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Transcript
```9.4
Quick Review
Find the limit of the expression as n  . Assume x remains fixed as n changes.
nx
n 1
n x 3
2.
n  n  1
1.
2
x
n!
 n+1 x
4.
 2n 
3.
4
2
4
2 2x 1
5.
2 2x 1
n
n 1
n 1
n
Quick Review
Let a be the nth term of the first series and b the nth term of the
n
n
second series. Find the smallest positive integer N for which
a  b for all n  N . Identify a and b .
n
n
n
6.  5n,  n
7.  n ,  5
5
2
n
8.  ln n,  n
1
9. 
, n
10
-3
n
n
Quick Review Solutions
Find the limit of the expression as n  . Assume x remains fixed as n changes.
nx
|x|
n 1
n x 3
2.
| x 3|
n  n  1
1.
2
x
0
n!
 n+1 x
4.
 2n 
3.
4
2
2
x /16
4
2 2x 1
5.
2 2x 1
n
n 1
n 1
n
| 2x 1 | / 2
Quick Review Solutions
Let a be the nth term of the first series and b the nth term of the
n
n
second series. Find the smallest positive integer N for which
a  b for all n  N . Identify a and b .
n
n
n
n
6.  5n,  n a  n , b  5n, N  6
7.  n ,  5 a  5 , b  n , N  6
2
2
n
5
n
n
n
n
8.  ln n,
1
9. 
,
10
1
10.  ,
n
n a 

n
1
n!
n

2
n
-3
5
n
n , b  ln n, N  1
n
1
1
, b  , N  25
10
n!
1
a 
, b n , N 2
n
a 
n
n
n
3
n
2
n





Convergence
nth-Term Test
Comparing Nonnegative Series
Ratio Test
Endpoint Convergence
… and why
It is important to develop a strategy for finding the
interval of convergence of a power series and to obtain
some tests that can be used to determine convergence of
a series.
The Convergence Theorem for
Power Series
There are three possibilities for  c  x  a  with respect to convergence:

n 0
n
n
1. There is a positive number R such that the series diverges for x  a  R
but converges for x  a  R. The series may or may not converge at either
of the endpoints x  a  R amd x  a  R.
2. The series converges for every x ( R  ).
3. The series converges at x  a and diverges elsewhere ( R  0).
The nth-Term Test for
Divergence

 a diverges if lim a fails to exist or is different from zero.
n 1
n
n 
n
The Direct Comparison Test
Let  a be a series with no negative terms.
n
(a)  a converges if there is a convergent series  c with
a  c for all n  N , for some integer N .
n
n
n
n
(b)  a diverges if there is a divergent series  d with
a  d for all n  N , for some integer N .
n
n
n
n
Example Proving
Convergence
by
Comparison
x
Prove that 
converges for all real x.
2n

n 0
 n !
2
2n
x
Let x be any real number. The series 
has no negative terms.
 n !

n 0
x 
x 
x
x


. Recognize 
as the Taylor series
n!
 n ! n ! n !
2n
For any n,
2
2
2n
n
2
n

2
n 0
x2
x2
for e , which we know converges to e for all real numbers. Since
2n
x
the e series dominates 
term by term, the latter series must also
n
!
 
x2

n 0
2
converge for all real numbers by the Direct Comparison Test.
Absolute Convergence
If the series  a of absolute values converges, then  a
n
converges absolutely.
n
Absolute Convergence Implies
Convergence
If  a converges, then  a converges.
n
n
Example Using Absolute
Convergence
 sin x 
Show that 
converges for all x.
n

n!
n 0
n
sin x
Let x be any real number. The series 
has no negative terms,
n!
1
and it is term-by-term less than or equal to the series  , which converges
n!
sin x
to e. Therefore, 
converges by direct comparison.
n!
 sin x 
Since 
converges absolutely, it converges.
n!

n 0

n 0
n

n 0
n

n 0
The Ratio Test
Let  a be a series with positive terms, and with lim
n
n 
a
 L.
a
n 1
n
Then,
(a) the series converges if L  1,
(b) the series diverges if L  1,
(c) the test is inconclusive if L  1.
Convergence
nx
Find the radius of convergence of 
.
n

n 0
10
n
Check for absolute convergence using the Ratio Test.
a
 n  1 x
lim
 lim
a
10
n 1
n 
n 
n
n 1
n 1
n
10

nx
n
x
 n 1 x
 lim 
 
 n  10 10
x
Setting
 1, we see that the series converges absolutely for  10  x  10.
10
The series diverges for x  10 and x  10.
n 
The radius of convergence is 10.
Example Determining
Convergence
of
a
Series
2
Determine the convergence or divergence of the series 
.
n

n 0
Use the Ratio Test:
2
n 1
a
 2  3  1 
lim
 lim 3  1  lim 


2
a
 3  1  2 
3 1
 3 1 
 lim 2 

 3 1
1
1
 lim 2 3
1
3
3
2
2

The series converges because  1.
3
3
n 1
n 1
n
n 1
n 
n
n 
n 
n
n
n
n 
n 1
n
n 
n
n 1
n
3 1
n
Pg. 386, 7.1 #1-25 odd
```
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