§ 11.1 Sequences
A sequence is an ordered arrangement of numbers, one for each positive integer. An infinite sequence is a function
whose domain is the set of positive integers.
A sequence can be specified explicitly or recursively. For our purposes, an explicit formula is preferable.
Examples:
1.
1
an  1  , n  1
n
2.
bn  1  (1) n
3.
1
n
c n   1  , n  1
n
4.
d n  .999, n  1
1
,n 1
n
Convergence: we say that a sequence converges to L , or has limit L , and write lim a n  L , if for each
n 
there is a corresponding
M  0 such that an  L  
whenever
  0,
nM.
Sequences that have limits are said to converge, sequences that fail to have limits diverge.
Consider the 4 examples above, which converge? Diverge?
Theorem: If lim f  x   L (x
x 
Î ¡ ) , then lim f n   L (n Î + ¢ )
n 
This theorem allows us to use L’Hopital’s rule in difficult cases to determine the convergence/divergence of a
sequence.
Theorem: If
a n  is a sequence such that f n  an
(the nth term), then lim a n  L .
n 
Limit laws for sequences (p. 696)
Squeeze Theorem: Suppose
a n  and cn  both converge to
L.
Example:
Prove:
sin 3 n
0
n 
n
lim
L and an  bn  cn . Then bn  also converges to
This leads to the following theorem: If
lim an  0 , then lim a n  0 .
n 
n
Examples: Determine if the sequence converges/diverges. If it converges, find lim a n .
n
1.
4.
an 
n
2n  1
a n   1
n
n
n 1
ln n
7. a n 
n
10. an =
(-
2.
an 
4n 2  1
n 2  2n  3
 n 
sin 

2 

5. a n 
n

8. a n  1 

1

n
3.
an 
n4
2n 2  1
6.
an 
en
n2
9.
an =
en
2n
n
p )n
4n
A sequence is monotonic if its terms are non-decreasing,
a1  a2  a3  ....  an , or non-increasing,
a1  a2  a3  ....  an .
In order to determine whether a sequence is monotonic, you must examine the relationship between
a n and a n 1
2
Example: Determine whether the sequence
an   nn
2
is monotonic.
Sometimes it is helpful to use math induction to show monotonicity.
A sequence is bounded above if there is a number M such that
an  M for all n . The number M is the upper
bound of the sequence.
A sequence is bounded below if there is a number
N
such that
a n  N for all n . The number N is the lower
bound of the sequence.
A sequence is bounded if it has an upper & lower bound.
Consider the original 4 sequences again. What are the upper/lower bounds for each?
Theorem: If a sequence is bounded and monotonic, then the sequence converges.
an  1 
Prove:
1 1
1
  ...  converges using the theorem above.
2! 3!
n!
§ 11.2 Infinite Series

a
n
 a1  a 2  a3  ...  a n  ... the sum of an infinite sequence is an infinite series.
n 1
To find the sum of an infinite series (the limit) we examine the sequence of partial sums
1 1 1
1
1
+ + +
+
+ ...an
2 4 8 16 32
Consider the series
The first partial sum is the sum of the first term. The second partial sum is the sum of the first 2
terms. The nth partial sum is the sum of the first n terms.
S1 
1
2
S3 =
S2 =
What is S n ?
S4 =
What is lim S n ?
n 

The infinite series
a
n
converges and has a sum S , if the sequence of partial sums S n 
n 1
converges to S . If S n  diverges, the series diverges.
Special Infinite Series
Telescoping Series: b1  b2   b2  b3   b3  b4   ...

Example:
 n  2n  3
1
n 1
fractions

  n  2  n  3 
n 1
1
1
HINT:
decompose
using
partial
Theorem: The sum of a telescoping series is S n  b1  lim bn 1 .
n 
Note: This is NOT the first term minus the limit of the last term. You must write out the first
few terms to decide which terms cancel and which terms stay.

Find the sum: a)

n 1

b)

k 1
1 
1
 

 n n 1
 3
3 



2
2 
 k  1
k


æ1
ö
1 ÷
çç ÷
ç
ø
n + 2÷
n = 1 èn
¥
c)
å

Geometric Series:
 ar
n 1
 a  ar  ar 2  ...  ar n  ...
a0
n 1
Recall from previous coursework, that a geometric series converges if r  1 , it diverges if r  1
, and S n 
a
.
1 r
Examples: Tell whether the geometric series converges or diverges. If it converges, find the
sum.
1.
4 4 4
4
 
  ...
3 9 27 81
1 1 1
1
  ...
2. 1   
3 9 27 81
3.
4 16 64 256
 

 ...
3 9 27 81
4. .51515151...
Properties of infinite series: Let
a
n
A ,
b
n
B , and c Î ¡ , then



1.
ca n  cA
2.
Find

n 1

3.
n 1
n 1

 an  bn   A  B
 a
n
 bn   A  B
n 1
n
  1 n
1 
3   2  
 5  
  4 

 n  1  2  3  4  5  ...
Harmonic Series:
1
1
1
1
1
n 1
We know the harmonic sequence converges to 0, but what about the harmonic series?
1
We know that the harmonic sequence is a subset of the function f  x   . Extend that.
x
The sum of the harmonic sequence is a subset of the sum of the values of the function f  x  
How do we find that sum?

b
1
1
dx  lim
dx  lim ln x
b  1 x
b 
1 x


1
So does its subset  .
n
b
1
1
.
x
 lim ln b  ln 1   . Therefore the sum diverges.
b 

n 1
We can now utilize the fact that the harmonic series diverges in other comparisons.
Nth term test for divergence: For sequences, lim a n  0 is the limit of the sequence. But for
n 
series, it is not the nth term that gives us the limit, but the limit of S n . The nth term can tell us,
however, if the series diverges. Unfortunately, it cannot tell us if a series converges.

For a series
a
n
, if lim a n  0 the series diverges. If lim a n  0 , the series may or may not
n 
n 1
converge.

Examples:
1.

n 1
4
 
3
n

2.
2
n 1

3.
  n 
1
n 1
Use the nth term test to determine if the series diverges:
n 

1.

n 1

2.
n3
n
n
2
n 1

3.
  n  n  2 
1
1
n 1

4.
 cos n
n 1
§ 11.3 The Integral Test & P-Series
1
Suppose you wish to sum å . 2
i= 1 i
n® ¥
1 1 1
1
1
åi = 1 i 2 = 1 + 22 + 32 + ... + n 2 + ...
¥
1
Now consider the curve y = 2 . The area under this curve is
x
¥
1
òx
2
dx .
1
We know that the area of the curve can be approximated with rectangles. Take rectangles of width 1
inscribed under the
curve.
What is the height of the 1st rectangle? 2nd rectangle? 3rd rectangle? Write a Riemann sum to express the
sum of these areas.
We know that the area of these inscribed rectangles (our series) is less than the area under the curve.
Do you think this is true for all series and their corresponding functions?
The Integral Test: If f x  is positive, continuous, and decreasing for x  1 and a n  f n  , then

a
n
n 1

 f xdx converge together or diverge together.
and
In order to use the integral test, you must first
1
SHOW that the conditions are met.
Apply the integral test to:

1.

n 1

1
n
2.

n 1

1
n3
3.

n 1
1
n
Each of these examples is a power series or p-series. The general form of a p-series is

n
1
n 1
p

1
1
1
 p  p  ... , where p is any positive constant.
p
1
2
3
Note: If p  1 , then the p-series converges; if 0  p  1 , then the p-series diverges.
¥
If
å
an has been found to converge by the INTEGRAL TEST using f(x), then we can say the series
n=1
converges to a sum S. The remainder or truncation error, Rn , between the actual series sum S and the nth
partial sum Sn is bounded below by
ò
¥
f (x )dx and bounded above by
n+1
ò
¥
n+1
f (x )dx £ R n £
ò
¥
f (x )dx . So Rn = S – Sn
n
ò
¥
f (x )dx
n
Since R N is itself an infinite series, which is as difficult to evaluate as the original series, you might think
that there would be no advantage to finding R N . But sometimes R N can indicate an estimate of L that is
closer to L than S N .
¥
Example 1: Find the 6th partial sum of
1
, then estimate the remainder or error of this sum.
4
n=1 n
å
¥
1
within 0.01?
n=1 n + 1
å
Example 2: How many terms are necessary to find the sum of
2
§ 11.4 Comparison Tests
Direct (ordinary) Comparison Test: Let 0  a n  bn for all n .

a. If
b

n
converges, then
n 1
a
n
also converges.
n 1

b. If
a

n
diverges, then
n 1
b
n
also diverges.
n 1
In other words, if each term of an unknown series is less than the corresponding term of a
convergent series, then the unknown series converges.
If each term of an unknown series is greater than the corresponding term of a divergent series,
then the unknown series diverges

Determine the convergence of
23
1
n
n 1
3 n  3 n  2 for all n  1.
1
1
 n
(reciprocal property)
n
3
3 2

1
1
converges because it is a geometric series with r   1
n
3
3
n 1
SOLUTION:


By the direct comparison test
23
1
n
also converges.
n 1

Practice:

n 1
1
2
n 1


n 1
n
5n  4
2
a
Limit Comparison Test: Suppose that given an  0 , bn  0 and lim  n
n  b
 n
and positive. Then the two series
a
n
and


  L , where L is finite

bn either both converge or both diverge. (The
trick is to look at the dominant terms and find a simple, similar series to compare with)
The Limit Comparison Test works well with “messy algebraic” series compared with p-series.
Find the dominant term in the numerator and denominator, set up the ratio, reduce and use this as
a comparison.
¥
5n 2 - 2n + 3
Determine the convergence of å
.
3
2
n = 1 7n + 2n + 1
SOLUTION: The dominating term in the numerator is n 2 , the dominating term in the
n2
1
denominator is n 3 , so compare to å 3 = å
.
n
n
5n 2 - 2n + 3
3
2
5n 2 - 2n + 3
5n 3 - 2n 2 + 3n
lim 7n + 2n + 1 Þ lim 3
·
n
Þ
lim
=
1
n® ¥
n ® ¥ 7n + 2n 2 + 1
n ® ¥ 7n 3 + 2n 2 + 1
n
¥
1
This is positive and finite. Since å
is the harmonic series and divergent, å
n
n=1
is also divergent by the Limit Comparison Test.
¥
Practice:
å
n=1


n 1
n 2  10
4n 5  n 3
1
3n - 4n + 5
2
5
7
5n 2 - 2n + 3
7n 3 + 2n 2 + 1
¥
1
å
2
n +1
n=1
¥
1
å
n=1
n n2 + 1
Practice – Tell whether each series converges or diverges:
¥ æ2 ön
ç ÷
÷
n = 0 çè 3 ø
¥
1
k = 4 k 1.001
2. å
1. å
3. 1 +
5.
1
1
1
+
+
+ ...
2
3
4
1 1
1
1
1
+ +
+
+
+ ...
2 5 10 17 26
4. 1 +
1
2 2
+
1
3 3
+
1
4 4
+ ...
¥
6. Find a good upper bound for the error in using the first 5 terms of the convergent series å
n2
n = 1 (n + 1)2
.
§ 11.5 Alternating Series
An alternating series is of the form a1  a2  a3  a4  ... OR  a1  a2  a3  a4  ...

Alternating Series Test: Given an alternating series
  1
n 1
a n , the series converges if both of
n 1
the following conditions are met:
2. lim a n  0
1. an ³ an + 1 > 0
n 
If either condition is not met, the series diverges. Moreover, the error made by using the nth
partial sum, Sn, to estimate the sum, S, is not more than an + 1 .
Examples:
1. Does the series 1 
1 1 1
   ... converge or diverge?
2 3 4

Solution: First, write the series in the form
  1
n 1

an 
n 1
whether the two conditions are met. (1) Is
  1
n 1
n 1
1
 .
n
Then decide
1
1
³
? Cross multiply - n 1  n ?
n
n+1
1
 0?
n  n
(2) lim
(1) & (2) are both true, therefore the series converges.
harmonic series.
2. Do the following series converge or diverge?
This is called the alternating

A)
  1


1


 ln n  1 
n 1
n 1

B)
  1
n 1
n 1
 n 


 10n  1 
¥
C)
å
n- 1
(- 1)
n=1
æln n ö
÷
ç
÷
÷
çè n ø

WS conclusion: If
  1
n 1
R n £ an + 1
n 1
a n is an alternating series with an ³ an + 1 > 0 and lim a n  0 , then
n 
§ 11.6 Absolute and Conditional Convergence and The Ratio & Root Tests
An alternating series can be classified as absolutely convergent, conditionally convergent, or
divergent.
If
a
n
converges, then
a

Example:
Show that
n
converges ABSOLUTELY.
  1
n 1
n 1
If the absolute value (å a n
 1 
 3  is absolutely convergent.
n 
) does not converge but the alternating version does, then the series
is conditionally convergent.

  1
 1 

 is conditionally convergent (HINT: you must show
3
n

1


n 1
that it is not absolutely convergent and then use the alternating series test to show that it is
convergent)
Example:
Show that

Show that
  1
n 1
n 1
n 1
 n 
 2
 is convergent and then estimate the error made by using the
 n 1
partial sum S 9 as an approximation of the sum.
Is each of the following absolutely convergent, conditionally convergent, or divergent?
n æ1 ö
(- 1) ççç ÷
÷
÷
è nø
n=1
¥
A)
å
¥
B)
å
n+1
(- 1)
n=1
¥
C)
å
n+1
(- 1)
n=1

D)
  1
n 1
n 1
æ 1
çç 2
çè3n +
ö
÷
÷
÷
1ø
æ2n 2 - 1ö
÷
çç
÷
çè 1 - n 2 ÷
ø
 cos n 


 n 
¥
The Ratio Test: If
å
an is a series with non-zero terms, then you perform the ratio test by
n=1
finding
lim
n® ¥
an + 1
= R
an
1) If R  1 , then the series converges absolutely
2) If R  1 , then the series diverges
3) If R  1 , then the test is inconclusive
¥
2n
Examples: å
n=1 n !
¥
nn
ån = 1 n !
¥
å
n=1
n 2 (2n + 1 )
3n
¥
The Root Test: If
å
an is a series, then you perform the root test by finding lim n an = M
n=1
1) If M  1 , then the series converges
2) If M  1 , then the series diverges
3) If M  1 , then the test is inconclusive

Examples:

n 1
¥
n
æln n ö
ån = 1 ççè 2n ø÷÷÷
e 2n
nn
n® ¥


n 1


n 1
n3
3n
 n 


 2n  1 
n