Download Alternating Series and Absolute Convergence

Alternating Series and Absolute Convergence
Alternating Series Test:

The alternating series
 (1)
n 1
n 1
a n  a1  a 2  a3  a 4      (1) n 1 a n    
is convergent if the following two conditions are satisfied:
(i) a k  a k 1  0 for every k
(ii) lim n an  0
Example 1: Determine whether the alternating series converges or diverges.

(a)
 (1) n1
n
Definition: A series
| a
n

2n
4n 2  3
a
(b)
 (1)
n
n
n 1
2n
4n  3
is absolutely convergent if the series
|| a1 |  | a2 |     | an |     is convergent.
Example 2: Show that the following alternating series is absolutely convergent.
1
1 1 1
1
 1         (1) n 1    
n
2 3 4
n
Show that this series is (a) convergent (b) not absolutely convergent
Example 3: The alternating harmonic series is
Definition: A series
divergent.
a
Theorem: If the series
n
 (1)
is conditionally convergent if
a
n
1
n
a
is absolutely convergent, then
1
1
1
2
3
4
 a be the series 2  2  2  2
Determine whether  a converges or diverges.
Example 4: Let
n 1

n
a
is convergent and
n
is convergent.
1
1
1
1
 6  7  8  
5
2
2
2
2
n
Example 5: Determine whether the following series is convergent or divergent:
sin 2 sin 3
sin n
sin 1  2  2      2    
2
3
n
| a
n
| is
A series may be classified in exactly one of the following ways:
(1) absolutely convergent
(ii) conditionally convergent
Ratio Test For Absolute Convergence: Let
lim n
(i)
(ii)
(iii)
a
n
(iii) divergent
be a series of nonzero terms, and suppose
a n 1
L
an
If L<1, the series is absolutely convergent
a
If L>1, or lim n n 1   , the series is divergent.
an
If L = 1, apply a different test; the series may be absolutely convergent, conditionally
convergent, or divergent.
Example 6: Determine whether the following series is absolutely convergent, conditionally
2

n n 4
convergent, or divergent:  (1)
2n
n 1
Power Series and Interval of Convergence:
Example 1: Find all values of x for which the following power series is absolutely convergent
(Interval of convergence):
1
2
n
1 x  2 x2    n xn  
5
5
5
Example 2: Find all values of x for which the following power series is absolutely convergent:
1
1
1
1 x  x2    xn  
1!
2!
n!
Example 3: Find all values of x for which
 n! x
n
is convergent:
Theorem:
(i)
If a power series  an x n converges for a nonzero number c, then it is absolutely
convergent whenever | x || c | .
(ii)
If a power series  an x n diverges for a nonzero number d, then it diverges whenever
| x || d | .
Theorem:
(i)
(ii)
(iii)
If  an x n is a power series, then exactly one of the following is true:
The series converges only if x  0.
The series is absolutely convergent for every x.
There is a number r>0 such that the series is absolutely convergent if x is in the open
interval (-r,r) and divergent if x<-r or x>r.
The number r is the radius of convergence of the series. Either convergence or divergence may
occur at –r or r, depending on the nature of the series. The totality of numbers for which a power
series converges is called its interval of convergence. If the radius of convergence r is positive,
then the interval of convergence is one of the following
(-r,r), (-r,r], [-r,r), [r,r]
To determine which of these intervals occurs, we must conduct separate investigations for the
numbers x  r and x  r . The radius of convergence can be denoted 0 or  (Ex 2 and 3
respectively).
(note: save this for next class) Example 4: Find the interval of convergence and the radius of

1 n
x .
convergence of the power series 
n
n 1
k
1 n
x for k = 3, 4, 5, and 6.
Use a graphing utility to graph the polynomials Pk ( x)  
n
n 1
Example 5: Find the interval of convergence and radius of convergence of the series
1
1
1
1  ( x  3)  ( x  3) 2      (1) n
( x  3) n    
2
3
n 1