Transcript Document 7716930

Alternating Series;
Absolute and Conditional
Convergence
Lesson 8.6
Alternating Series
Two versions
• When odd-indexed terms are negative

k
(

1)
 ak   a1  a2  a3  ...

k 1
• When even-indexed terms are negative

 (1)
k 1
k 1
 ak  a1  a2  a3  ...
Alternating Series Test
ak  0
• Recall lim
does not guarantee
k 
convergence of the series
In case of alternating series …
• Must converge if
 lim ak  0
k 
 { ak } is a decreasing sequence
(that is ak + 1 ≤ ak for all k )
Alternating Series Test
• Text suggests starting out by calculating
lim ak
k 
• If limit ≠ 0, you know it diverges
• If the limit = 0
 Proceed to verify { ak } is a decreasing sequence
(1)k 1  k

2
k
1
k 1

• Try it …
(1) k

k
k 1

Using l'Hopital's Rule
ak
• In checking for lim
k 
be useful
(1)k 1  ln k

k
k 1

• Consider
• Find
lim ak  ?
k 
ln x
1/ x
lim
 lim
0
x  x
x  1
l'Hopital's rule may
Absolute Convergence
• Consider a series  ak
where the
general terms vary in sign
 The alternation of the signs may or may not be
any regular pattern
• If
a
k
converges … so does
a
• This is called absolute convergence
k
Absolutely!
• Show that this alternating series converges
absolutely
(1)k 1

3/ 2
k 1 k

• Hint: recall rules about p-series
Conditional Convergence
a
• It is still possible that even though
diverges …

a
k
k
can still converge
• This is called conditional convergence
• Example – consider
(1) k

k
k 1

vs.

1

k 1 k
Generalized Ratio Test
• Given
a
k
 ak ≠ 0 for k ≥ 0 and
 where L is real or

ak 1
lim
L
k  a
k
• Then we know
 If L < 1, then  a converges absolutely
k
 If L > 1 or L infinite, the series diverges
 If L = 1, the test is inconclusive
Apply General Ratio
• Given the following alternating series
 Use generalized ratio test
2 k 1
2
k 1
(1)

k!
k 1

Assignment
• Lesson 8.6
• Page 542
• Exercises 5 – 29 EOO
Note guidelines, Table 8.1, pg. 540