Transcript MTH 251 Differential Calculus

MTH 253
Calculus (Other Topics)
Chapter 11 – Infinite Sequences and Series
Section 11.5 – The Ratio and Root Tests
Copyright © 2009 by Ron Wallace, all rights reserved.

a
k 1
k
Does the Series Converge?
10 Tests for Convergence
Geometric Series
N-th Term Test (Divergence Test)
Integral Test
p-Series Test
Comparison Test
Limit Comparison Test
Ratio Test
The test
tells you
nothing!
Root Test
Alternating Series Test
Absolute Convergence Test
Each test has it limitations (i.e. conditions where the test fails).
The Ratio & Root Tests
NOTE:
For all series in this section, it will be
assumed that each term is positive.
That is, given
then
a
k
ak  0, k
The Ratio Test
ak 1
Let: lim
L
k  a
k

ak 1
 r where L    r k  K
ak
(from the limit definition)

ak 1  rak  r 2ak 1...  r k a1
  ak 1   r a1
k
Geometric Series!
Convergent if |r|<1
Comparison Test  If L < 1, the series converges.
The Ratio Test
ak 1
Let: lim
L
k  a
k

ak 1
 s where s  L   k  K
ak
(from the limit definition)

ak 1  sak  s2ak 1...  sk a1
  ak 1   s a1
k
Geometric Series!
Divergent if |s|>1
Comparison Test  If L > 1, the series diverges.
The Ratio Test
ak 1
Let: lim
 L 1
k  a
k
1
 k diverges
1 (k  1)
k
lim
 lim
1
k 
k  k  1
1k
1
 k 2 converges
1 (k  1)2
k2
lim
 lim 2
1
2
k 
k  k  2k  1
1k
If L=1, the test fails!
The Ratio Test
Given  ak where ak  0 k
ak 1
Let: lim
L
k  a
k
• If L < 1, the series converges.
• If L > 1, the series diverges.
• If L = 1, the test fails.
Example w/ the Ratio Test
k
4
 k2
4k 1
lim
k 
4k 1
k2
4k 2
(k  1)2
 lim
 k  lim 2
 4 1
k
2
k  (k  1)
k  k  2k  1
4 2
4
k
 Divergent!
The Root Test
Given  ak where ak  0 k
Let: lim k ak  L
k 
• If L < 1, the series converges.
• If L > 1, the series diverges.
• If L = 1, the test fails.
Proof is similar to the ratio test!
Example w/ the Root Test
 1 e
 2

k



k
k
k
 1 e 
1

e
1
k
lim 
 lim
 1

k 
k 
2
2
 2 
k
 Convergent!