Transcript Calculus 9.4 - University of Houston

9.4
Photo by Vickie Kelly, 2008
Radius of Convergence
Abraham Lincoln’s Home
Springfield, Illinois
Greg Kelly, Hanford High School, Richland, Washington
Convergence
The series that are of the most interest to us are those
that converge.
Today we will consider the question:
“Does this series converge, and if so, for what values of
x does it converge?”

The first requirement of convergence is that the terms
must approach zero.
nth term test for divergence


diverges if lim an
a
n
n 1
n 

fails to exist or is not zero.

n
n
Ex.Note
2: that thisncan
If
then
grows without

1
n
!
x
!
x
prove
that
a
series
diverges,
but can
x
n 0
not prove that a series
converges.
bound.
n!
If 0  x  1 then lim n ! x  lim n  
n 
n   1 
n
1
, therefore
n  The series diverges. (except
when
x
the numerator grows faster than the denominator.
As

, eventually n is larger than
 
x
x=0)

There are three possibilities for power series convergence.
1
The series converges over some finite interval:
(the interval of convergence).
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 the endpoints
of the interval.
2
The series converges for every x. ( R   )
3
The series converges at x  a and diverges
everywhere else. ( R  0 ) (As in the previous example.)
The number R is the radius of convergence.

Direct Comparison Test
For non-negative series:
If every term of a series is
less than the corresponding
term of a convergent series,
then both series converge.
If every term of a series is
greater than the
corresponding term of a
divergent series, then both
series diverge.
This series converges.
So this series must
also converge.
So this series must also diverge.
This series diverges.


Ex. 3:
x2n
Prove that 
2 converges for all real x.
n  0  n !
There are no negative terms: 
x
2n
 n!
2

x 
2 n
n!
larger denominator


n 0
 x 2 n
n!
is the Taylor series for
e
x2
, which converges.
 The original series converges.
The direct comparison test only works when the terms are
non-negative.

Absolute Convergence
If
a
n
converges, then we say
a
n
converges absolutely.
The term “converges absolutely” means that the series
an converges,
formed
the absolute
value aof
term
If by taking
then
converges.
n each
converges. Sometimes in the English language we use
the word “absolutely” to mean “really” or “actually”. This is
If
the
series
not
the
caseformed
here! by taking the absolute value of each
term converges, then the original series must also
converge.


“If a series converges absolutely, then it converges.”

Ex. 4:


 sin x 
n
n!
n 0
We test for absolute convergence:
sin x
n
n!
1

n!
2
3
n
x
x
x
Since e x  1  x 
,
   
2! 3!
n!

1
converges to

n0 n !


n 0
sin x
n!

Since

n 0
e e
1
n
converges by the direct comparison test.
 sin x 
n!
n
converges absolutely, it converges.

Ratio Technique
We have learned that the partial sum of a geometric series
is given by:
1 r
Sn  t1 
1 r
n
When
where r = common ratio between terms
r  1 , the series converges.

Geometric series have a constant ratio between terms.
Other series have ratios that are not constant. If the
absolute value of the limit of the ratio between
consecutive terms is less than one, then the series will
converge.

For
t
n 1
n
tn 1
, if L  lim
n  t
n
then:
if
L 1
the series converges.
if
L 1
the series diverges.
if
L 1
the series may or may not converge.

Ex:
x 2 x3 x 4
ln 1  x   x      If we replace
2 3 4
1
1
1
2
3
4
ln  x    x  1   x  1   x  1   x  1   
2
3
4
1  x  1

L  lim
n2
n
n 1

n 1
 x  1  x  1
n
 lim
n 
n 1
x  1 n

 lim
n 
n 1

n
 1  x  1
n 1
n
x with x-1, we get:

  1
n 1
n 1
1
n
  x  1
n
an 1
1
 an 1 
an
an
n
 x  1
 x 1
n
If the limit of the ratio
between consecutive terms
is less than one, then the
series will converge.

If the limit of the ratio between consecutive terms
is less than one, then the series will converge.
x 1  1
1  x  1  1
0 x2
The interval of convergence is (0,2).
The radius of convergence is 1.

Ex:

n
n
x  5

n 
n 1 3
1
2
3
2
3
 x  5   x  5   x  5  
3
9
27
L  lim
n 
 n  1 x  5
n 1

n 1
3
3n
n  x  5
n
n
n

1
x

5
x

5

3
    
n
L  lim
n 
3  3  n  x  5
n
n
n  1 x  5

L  lim
n 
3n

Ex:

n
n
x  5

n 
n 1 3
n  1 x  5

L  lim
n 
3n
n 1
L  x  5 lim
n  3n
1
L  x5 
3
1
x 5 1
3
x 5  3
3  x  5  3
2 x8
The interval of convergence is (2,8).
82
The radius of convergence is
3 .
2

Ex:

n!
n
x  3

4 
n 1 n
1
2
3
2
3
4
 x  3   x  3   x  3   x  3  
8
27
32
L  lim
n 
 n  1! x  3
 n  1
n 1

4
n4
n ! x  3
n ! n  1 x  3  x  3
n
L  lim
n 
 n  1
4

n
n4
n ! x  3
n
1
 n 
L  x  3 lim  n  1 

n 
n

1


4

Ex:

n!
n
x  3

4 
n 1 n
1
 n 
L  x  3 lim  n  1 

n 
 n 1 
L
for all
x 3.
4
Radius of convergence = 0.
At x  3, the series is 0  0  0   , which converges to zero.
Note: If R is infinite, then the series converges for all values of x.

Another series for which it is easy to find the
sum is the telescoping series.

Ex. 6:
1
 n  n  1
n 1

1
1


n 1
n 1 n
 1    1 1    1 1   
1        
 2  2 3 3 4
1
S3  1 
4
1
Sn  1 
n 1
lim S n  1
n 
Using partial fractions:
1
A
B
 
n  n  1 n n  1
Telescoping
1  A n 1 Series
Bn

1  An  A  Bn
  bn  bn1 
n 1
1  A 0  A B
bn 1
converges to b1  lim
n 
0  1 B
1  B
p