1 2

| 1 1

| 1 | 1

| 2 Let

.

| 1

| 1 2

1 lim 2

 

 



 



















 

 





x x

x

R x n x

n x

n

 Check end points：

 

2 . : 1 e convergenc of

radius

2 . , 3 2 : 1 e convergenc of

interval

. divergence

1 2

3

e.

convergenc 1

2 1

1 1



 

 

 









 

















r x x n

x n

n n

n

Example 6：

Find a value of b that will make the radius of convergence of power series



^

2

ln

n n n

n x

b

equal to 5.

Solution：

5 . 5 1

1 1 1

1 1

|

| Let Test, Ratio By the

| ) | 1 ln(

lim ln

| ln |

) 1 lim ln(

lim ln

1 1 1











 













 



 

















b r

x b bx b

bx

n bx bx n

x b

n n

x b a

a n x a b

n n n n n n n

n n

n n n



Example 7：

For which α is the interval of convergence of 

^

1

 1

n

x

n

n n

^

equal to   1 , 1 

Solution：



^

1

 1

n

x

n

n

n

^

Example 8：

If 

^

1 n

n n

x

c converges with x = 4 and diverges when x = 6, then determine the convergence of the following series.

Solution：

answers reasons

i. 

^

1 n

c

n

convergence

x = 1

ii. 

^

1

8

n n

c

n

divergence

x = 8

iii.  

ⁿ

n

c

n

3

1



^





convergence

x = -3

iv. 

^

 





1

9 1

n

n n

n

c divergence

x = 9

＊註： If 4  x | |  6 , then can' t determine whether it is convergent ot divergent.

Example 9：

The power series 

^

 





1

2

n

n n

x

a and 

^

 





1

3

n

n n

x

b both converge at x = 6.

Find the largest interval over which both series must converge.

Solution：

Theorem：

For series 

^

 





n 1

n

n

x a

c , either one of the following 3 possibilities holds：

(i). r = 0：只有一點收斂

(ii).

r

  ：兩端點須被檢驗才可知其收斂與否

(iii).

r

  ：每一點接收斂