# Sequences and Series** ```Sequences and Series (BC Only)
Sequences and series is one of the most important topics to be tested on the BC Exam. It
is not listed in the topics for the AB Exam. There is a great deal of material in this
section. The Test Development Committee always includes a good selection of topics
about sequences and series in the BC portion of the AP Exam. Because of the amount of
material, not all topics are tested each year. However, as a student, you don't know which
items will be covered and which one won't. So, you have the responsibility of being
familiar with entire section.
In this chapter we will review sequences and explain how series differ from
sequences. Then we will look at Power Series, Taylor Series, and Maclaurin Series. We
will next discuss Taylor Polynomials along with their error bounds. Finally, we will
review a variety of techniques to determine if a series converges or diverges. We will
place the emphasis on the understanding rather than the manipulation.
Sequences
A sequence is a list of numbers. You have worked with sequences throughout your math
classes. We speak of the even numbers {2, 4, 6,…}, alternating powers of 2
1 1 1
2, 4,8, 16,32,... , and decreasing fractions  , , ,... . All of the lists follow some
2 3 4 
pattern. Sometimes the pattern is easy to spot; other times it is more difficult to
determine. The domain of a sequence is the set of positive integers. These integers are
not the value of the sequence. Rather, they are the means by which we order the
sequence. We identify the first term, the second term, the third term, and so on. We
label the terms for convenience using the notation, an . So a1 means the first term of
some sequence. a10 is the tenth term of a sequence. an is the nth term or general term of
a sequence. b3 would mean the third term of some sequence that is different from the
sequence a.
Limit or Convergence of a Sequence
As we work with sequences, we want to find out if the sequence converges or
diverges. We use the following definition to help us.
Limit of a Sequence
A sequence an  has a limit L, or converges to L, if
lim an  L or an  L as n   .
n 
If such a number L does not exist, the sequence has no limit
diverges.
or
To establish convergence of a series, we often look at functions based on x that
mimic the sequence. We then see if the function in x converges or diverges and apply the
results to the sequence that is based on n. The following rule summarizes this statement.
Let an  be a sequence with f (n)  an and suppose that f ( x) exists for
every real number x  1. Then the following cases apply.
Case 1. If lim f ( x)  L, then lim f (n)  L.
x 
n 
Case 2. If lim f ( x)   (or  ) , then lim f (n)   (or  ).
x 
n 
Example 1.
Let an  1 
1
. Does the sequence an  converge or diverge?
n
We use a function f(x) that mimics the sequence, so we choose
1
 1
f ( x)  1  for every real number x  1. We check lim 1   . From the chapter on
x 
x
 x
1
 1
limits, we recall that lim 1    lim1  lim  1  0  1. Since f ( x) converged to 1, we
x 
 x  x  x  x
know that the sequence an  must also converge to 1. We have shown that
an  converges.
Figure 1
The graph of the sequence supports this result. The values of the sequence as indicated
by the points seem to level off at y  1 .
Test Tip. Although a graphing calculator can be used to investigate convergence of a
sequence, the graph itself can never be considered as proof. As we stated in the Rule of
Three, we can investigate graphically, numerically, or analytically. But we can only
verify analytically.
Example 2.
 e3 n 
Does the sequence   converge?
 4n 
Rewriting the sequence as a corresponding function of x, we get f ( x) 
e3 x
.
4x
Applying the limit, we get
e3 x 
lim

(Apply L'Hopital's Rule.)
x  4 x

e3 x
3e3 x
lim
 lim

x  4 x
x  4
Since f ( x) is divergent, an  is also divergent.
Series
A series is a sequence of partial sums. Symbolically, we write Sn   an . Let's examine
a series term by term to see how the individual terms relate to each other.
Example 3
Suppose that an 
1
, write the first five terms of S n
n2
1
1

1 2 3
1
1
7
S2  a1  a2 


1  2 2  2 12
1
1
1
47
S3  a1  a2  a3 



1  2 2  2 3  2 60
1
1
1
1
19
S4  a1  a2  a3  a4 




1  2 2  2 3  2 4  2 20
19
1
153
S5  S4  a5 


20 5  2 140
S1  a1 
It is very easy to mix sequences with series. There is a lot of vocabulary that you
need to understand and keep straight in your mind. The following list is given to explain
the differences in vocabulary
Sequence
an infinite set of ordered numbers. 1, 4, 9, 16, 25,… is the
sequence of perfect squares.
Series
the sum of all the terms of a sequence. 1 + 4 + 9 + 16 + 25 + … is
the series of perfect squares.
Partial sum the sum of the first n terms of a sequence. S3  1  4  9 is the third
partial sum of the series of perfect squares.
n
the term index used to calculate the term value of a sequence or a series. It
is always an integer greater than or equal to 1.
Power Series
In some of your earlier math classes you worked with a geometric sequences and series
1 1 1
1 3 7
such as the sequence , , ,... and its related series , , ,... . In this section, we
2 4 8
2 4 8
want to review a geometric series with constants; then we will extend it to variables.
Example 4.
1
1
1


 ... converges.
10 100 1000
1
1
1
1
We can rewrite this series as 1  2  3  ...  n  ... which is a geometric
10 10 10
10
1
series with a common ratio of . We recall the sum of any geometric series can be
10
a
1
found by using the formula S 
. In this example r  . The sum is
1 r
10
1
1
10  10  1 .
1 9
9
1
10
10
Determine if the series
Geometric Series
A geometric series is written in the form a  ar  ar 2  ar 3  ...  ar n 1  ...
We can use   sigma  notation to write this in a short form,

a
.
1 r
n 1
This series converges if r  1 and diverges if r  1 .
 ar
n 1

Example 5.
Check the series
2 2 2
2
 
 ...  n  ... for convergence.
3 9 27
3
The first term is
2
1
and the common ratio is . By the formula,
3
3
2
2
3
S
 3 1 .
2
1 1
3
3
This is a finite number so the series converges.
Example 6.

Does the series
en
converge?

n
n 1 2
e
e
, and the common ratio is , which is greater than 1. Since
2
2
the ratio is greater than 1, the series diverges.
The first term is
Example 7.
Write the expansion of a geometric series whose ratio is x.
We replace r with x which gives us the sum
1
S
 1  x  x 2  x 3  x 4  ...  x n  ...
1 x
Representing Function by Series
Example 7 illustrates the reason we are so concerned about the geometric series. When x
was used in place of the constant, the right side became a polynomial expression that is
1
equivalent to
whenever x  1. This expression is very powerful. It lets us describe
1 x
a non-polynomial function as a polynomial function.
As the figure shows, the partial sum, S4  1  x  x 2  x 3 , of the polynomial series
1
converges very closely to y 
as long as 1  x  1 . It diverges quickly when x is
1 x
outside the interval. As we increase the number of terms in partial sums, the convergence
gets stronger.
Figure 2
Although we use a polynomial to approximate the rational function, the

expression
x
n 1
is not a polynomial. Polynomials have finite degrees and are not
n 1
subject to divergence.
Example 8.
Use the formula for the geometric series to find a power series representation for
1
f ( x) 
. Determine the interval of convergence.
1 x
We rewrite the function as
1
. The ratio is -x. The expansion is given by
1  ( x)
1
 1  x  x 2  x3  ...  ( x) n 1  ... .
1  (  x)
The figure shows the relationship between f(x) and S 4 on the interval 1  x  1 .
Figure 3
After we have a power series representation, we can manipulate using techniques
we have learned in algebra or calculus classes.
Example 9.
Since
for
1
.
(1  x)2
1
 1  x  x 2  x3  x 4  ...  x n  ... , find a power series representation
1 x
We get a little sneaky here. Notice that
of
1
just happens to be the derivative
(1  x)2
1
. We can take the derivative of both sides of the equation.
1 x
d  1  d
1  x  x 2  x 3  x 4  ...  x n  ...


dx  1  x  dx
1

 1  2 x  3 x 2  4 x 3  ...  nx n 1  ...
2
(1  x)


As a result, we have a power series representation for a new function,
1
.
(1  x)2
We can use integration as well as differentiation to manipulate both sides of an
equation.
Example 10.
1
is 1  x 2  x 4  x 6  ...  ( x 2 ) n  ... .
2
1 x
Integrate both sides to determine a new power series.
The power series representation for
1
= 1  x 2  x 4  x 6  ...  ( x 2 ) n  ...
2
1 x
 1 
 
dx   1  x 2  x 4  x 6  ...  ( x 2 ) n  ... dx
2 
1

x


2 n 1
x3 x5 x 7
n ( x)
 arctan x  x     ...   1
 ...
3 5 7
2n  1
We now have a power series representation for arctan x.


Example 11.
Find a power series representation for
x
.
1 x
This problem looks similar to Example 8. Unfortunately, it is not quite the same.
 1 
But we can rewrite our problem as x  
 . We already have a power series
 1 x 
1
representation for
. We can multiply both sides by x to get the new power series.
1 x
1
 1  x  x 2  x 3  ...  ( x) n 1  ...
1 x
 1 
2
3
n 1
 x 
  x  1  x  x  x  ...  ( x)  ...
 1 x 
x
n 1

 x  x 2  x 3  x 4  ...   1 ( x) n  ...
1 x


As you can imagine, there are many, many functions for which we could use one
of these techniques to develop a power series representation. Because you may be
wondering which one to use, consider this idea. You have only three tools at your
disposal. You can differentiate a power series, integrate a power series, or multiply (or
divide) a power series by a variable. Always try one of these techniques to the compact
side of a known expression to see if you can obtain an expression you need. If the left
side works out properly, do the same thing to the right side.
The power series that we developed in the preceding examples all had similar
features. They all started with increasing powers of x, the defining characteristic of a
power series.
Definition of a Power Series

An expression in the form
c x
n 0
n
n
, which expands as
P( x)  c0  c1 x1  c2 x 2  c3 x 3  ...  cn x n  ...
is a power series centered at x  0 . This is known as a Maclaurin Series.

An expression in the form
 c  x  a
n 0
n
n
, which expands as
P( x)  c0  c1  x  a   c2  x  a   c3  x  a   ...  cn  x  a   ...
1
2
3
n
is a power series centered at x  a . This is known as a Taylor Series.
These definitions are very powerful. They allow us to custom design any function we
want as a polynomial representation.
Taylor Series
Example 12.
Construct a polynomial P( x)  c0  c1 x  c2 x 2  c3 x 3  c4 x 4 with the following
characteristics at x  0 .
P (0)  2
P(0)  5
P(0)  4
P(0)  12
P  iv  (0)  8
We see immediately that P(0)  c0  c1 (0)  c2 (0) 2  c3 (0)3  c4 (0) x 4  c0 . Since
P(0)  2, c0  2 . The other terms are a little harder to find. First, let's take the first four
derivatives of P(x).
P( x)  c1  2c2 x  3c3 x 2  4c4 x 3
P( x)  2c2  6c3 x  12c4 x 2
P( x)  6c3  24c4 x
P iv  ( x)  24c4
We can evaluate each derivative at x = 0 and set it equal to the given value.
c1  5
2c2  4  c2  2
6c3  12  c3  2
24c4  8  c4  1
3
With the value for each constant now determined, we can write our customized
polynomial:
P( x)  2  5 x  2 x 2  2 x 3  1 x 4 .
3
Example 13.
Given f ( x)  ln x , construct a polynomial in the form
P( x)  c0  c1 x  c2 x 2  c3 x 3  c4 x 4 centered about x  1 .
Since we are centered about x  1 , we need to rewrite the polynomial as
1
2
3
4
P( x)  c0  c1  x  1  c2  x  1  c3  x  1  c4  x  1 . First, we evaluate the function
at x  1 . Then we find the first four derivatives and evaluate each at x  1 .
f ( x)  ln x
P(1)  ln(1)  0
1
f ( x) 
P(1)  f (1)  1
x
1
f ( x)  2
P(1)  f (1)  1
x
2
f ( x)  3
P(1)  f (1)  2
x
6
f  iv  ( x)  4
P  iv  (1)  f  iv  (1)  6
x
From our work in Example 12 we can find the derivatives of the polynomial
P( x)  c1  2c2 ( x  1)  3c3 ( x  1) 2  4c4 ( x  1)3
P( x)  2c2  6c3 ( x  1)  12c4 ( x  1) 2
P( x)  6c3  24c4 ( x  1)
P  iv  ( x)  24c4
When we evaluate each derivative at x  1 , each factor ( x  1) will become zero.
We can find each of the constants.
c0  P(1)  0
c1  1
2c2  1  c2  1
2
6c3  2  c3  1
3
24c4  6  c4  1
4
We complete the problem by substituting the values for the constants. The answer is
P( x)  0  1x  1 x 2  1 x3  1 x 4 .
2
3
4
This polynomial does not equal ln x . It is a very close approximation over the
interval of convergence (which we have not yet found).
Examples 12 and 13 are examples of Taylor Polynomials. Since both
polynomials are written to the fourth power, they are referred to as fourth order Taylor
Polynomials. Example 13 was expanded about x  1 . Example 12 was expanded about
x  0 . Any Taylor polynomial expanded about x  0 is called a Maclaurin polynomial.
In examples 12 and 13, our work was to find the value of each coefficient. We
could work a problem step by step each time. However, mathematicians long ago
discovered the relationship between the coefficient and the degree of the term.
Taylor Series at x = a
Let f ( x) be a function with derivatives of all orders on some open interval
containing a. The Taylor Series expansion about x  a is
f (a)  f (a)( x  a) 

which is equal to

n 0
f (a)
f (a)
f  n  (a)
( x  a) 2 
( x  a)3  ... 
( x  a ) n  ...
2!
3!
n!
f  n  (a)
( x  a)n .
n!
The partial sum given by
k
f  n  (a)
Pk ( x)  
( x  a) n
n!
n 0
is called a Taylor polynomial of order k at x  a .
Example 14.
Find the Taylor Series expansion for sin x at x   .
2
First, we find all the coefficients by using the definition at x  
 2   sin  2   1
c  f     cos    0
2
2
f     sin   1
2 
2 
c 
c0  f 
2
1
2
c3 
c4 
2!
2!
2!
 2    cos  2   0
f  
3!
3!
 2   sin  2   1
f  iv  
4!
4!
4!
You can see that the higher-order derivatives of the sin x will repeat in the preceding
pattern. The Taylor Series expansion of sin x at x   is given by
2
x    x   x  

2 
2 
2
sin x  1 
2
2!
4
4!
6!
6
 ...  (1)n
 x  2
(2n)!
2n
 ...
Test Tip. It is easy to mix up the position of the terms. The first term is c0 . We
always begin our counting at n = 0.
Earlier in the chapter, we talked about the many different functions that could be
represented by a power series expansion. There are seven power series expansions that
you should know. Many AP questions have their beginnings in these power series. We
will present them centered about x  0, which makes them Maclaurin Series. To convert
to a general Taylor Series about x  a , substitute ( x  a) in place of x . The first term
will no longer be f (0) but f (a) .
Seven Well-known Power Series (with Intervals of
Convergence)

x 2 x3
xn
xn
  ...   ...  
(all real x)
2! 3!
n!
n 0 n !
ex  1  x 
sin x  x 

x3 x5
x 2 n1
x 2 n1
  ...  (1)n
 ...   (1)n
(all real x)
3! 5!
(2n  1)!
(2n  1)!
n 0
2n
2n

x2 x4
n x
n x
cos x  1    ...  (1)
 ...   (1)
(all real x)
2! 4!
(2n)!
(2n)!
n 0

1
2
3
4
n
 1  x  x  x  x  ...  x  ...   x n (1  x  1)
1 x
n 0

1
 1  x  x 2  x3  ...  ( x) n 1  ...   (1) n x n (  1  x  1)
1 x
n 0
n
n

x 2 x3
n 1 x
n 1 x
ln(1  x)  x    ...  (1)
 ...   (1)
(1  x  1)
2 3
n
n
n 0
arctan x  x 
2 n 1
2 n 1

x3 x5
n ( x)
n ( x)
  ...   1
 ...    1
3 5
2n  1
2n  1
n 0
(1  x  1)
Example 15.
Find a power series representation for cos(2 x) .
We will use the power series for cos x , substituting 2 x in place of x .
cos 2 x  1 
 1
(2 x) 2 (2 x) 4
(2 x) 2 n

 ...  (1) n
 ...
2!
4!
(2n)!
4 x 2 16 x 4
4n x 2 n

 ...  (1) n
 ...
2!
4!
(2n)!
Taylor Series Formula with Remainder
We can write any Taylor Series as a two-part definition. It is the nth-degree
Taylor polynomial plus any remainder after the nth-degree term. The remainder is
actually the error that occurs when using a Taylor polynomial to approximate a function.
We are able to connect both the polynomial and the error involved in using the Taylor
polynomial as an approximation for the series.
Taylor's Formula with Remainder
Let f ( x) have all order of derivatives in an interval between a and x. If x is any number
in the interval, then there is a number c between a and x such that
f ( x)  f (a)  f (a )( x  a ) 
f (a )
f (a )
f  n  (a )
2
3
( x  a) 
( x  a)  ... 
( x  a) n
2!
3!
n!
 Rn ( x)
f  n 1 (c)
where Rn ( x) 
( x  a ) n 1
(n  1)!
The function Rn ( x) is the error term or the remainder of order n. Some texts refer to this
as the Lagrange form of the remainder which can be used to find the Lagrange error
bounds.
This formula is especially useful if we are working with an alternating convergent
series. Since the remaining terms converge and always switch between positive and
negative values, the remainder will always be less that the value of the (n+1)th term.
Example 16.
Use the first three nonzero terms to estimate the value of sin(0.1) . Estimate the
error bound in the approximation.
Using only the first three terms means we want a third order Taylor polynomial
x3 x5
which gives us sin x  P3 ( x)  x   . Evaluating the polynomial at x = 0.1, we get
3! 5!
3
5
0.1  0.1

sin(0.1)  P3 (0.1)   0.1 

0.0998334
3!
5!
x7
The next term in the series, the error bound is less than
evaluated at x  0.1 . The
7!
maximum Error Bound is less than 1.98 1011 .
Example 17.
Use a power series to find
1
 sin( x ) dx correct to four decimal places.
2
0
Determine
the maximum error bound.
We aren't told how many terms to use. For convenience, let's use the first four
non-zero terms to see what happens. If needed, we can take additional terms to improve
We can the power series representation for sin(x) to write
x   x   x 

2
 
sin x 2  P4 ( x 2 )  x 2
6
10
3
2
3!
5
5!
2
7
7!
14
x
x
x


3! 5! 7!
3
5
7
2
2
2


x
x
x
1
1
 dx
  sin x 2 dx    x 2 


0
0
3!
5!
7! 


 x2 
 
     
1
 x3 x7
x11
x15 
  


 x 42 1320 75, 600  0
1
1 1
1
1
  sin x 2 dx   

0
3 42 1320 75, 600
 
2867
 0.31028 . The answer to
9240
1
or 0.000013 .
four decimal places is 0.3103. The error bound is less than
75, 600
Summing the first three terms of the answer, we get
Power series can give very accurate approximations within just a very few terms.
To reach the same degree of accuracy in Example 17 using the Riemann Sum would
require dividing the interval into many, many subintervals which would require more
calculations than we did.
Tests for Convergence
Power series are useful when we know they converge. They allow to us decide the
degree of accuracy we want. Series that do not converge are of little use to us. We need
to be able to check for convergence for any functions that can be represented by power
series. This section will explain and illustrate the most useful tests. There is no special
order to the presentation of the tests. We are showing them in a convenient manner.
The Integral Test
Consider the series

1 1 1 1 1 1
 3  3  3  3  3  ... .

3
1 2 3 4 5
n 1 n
There is no convenient formula that gives the sum like the Geometric Series formula did.
We can generate a table of values to see if convergence seems likely.
n 5
50
500
5000
Sn 1.1857… 1.2019… 1.2020549… 1.2020569…
We remember that a table is not proof of convergence, but it seems reasonable in this
case that convergence occurs in this series. We still need to verify the result analytically.
We will use the Integral Test to do this.
The Integral Test for Convergence
Suppose that f is a continuous, positive, decreasing function on the interval [1, ),
and let an  f (n) . Then the series defined by
improper integral
If




1


1

a
n 1
n
is convergent if and only if the
f ( x) dx is convergent. In other words,

f ( x) dx is convergent, then
a
n 1
n
is convergent.
or
If
1

f ( x) dx is divergent, then
a
n 1
n
is divergent.
We apply the Integral Test to our example. The improper integral associated with
 1
1
is  3 dx, which we know converges by the p-series rule we developed in the

3
1 x
n 1 n

1 1 1 1 1 1
last chapter. Therefore, the series  3  3  3  3  3  3  ... converges.
1 2 3 4 5
n 1 n

Once again we raise the warning. Although we know the series converges, we
don't know the actual value for the limit of the series.
As we learned in the review of the p-series rule, any improper integral in the form

1
dx converged if p  1 and diverged if p  1 . The Integral Test for Convergence
xp
allows us to formalize the rule for series.

1
p-series Test for Convergence

The p-series
1
n
n 1
p
converges if p  1 and diverges if p  1 .
Example 18.

Does
5
 ( x  3)
n4
2
converge?
Beginning the sequence at n=4 is okay. If a series converges on the interval
[1, ] , it will surely converge on some smaller interval that starts to the right of n = 1. It
is not really necessary for f to be always decreasing as long as f is ultimately decreasing
on the infinite interval. This example fits the p-series format. Since p  1 , we can safely
say the series will converge.
The next test is used to confirm divergence.
The nth Term Test for Divergence
The nth Term Test for Divergence

The series
a
n 1
n
diverges if lim an is not equal to zero of fails to exist.
n 
This statement is logical. If the terms of a sequence do not converge, it is impossible for
the sequence of partial sums generated by that sequence to converge. Usually, we check
for convergence, not divergence. But if we know that a series is clearly divergent, we do
not have to apply any of the tests for convergence.
Example 19.

Verify that
 ln n diverges.
n 1
As n gets larger, ln( n) increases without bound. Since the sequence does not limit
to zero, the associated series must diverge.
Direct Comparison Test
The Direct Comparison Test for series is similar to the Direct Comparison Test for
improper integrals we studied in the last chapter.
The Direct Comparison Test for Series
Let
a
n
be a series with no negative terms
Case 1. If an  cn for all n  N and if
converges.
c
converges, then
Case 2. If an  cn for all n  N and if
diverges.
c
diverges, then
As before, if the series
c
n
n
n
a
n
a
n
also
also
converges to a limit, it &quot;squeezes&quot; all series smaller
that it to the same limit. If the series
away from a limit.
c
n
diverges, it &quot;pushes&quot; all series that are larger
Example 20.

Show that
xn
converges for all real values of x.

n 0 (n  1)!
There are no negative terms in the series and we know that
xn
xn
 . We
(n  1)! n!

recognize
xn
as the Taylor Series for e x , which we know converges. Therefore,

n
!
n 0

xn
also converges by the Direct Comparison Test.

n 0 (n  1)!
Test Tip. When verifying convergence of a series, always state the test you are
using. It helps the reader of the question determine if you understand what you are doing.
Absolute Convergence
 a  a  a  a  ...  a  ... converges, then  a
converge and  a is said to converge absolutely.
If the series
n
1
n
The Ratio Test
2
3
n
n
must also
The Ratio Test is one of the most powerful tests we have to check convergence of series.
It is often use in conjunction with absolute convergence.
The following form of the Ratio Test can be used to check for absolute convergence.
The Ratio Test for Absolute Convergence
Let
an 1
 L.
n  a
n
Case 1. If L &lt; 1, the series is absolutely convergent.
a
n
be a series with positive terms and let lim
Case 2. If L &gt;1, the series is divergent.
Case 3. If L = 1, the test is inconclusive. Apply a different test.
Example 21.
n2  3
is convergent.
3n
n 1

Determine if the series

We see that for all values of n, each term will be positive. We will apply the
Ratio Test. We will take this explanation slowly, so you can follow the algebra.
n2  3
an  n
3
an 1
 n  1

2
n 1
3
3
Find the limit of the ratio. We did not use absolute value signs here, since all terms were
 n  1
2
3
n 1
lim
n 
3
n2  3
3n
  n  12  3 n 2  3 
 lim 
 n 
n  
3n 1
3 


  n  12  3 3n 

 lim 
 2
n  
3n 1
n 3


2
1  n  2n  4 
 lim 

2
n  3
 n 3 
1
1
1 
3
3
Since L &lt; 1, the series converges absolutely, so it is convergent.

The Limit Comparison Test
We found the p-series Test to be very useful in checking convergence of series in the
1
form  p . Unfortunately, not all series fit this pattern exactly. We can test series with
n
p-series of similar dimension. We use the Limit Comparison Test to do this.
The Limit Comparison Test (LCT)
Let an  0 and bn  0 for all n  N .
an
 c, where c  0, then
bn
diverge.
Case 1. If lim
n 
an
 0 and
n  b
n
Case 2. If lim
an
  and
n  b
n
Case 3. If lim
b
n
b
n
a
n
and  bn both converge or both
converges, then
diverges, then
a
n
a
n
converges.
diverges.
The Limit Comparison Test is not quite the same as the Direct Comparison Test. That
test compares one series with a bounding series. The Limit Comparison Test checks the
limit of the ratio between two series.
Example 22.
Determine if

2 5 8
3n  1
   ...  
2
9 16 25
n 1  n  2 
All the terms are positive and we see that the function is somewhat like a p-series.
3n  1 3n 3
1
As n grows very large,
 2   3   . So we will compare our series with
2
 n  2 n n  n 
1
n.
the series
3n  1
 n  2
L  lim
1
n
3n  1
n 
 lim
n 
2
 n  2
2
 lim
n 
3n  1
 n  2
2

1
n
n
3n 2  n
  lim 2
3
1 n n  4n  4

Since the limit is greater than one and the series 
1
1
diverges, the series
n


n 1
3n  1
 n  2
must also diverge.
Example 23.

1 1
1
1
Check the convergence of    ...   n
2 14 62
n 1 4  2

Ignoring the constant in the denominator, we see that
4
n 1

similarly to
1
4
n 1
n
n
1
behaves
2
. We will try the LCT.
1
1
4n
4

2
L  lim
 lim n

n 
n  4  2 1
1
4n
4n
1
 lim n
 lim
1
n  4  2
n 
2
1 n
4

1
The limit is a constant and  n converges, so
n 1 4
n
The Alternating Series Test

4
n 1
n
1
converges.
2
2
When the terms of the sequence alternate between positive and negative values, we can
use the Alternating Series Test to check for convergence.
The Alternating Series Test

If the series
  1
n 1
n 1
an  a1  a2  a3  a4  ...
satisfies all three of the following conditions:
1.
2.
3.
Each term an is positive
an  an1 for all n  N
lim an  0
n 
then the series is convergent.
We can visualize the reason this test works. Suppose we have the alternating
1 1 1 1
1
series     ... . As the figure shows, the first sum moves to right , then
2 3 4 5
2
1
1
1
moves back , then moves right , back , and so on. As this process continues
3
4
5
forever, the sums will congregate about a specific value.
&lt;INSERT FIG 4, Alternating series going back and forth&gt;
We can also see this in a series plot.
&lt;insert figure 5&gt;
Figure 5
Example 24.

Determine if
  1
n 1
n 1
3n
converges or diverges.
4n  2
This is an alternating series, so we try the Alternating Series Test. Checking each
condition, we get
1.
2.
3.
Are all terms positive? - yes
Is an  an1 ? - yes
3n
 0 ? - no, the limit is 3 . By the nth term test, we see
Does lim
4
n  4n  2
the series diverges.
Example 25.

Does
  1
n 1
n 1
n3
converge or diverge? Show your reasoning.
n4  1
You should recognize this as an alternating series. The Alternating Series Test is
reasonable to use here. We check each of the conditions
1.
Are all terms positive? - yes
2.
Is an  an1 ? - This is difficult to tell by inspection. So we get sneaky.
x3
. If this function decreases, we
x4  1
know the sequence also decreases. We can check for decreasing function by looking
We consider the related function f ( x) 
at the derivative of f ( x).
f ( x) 
3x 2  x6
x
4

1
2
.
The denominator is always positive. We need to see when 3x 2  x 6  0. Factoring, we get
x 2 (3  x 4 ). x 2 is always positive. (3  x 4 )  0  x  4 3. Therefore, for all n  2, f ( x) is decreasing.
The answer to the second condition is yes.
n3
3.
Is lim 4
 0 ? -- yes.
n  n  1
Since all three conditions have been met, the series is convergent by the Alternating
Series Test.
Intervals of Convergence on Power Series
In the list of power series representation for seven common functions, we also listed the
interval of convergence. In this section we want to see how those intervals relate to a
specific power series and whether the interval is open or closed. We will demonstrate a
series of steps you can follow to get the solution. This method uses the Ratio Test.
The Ratio Test for Absolute Convergence
Let
an 1
 L.
n  a
n
Case 1.: If L &lt; 1, the series is absolutely convergent.
a
n
be a series with positive terms and let lim
Case 2. If L &gt;1, the series is divergent.
Case 3. If L = 1, the test is inconclusive. Apply a different test.
Notice that the series converges when the limit is less than one, diverges when the
limit is greater than one, and is inconclusive when the limit is equal to one.
How to Test a Power Series for Convergence
a
1.
Use the Ratio Test to find lim n 1
n  a
n
2.
Set the absolute value of the limit less than 1 and solve for the open interval.
3.
Evaluate the series when x equals the right endpoint of the open interval. If
this is a finite value, the series converges at the right endpoint. Otherwise, the
interval is open on the right side.
4.
Evaluate the series when x equals the left endpoint of the open interval. If this
is a finite value, the series converges at the left endpoint. Otherwise, the
interval is open on the left side.
Example 26.

Given the series
n
 5  x  2
n 1
convergence.
Use the Ratio Test.
n
an  n ( x  2) n
5
n 1
an 1  n 1 ( x  2) n 1
5
Find the limit.
n
n
, find the interval of convergence and the radius of
n 1
( x  2) n 1
n 1
an 1
5
L  lim
 lim
n  a
n 
n
n
( x  2) n
5n
  n  1  ( x  2) n 1  

5n
n 1
L  lim 
 lim
( x  2)
  
n 1
n 
n 
n

5
5n

  (n)( x  2) 
L  x  2 lim
n 
L
n 1
5n
(Remember that n is the variable here. x - 2 is a constant)
1
x2
5
Set the absolute value of the limit less than 1.
1
x  2 1
5
 5  x  2  5
 3  x  7 as the interval of convergence.
We still need to check the endpoints of the interval for convergence. We
substitute these values into the series and check for convergence.
Case 1. Check right endpoint x = 7. When x = 7, the series becomes

n
n
7  2

n 
n 1 5

n
n
5
n  
n 1 5


  n, which does not converge
n 1
Case 2. Check left endpoint x = -3. When x = -3, the series becomes

n
n
3  2 

n 
n 1 5

n
n
5
n 
n 1 5


  (1) n n, which does not converge
n 1
Since the power series does not converge at either endpoint, the interval of
convergence is 3  x  7 . The radius of convergence is the distance from the center of
the interval to one endpoint. In this example the center is at x  2 . The distance from
x  2 to x  7 is 5. So, 5 is the radius of convergence centered at x  2 .
Example 27.

Find the interval of convergence for 
n 1
 4x
n!
n
.
Find the limit using the Ratio Test.
n
4x

an 
n!
an 1
 4x

n 1
(n  1)!
 4x
L  lim
n 
n 1
(n  1)!
 4x
n
  4 x n 1   n ! 


 lim 
n
n   ( n  1)!  

4
x



 

n!
L  lim
n 
 4x
n 1
0
Since the limit is zero, the series converges absolutely for all values of x. The
interval of convergence is  ,   . It doesn't make sense to talk about the radius of
convergence for this interval.
Example 28.
x 
Find the interval of convergence for the series defined by   1
n

n 1
2
n
n 1
Use the Ratio Test to establish the limit, then check the limit against the intervals
of convergence.
an
x 

an 1
x 

2
n
n
2
n 1
n 1
 
 x 2 n 1

L  lim  n  1n
n 
2
 x

 n
 


 2 n 1  
  lim  x
 n
 n  n  1   2


  x


 
 


n


  n 
2
L  lim  x 2 
  x
n 
n

1

 
The series will converge when x 2  1  1  x  1 . Next we check endpoints for
convergence.
Case 1. x = 1. When x = 1, the series becomes
1 
  1 n

n 1
2
n
n 1


 1
n 1
n
This series converges by the Alternating Series Test.
n 1
Case 2. x = -1. When x = -1, the series becomes

  1
 1
n 1 
n
n 1


 1

2 n
n 1
n
This series also converges by the Alternating Series Test.
n 1
Therefore, the interval of convergence is [1,1] . The radius of convergence is 1.
Strategies for Testing Series
We have reviewed several tests for verifying convergence and divergence of series. The
difficult job you have as a student is deciding which one to use. There are no specific
rules to follow, but the following guidelines can help. As you look at the strategies, don’t
try each test until you find one that works. Instead, try to classify the series according to
its form. Then find the guideline that fits the form.
1.
If you can see quickly that lim an  0 , check for divergence.
2.
If the series fits the form
n 
1
n
, it is a p-series. If p  1 , the series
p
converges. If p  1 , the series diverges.
3.
If the series fits the form
Series Test.
If the series fits the form
4.
  1 a
n
n
 ar
n
or
  -1
n 1
an , then try the Alternating
or  ar n1 , then it is a geometric series. If
r  1 , the series converges. If r  1 , the series diverges.
If a series that has factorials or other products, try the Ratio Test.
If a series fits a form similar to a p-series or geometric series, then try one of
the comparison tests (Direct or Limit). If an is a rational function or an
algebraic function of n, try to compare it with a p-series. The comparison
tests apply to series with positively valued terms. If any terms are negative,
we can use a comparison test with  an
5.
6.
If there is a function f ( x) that corresponds to an  f (n) , then try the Integral
7.
Test on the improper integral


1
f ( x) dx if the antiderivative is easy to do.
Not all situations can be described in exactly one of these guidelines. In some ways, it is
like factoring or integrating. You may need to do some algebra manipulations on the
expressions to write them in a more convenient form. Don’t get too stressed about these
rules. The questions on the AP Exam tend to be fairly standard. You are being tested to
see if you know the various tests for series, not if you know some obscure tricks that just
happen to make the problem work.
For the following examples, indicate which test would be most appropriate. Do not work
the problems out.
Example 29.

5n 2  3

2
n 1 8n  3n
5n 2  3 5

 0 , check for divergence.
n  8n  3n 2
3
Since the lim
Example 30.
n3  3

2
n 1 4 n  8n  7

4
an is an algebraic function; we can compare it to a convenient p-series. But first
n3  3
n4
1
 lim 2  lim 5 . The comparison
we need to manipulate it. lim  2
n 
n

n

4n
4n 4
n 1 4n  8n  7
1
series should be bn  5 .
n4

3
4
Example 31.

  1
3n2  5
en
n
n 1
This is an alternating series; use the Alternating Series Test.
Example 32.

3n

n 1 (2n)!
The presence of factorials suggests that we try the Ratio Test.
Example 33.

n
2
 
sin n3
n 1
This series fits very nicely with


1
x 2 sin( x 3 ) dx; use the Integral Test.
Example 34.

7
 6  13
n 1
n
This series is a rational function that is similar to a p-series. Use the Direct
Comparison Test.
Now You Try It
Determine if the sequence converges or diverges. If the sequence is convergent, give the
limit
1.
 ln(n3  5) 


n


2.
an 
3.
n
2
   3  
n

4.
1

bn   4  
n

7n
n7
2n
Check the following series for convergence. If the series contains negative terms,
determine if it is absolutely convergent, conditionally convergent, or divergent.

5.
1
(n  1)(n  5)

1

6.
 3n  8
 (n  2)
n 1

7.
2
n 0
n

8.
n
 n  1!
 ln(n  1)
n 1

10.
3
1
5
 3

 
4
n 0 

9.
2
n2

n
n 1 2

11.


cos 2 n
n
n 1

12.
2

3
3
cos n
 1 n
n 1
2

13.
n2

n
n 1 3
14.


n 1
1
 5n  2 

15.
n
n 3

16.
n 1
1
 ln n 
n!
5
n
2
x5n

17.
4
  1
n 1
 3x  2 
n
n2
n 1

18.
 n! x
n
n 1
Write the first four terms and the general term for a Maclaurin Series representation for
the following functions.
1
1  x3
19.
f ( x) 
20.
g ( x)  cos x 2
21.
f ( x)  ln(1  3x)
22.
h( x)  xe x
23.
f ( x)  arctan(4 x)
 
2
2
1  3x
24.
t ( x) 
25.
g ( x) 
1
4  x
2
Find the first four nonzero terms and the general term for the Taylor Series generated by
f ( x) at x  a.
26.
f ( x)  x 4  2 x3  4 x 2  5x  3 at a  2
27.
f ( x)  x 2 at a  4
28.
f ( x)  ln x at x  2
29.
f ( x)  cos x at x 
3

3
30.
Let P4 ( x)  8  2( x  3)  7( x  3) 2  14( x  3)3  5( x  3) 4 be the fourth-order
Taylor polynomial for the function f ( x) centered at x  3 .
(a)
Find f (3) and f (3)
(b)
Write the third-order Taylor polynomial for f ( 3) and use it to
approximate f (2.8) .
(c)
Is it possible to find the exact value of f (4) from the information given.
Explain.
(d)
x
Let g ( x)   f (t ) dt. Write the fourth-order Taylor polynomial
3
representation for g ( x ) at x = -3.
1
x4
(a)
Write the first four terms and the general term of the Taylor Series
generated by f ( x) at x  5 .
(b)
Use the results from part (a) to write the first four terms and general term
for ln( x  4) at x  5 .
31.
Let f ( x) 

x2
4
Let f ( x)  e .
(a)
Write the first four terms and the general term of the Taylor Series
generated by f ( x) at x  0 .
(b)
What is the interval of convergence of the series found in part (a). Show
32.
33.
Let f be a function with derivatives of all orders with the following properties:
f (6)  1, f (6)  3, f ( x)  2, and f (6)  18 .
(a)
Write a third-order Taylor polynomial for f at x = 6.
(b)
Use the polynomial found in part (a) to approximate f(6.15).
(c)
What is the second-order Taylor polynomial for f '( x) at x = 6?
x2
. For what values of x
2!
will the error bound be less than 4  10 6 . Explain your reasoning.
34.
The first two terms for the power series for cos x are 1-
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