9.5 Part 1 Ratio and Root Tests
Absolute Convergence


You’ve all seen the Harmonic Series
n 1
1
n
converges
which
?
despite the fact that lim
n 
diverges
1
0
n
which only proves that a series
might converge.
What if we were to alternate the signs of each term?
1
1

1
2

1
3

1
4

1
5

1
...
6
How would we write this using Sigma notation?
This is called an
Alternating Series
The signs of the terms alternate.
Does this series converge?.

  1
n 1
n 1
1
n

1
1

1
2

1
3

1
4

1
5

1
...
6
1 1
Notice that each
  
addition/subtraction is
3 4 1 1
partially canceled by
  
the next one.
5 6
…smaller and smaller…
The series converges as do all Alternating Series whose
terms go to 0
Alternating Series
The signs of the terms alternate.
This series converges by the Alternating Series Test.

  1
n 1
n 1
1

n
1

1
1
2

1
3

1

4
1
5

1
...
6
Alternating Series Test

For the series:
  1
n 1
n 1
an
If a n 1  a n and lim a n  0
n 
Then the series converges
This is called the Alternating Harmonic Series which as
you can see converges…unlike the Harmonic Series
Since each term of a convergent alternating series
moves the partial sum a little closer to the limit:

 1
  3

n 1 
n 1
is what kind of a series?
What is the sum of this series?
3
Very good! A
geometric
series.
 0 . 75
4
Try adding the first four terms of this series:

 1
  3

n 1 
n 1
 1
1
3

1
9

1
27

20
 . 740741
27
Which is not far from 0.75…but how far?
Since each term of a convergent alternating series
moves the partial sum a little closer to the limit:

 1
  3

n 1 
n 1
 1
1
1

3
1

9
20

 . 740741
27
27
Which is not far from 0.75…but how far?
1
. 740741  . 75 
Which means what?
108

 1
   
108
3
n5 
1
n 1
1

81
1
Can we agree that
81

1

243
1
243

1

...
729
1
729
... 
1
81
?
1

81
1

243
1
... 
729
1
leads us to this:
81
Alternating Series Estimation Theorem
For a convergent alternating series, the truncation
error is less than the first missing term, and is the
same sign as that term.
Which means that for an alternating series…
If k terms were generated then the error
would be no larger than term k + 1

  1
k
n
an  S
and
an  S k
where k < 
n 1
n 1
then
  1
n
S k  S  a k 1
in which S k  S is the truncation
error
1
1

81
243

1
... 
729
1
81
leads us to this:
Alternating Series Estimation Theorem
For a convergent alternating series, the truncation
error is less than the first missing term, and is the
same sign as that term.
For example, let’s go back to the first four terms of our
series:

 1
  3

n 1 
Answer:
n 1
 1
1
3

1
9

1
27
What is the maximum error
between this approximation
and the actual sum of infinite
terms?
1
81
1
Which is bigger than the actual
error which we determined to be 108
Since each term of a convergent alternating series
moves the partial sum a little closer to the limit:
Alternating Series Estimation Theorem
For a convergent alternating series, the truncation
error is less than the first missing term, and is the
same sign as that term.
Answer:
1
81
So what does this answer tell
us? It simply means that the
1
error will be no larger than
81
That is why it is commonly referred to as the
Error Bound

Since each term of a convergent alternating series
moves the partial sum a little closer to the limit:
Alternating Series Estimation Theorem
For a convergent alternating series, the truncation
error is less than the first missing term, and is the
same sign as that term.
This is also a good tool to remember because it is
easier than the LaGrange Error Bound…which you’ll
find out about soon enough…
Muhahahahahahaaa!
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. This is called the RATIO TEST

For the series
a
n
n
if
lim
n 
a n 1
 L
then
an
if L  1 the series converges.
if L  1 the series diverges.
This test is ideal
for factorial
terms
if L  1 the series may or may not converge.
This also works for the nth root of the nth term. This is
called the ROOT TEST(Ex. 57 pg. 508)

For the series
a
n
n
if lim
n 
n
an  L
then
if L  1 the series converges.
if L  1 the series diverges.
if L  1 the series may or may not converge.
Absolute Convergence
If

an
converges, then we say
If

an
a
converges, then
converges absolutely.
n
a
n
converges.
If the series formed by taking the absolute value of each
term converges, then the original series must also
converge.
“If a series converges absolutely, then it converges.”
Conditional Convergence
If
a
n
a
n
converges but

a n diverges, then we say that
converges conditionally
If
a
n
diverges, then

  1
n 1
n 1
1
n


a n diverges.
1
1

1
2

1
3

1
4
The Alternating Harmonic Series is a perfect
example of Conditional Convergence

1
5

1
6
...