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Chapter 6 Overview: Sequences and Numerical Series In most texts, the topic of sequences and series appears, at first, to be a side topic. There are almost no derivatives or integrals (which is what most students think all Calculus is comprised of) and those that do appear do so at the end of the chapter. The sense of disconnectedness is heightened by the fact that most Calculus students have not seen series since freshman year (and, then, only briefly). For the purposes of AP, this topic is broken into four basic subtopics:

• Numerical sequences and series • Radius and interval of convergence for a Power Series • Taylor polynomials • Creating a new series from an old one

This is the order most texts use, and all topics are in one chapter, which can be overwhelming. The final subtopic is where the derivatives and integrals reappear, built on the theoretical foundations laid by the first three. We are opting to break the topic of sequences and series into two chapters. This chapter covers convergence and divergence of numerical sequences and series. The other chapter will cover the material related to Power Series and occurs at the end of the curriculum. What we will consider in Calculus is the issue of whether a sequence or series converges or not. There are several processes to test if a series is convergent or not. There are seven Tests to prove convergence or divergence of a numerical series.

1. The Divergence (nth Term) Test 2. The Comparison Test 3. The Limit Comparison Test 4. The Integral Test 5. The Ratio Test 6. The nth Root Test 7. The Alternating Series Test

Each may be easier or more difficult depending on the series being tested. Obviously, we will not cover #4 in this chapter, since we have not learned Integration yet. The rest rely on Limits-at-Infinity, and they provide a good

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context within which to practice that material. Most of these Tests apply to series comprised of positive values. The Alternating Series Test applies to series in which the terms alternate signs. Though each section introduces a new test, the homework in each section will apply to all the tests that were introduced previously.

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6.1: Sequences—Algebra Review Let’s have a quick review from Algebra 2: A sequence is simply a list of numbers. Sequences are only interesting to a mathematician when there is a pattern within the numbers. For example:

1, 3, 5, 7, …

14, 7, 0, -7, -14, …

1, 3, 9, 27, …

4, 2, 1, 12 , 1

4 , ...

1, 1, 2, 3, 5, 8, 13, …

In Algebra, a sequence is written as

a1, a2, a3, a4, a5, ..., an, ...

The general terms in the sequence have a subscript indicating the cardinality--that is, which place they are in the sequence. a1 is the first number in the sequence, a2 is the second, and so on. an is the nth number. Note that n was always a counting number. In calculus and physics, this will change in that the subscript may be tied to the exponent and the series will generally be written

a0, a1, a2, a3, a4, ..., an, ...

Vocabulary Sequence-- a function with the Natural Numbers as the domain. Arithmetic Sequence-- A sequence in which one term equals a constant added to the preceding term, i.e. an+1 = an + d .

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Geometric Sequence-- A sequence in which one term equals a constant multiplied to the preceding term, i.e. an+1 = anr . Series--the sum of a sequence. Partial Sum--Defn: the sum of the first n terms of a sequence.

Much of what we did with sequences in Algebra 2H involved finding an (the nth term in the sequence) or Sn (the sum of the first n numbers in the sequence) directly without having to find all the numbers in the sequence. The pattern of the sequence leads to an equation where an is determined by n, and that pattern is determined by the type of sequence it is. It was basically a context for arithmetic. OBJECTIVES Identify Sequences and Series Find Partial Sums of a given Series. Find the terms, partial sums, infinite sums, or n in a geometric sequence.

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Ex 1 Find the next three terms and the general term in each of these sequences and classify them as an arithmetic sequence, a geometric sequence or neither. a) 1, 3, 5, 7, … b) 14, 7, 0, -7, -14, …

c) 4, 2, 1, 12, 1

4 , ... d) 1, 1, 2, 3, 5, 8, 13, …

e) 1, 14 , 1

9 , 116 , 1

25 , ... (f) 1, − 12 , 1

3, − 14 , 1

5 ,...

a) 9, 11, 13, …,1+ 2 n −1( ) . It is arithmetic because we add 2 each time. b) -21, -28, -35, …, 14 − 7 n −1( ) . It is arithmetic because we add -7 each time.

c) 18 , 1

16 , 132 ,..., 2 1

2⎛⎝⎜

⎞⎠⎟

n−1

. It is geometric because we multiply by 12

each time. d) 21, 34, 55. This is neither arithmetic nor geometric. We do not multiply by nor add the same amount each step. It is called the Fibonacci Sequence.

e) 136 , 1

49 , 164 ,..., 1

n2 . This is a p-Series with p=2.

f) − 16, 1

7, − 18,..., 

−1( )n+1

n . This is the Alternating Harmonic Series.

Sequences are generally written Set Notation:

an{ } Series are generally written with sigma notation:

akk=1

n

∑

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where ak is the kth term in the sequence and we are substituting the integers 1

through n in for k, getting the terms and adding them up. ∑ means sum. There

is a symbol for the product of a sequence Π( ) , but we will not go into that here. The first two examples at the beginning of the section are arithmetic sequences. In 1, 3, 5, 7, …, the difference (d) between any two consecutive terms is 2. In 14, 7, 0, -7, -14, …, the difference is -7. Since the difference is constant, we can jump from the first term to any term in the sequence by adding one less d than the number of terms. The fifth term is the first plus four d’s (or plus 4d). The fifth term is four steps away from the first.

Arithmetic Sequence

an = a1 + n −1( )d

Ex 2 Find a32 for an arithmetic sequence with a1 = 6 and d = 3

an = a1 + n −1( )da32 = 6 + 32 −1( )3

= 99

The arithmetic sequence formula an = a1 + n −1( )d clearly has four variables in it: an, a1, n, and d . Given any three, we can find the fourth.

Ex 3 Find the first term in the arithmetic sequence where d = − 43 and the 15th

term is -65.

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an = a1 + n −1( )d−65 = a1 + 15 −1( ) −4

3⎛⎝⎜

⎞⎠⎟

−65 = a1 +−563

a1 = −4613

Ex 4 If 2 is a term in the arithmetic sequence with a1 = 51 and d = -7, find which term it is.

an = a1 + n −1( )d2 = 51+ n −1( ) −7( )2 = 51− 7n + 77n = 56

= 8

So 2 is the eighth term in the sequence.

Note than we would also know if 2 was not in the sequence by the nature of the n in our solution. If n had not been a whole number, 2 would not be in the sequence. We can find several terms in a sequence if we know the first and last and how many there are. These terms are sometimes called the “means” because they are evenly spaced between the known values. Ex 5 Find six arithmetic means between 27 and 44.

This means the sequence is 27, , , , , , , 44, or

a1 = 27 and a8 = 44

This gives us enough information to find d:

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an = a1 + n −1( )d44 = 27 + 8 −1( )d17 = 7d

d = 177 = 2 37

d, of course, can be a fraction. The sequence, then, must be

27, 29 37 , 316

7 , 34 27 , 36 5

7 , 39 17 , 414

7 , 44

The third and fourth examples at the start of this section are geometric sequences. In 1, 3, 9, 27, …, the ratio (r) between any two consecutive terms is 3. In

4, 2, 1, 12 , 1

4 , ... , the ratio is 12 .

Since the ratio is constant, we can jump from the first term to any term in the sequence by multiplying one less r than the number of terms. The fifth term is the first times r four times (or r4 ).

Geometric Sequence

an = a1rn−1

As with the arithmetic sequences, the geometric sequence formula an = a1rn−1 clearly has four variables in it: an, a1, n, and r . Given any three, we can find the fourth.

Ex 6 Find a15 for an geometric sequence with a1 =1024 and r = − 12

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an = a1rn−1

a15 =1024−12

⎛⎝⎜

⎞⎠⎟

14

= −116

Ex 7 Find the first term in the geometric sequence where r = 3 and the 15th term is 28697814.

an = a1rn−1

28697814 = a1 3( )14a1 = 6

Ex 8 If 2 is a term in the geometric sequence with a1 = 51 and r = 12 , find which

term it is.

an = a1rn−1

2 = 51 12⎛⎝⎜

⎞⎠⎟

n-1

251 =

12

⎛⎝⎜

⎞⎠⎟

n-1

Solving for n in the exponent will require a logarithm.

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ln 251 = ln12

⎛⎝⎜

⎞⎠⎟

n-1

ln 251 = n−1( )ln 12

⎛⎝⎜

⎞⎠⎟

n−1=ln 251ln 12

⎛⎝⎜

⎞⎠⎟

n =ln 251ln 12

⎛⎝⎜

⎞⎠⎟

+1

= 5.672

Since this n is not an integer, 2 is not in the sequence.

Ex 9 Find four geometric means between 1 and 3125.

This means the sequence is 1, , , , , 3125, or

a1 =1 and a6 = 3125

This gives us enough information to find r:

an = a1rn−1

3125 =1r5

r = 31255

= 5

r, of course, can be a fraction. The sequence, then, must be

1, 5, 25, 125, 625, 3125

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6.1 Homework 1. Find the 12th term in 2, 5, 8, … 2. Find the 49th term in 2, -6, -14, … 3. Find a22 for an arithmetic sequence with a1 = 35 and d = -5. 4. Find a39 for an arithmetic sequence with a1 = −14 and d = .5.

5. Find the first term in the arithmetic sequence where d = 23 and the 11th term

is 243. 6. Find the first term in the arithmetic sequence where d = −2 and the 31st term is 451. 7. If 2 is a term in the arithmetic sequence with a1 = 42 and d = -6, find which term it is. 8. If 20 is a term in the arithmetic sequence with a1 = −22 and d = 7, find which term it is. 9. Find three arithmetic means between 42 and 70. 10. Find five arithmetic means between -107 and -86. 11. Find six arithmetic means between 47 and -11. Determine whether the sequence is arithmetic, geometric, or neither. If arithmetic or geometric, define d or r. 12. 8, 13, 18, 23, ...

13. 1, 13, 1

9 , 127 , ...

14. 25, 50, 100, 200, ... 15. 25, 75, 100, 125, ...

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16. 1, 12 , 1

3, 14 , ...

17. 5, 53 , 56 , 1, 156 , ...

Find the designated variable. 18. Find the 12th term in 2, 6, 18, … 19. Find the 49th term in 1, -2, 4, … 20. Find a37 for a geometric sequence with a1 =19 and r = .88.

21. Find a6 for a geometric sequence with a1 = −1 and r = 23 .

22. Find the first term in the geometric sequence where r = .95 and the 31st term is 214.64.

23. Find the first term in the geometric sequence where r = − 23 and the 4th term

is 24332 .

24. If 1536 is a term in the geometric sequence with a1 = 3 and r = 2, find which term it is.

25. If 1 is a term in the geometric sequence with a1 = 729 and r = 13 , find which

term it is. 26. Find two geometric means between 5 and 135.

27. Find three geometric means between 1525 and 2521 .

28. Find three geometric means between 5 and 45.

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6.2 Series

Vocabulary Series--the sum of a sequence. Partial Sum--Defn: the sum of the first n terms of a sequence.

Ex 1 Find 1− k2( )k=0

4

∑

1− k2( )k=0

4

∑ = 1− 02( )+ 1−12( )+ 1− 22( )+ 1− 32( )+ 1− 42( )=1+ 0− 3− 8−15= −25

Ex 2 Find 3k2k=2

5

∑

3k2k=2

5

∑ = 3 1k2k=2

5

∑

= 3 122 +

132 +

142 +

152

⎛⎝⎜

⎞⎠⎟

= 3 14 +19 +

116 +

125

⎛⎝⎜

⎞⎠⎟

=1.391

In an arithmetic sequence, the sum of the first and last terms is the same as the sum of the second and second to last, which is the same as the sum of the third and third to last. There are half as many pairs of numbers in a sequence as there are numbers in a sequence, so the partial sum is

Sn =n2 a1 + an( )

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Since we have a formula an = a1 + n −1( )d to get any term in the sequence from the first term, we can create a formula for the sum of the first n terms.

Arithmetic Partial Sum

Sn =

n2 2a1 + n−1( )d( )

Ex 3 Find S32 for an arithmetic sequence with a1 = 6 and d = 3

Sn =n2 2a1 + n−1( )d( )

S32 =322 2 6( )+ 31( )3( )

=1680

As with the arithmetic sequence formulaan = a1 + n −1( )d , Sn =

n2 2a1 + n −1( )d( )

has four variables in it: Sn, a1, n, and d . Given any three, we can find the fourth.

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Ex 4 Find the first term in the arithmetic series where d = − 43 and the 13th partial

sum is -65.

Sn =n2 2a1 + n−1( )d( )

−65 = 132 2a1 + 13−1( ) −43

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎞

⎠⎟

−10 = 2a1 −16a1 = 3

Ex 5 30705 is a partial sum of the arithmetic series with a1 =17 and d = 3. Which sum is it?

Sn =n2 2a1 + n−1( )d( )

30705 = n2 2 17( )+ n−1( ) 3( )( )61410 = n 34 + 3n− 3( )

0 = 3n2 + 31n− 61410

n =31± 312 − 4 3( ) −61410( )

6n =138 or −148 1

3

Since n must be an integer, n =138

Just as with the arithmetic series, there is a formula for the nth partial sum of a geometric series.

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Geometric Partial Sum

Sn = a11− rn1− r

As with the arithmetic sequences, the geometric sequence formula an = a1rn−1 clearly has four variables in it: Sn, a1, n, r . Given any three, we can find the fourth.

Ex 6 Find S15 for the geometric sequence with a1 =1024 and r = − 12

Sn = a11− rn1− r

S15 =10241− −1

2⎛⎝⎜

⎞⎠⎟

14

1− −12

⎛⎝⎜

⎞⎠⎟

= 682.625

Ex 7 Find the first term in the geometric sequence where r = −3 and the 9th partial sum is 24605.

Sn = a11− rn1− r

24605 = a11− −3( )91− −3( )

a1 = 5

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Ex 8 If Sn =1026 , a1 = 6 and r = −2 , find n.

Sn = a11− rn1− r

1026 = 6( )1− −2( )n1− −2( )

513=1− −2( )n

−512 = −2( )n

Since we know we cannot log a negative, the negatives in this equation must cancel, and

−512 = −2( )n

512 = 2( )nLn 512 = n Ln 2

n = Ln 512Ln2

n = 9

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Sum of an Infinite Geometric Series

S = a1− r ,

if and only if r <1 .

Ex 9 Find the infinite sum for the geometric sequence with a1 =1024 and

r = − 12 .

S = a11− r

S = 1024

1− −1 2

⎛⎝⎜

⎞⎠⎟

= 682 23

Ex 10 Find the infinite sum for the geometric sequence with a1 = 24 and r = 54 .

Since r = 54 >1 , the formula will yield an incorrect answer. This series is

divergent.

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6.2 Homework Find the sum.

1. 1kk=1

5

∑ 2. 3k − 5( )k=1

4

∑

3. −1( )k 2k +1( )k=1

5

∑ 4. 2kk=3

7

∑

5. −1( )k 23⎛⎝⎜

⎞⎠⎟k=0

5

∑k+2

6. 2+ k2( )k=3

7

∑

Find the designated variable. 7. Find S10 for an arithmetic series with a1 = 35 and d = -5. 8. Find S15 for an arithmetic series with a1 = −14 and d = .5. 9. Find S13 for an arithmetic series with a2 = 4 and d = -3. 10. Find n in the arithmetic series where a1 =15 , d = 4 and Sn = 3219 . 11. Find n in the arithmetic series where a1 = 8 , d = 5 and Sn = 4859 . 12. Find the first terms in the arithmetic series where d = 3 and S10 =175 . 13. Find the first term in the arithmetic series where d = −2 and S15 = 0 . 14. Find the sum of the first 12 terms of 2, 5, 8, … 15. Find the sum of the first 60 terms of 2, -6, -14, … 16. Find the sum of the first 12 terms of 2, 6, 18, … 17. Find the sum of the first 60 terms of 1, -2, 4, …

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18. In the geometric sequence from a1 = 6 to an = −196608 where r = -2, how many terms are there and what is the partial sum? 19. Find the sum of the infinite geometric series where a1 =15 and r = .4 .

20. Find the sum of the infinite geometric series where a1 = 8 and r = −3 7 .

21. Find the sum of the infinite geometric series where a1 = −42 and r = 2 .

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6.3: Convergence and Divergence of Infinite Series Vocabulary Infinite Sequence—Defn: a sequence with an infinite number of terms. Convergent Sequence --Defn: an infinite sequence where an approaches a

particular value when n goes to infinity. Divergent Sequence --Defn: an infinite sequence where an does not approach a

particular value when n goes to infinity. In some cases, a series of infinite terms can have a total, if the later terms get infinitely small. Much of the series work in Calculus is about finding whether an infinite series is convergent (has a total) or divergent (has total keeps getting bigger indefinitely). Geometric series are the easiest for which to find an infinite sum. In other words,

A sequence is convergent if and only if

n→∞lim an = c .

A sequence is divergent if and only if

n→∞lim an ≠ c .

Geometric and Arithmetic series are the ones we encountered in Algebra 2. There are several other kinds, some of which are important for our study of series. We will look at them more closely later.

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Kinds of Series Arithmetic Series: an+1 = a1 + n−1( )d Geometric Series: an = a1rn−1

p-series: 1npn=1

∞∑

Alternating series: −1( )n ann=0

∞∑ or cos πn( )an

n=0

∞∑ or any other series where the

signs of the terms alternate between + and -.

Harmonic Series: 1nn=1

∞∑ =1+ 12 +

13+

14 + ⋅⋅⋅ (Note that this is a p-series where p=1)

Alternating Harmonic Series: −1( )n+1nn=1

∞∑ =1− 12 +

13−14 + ⋅⋅⋅

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The one formula from freshman year that will be useful is for the sum of an Infinite Geometric Series:

Sum of an Infinite Geometric Series

S = a1− r ,

if and only if r <1 .

If r ≥1 , the series is divergent.

Ex 1 Find the infinite sum for the geometric sequence with a1 =1024 and

r = − 12 .

S = a11− r

S = 1024

1− −1 2

⎛⎝⎜

⎞⎠⎟

= 682 23

Ex 2 Find the infinite sum for the geometric sequence with a1 = 24 and r = 54 .

Since r = 54 >1 , the formula will yield an incorrect answer. This series is

divergent. Ex 3 Which of the following sequences converge?

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I. 4n3n −1

⎧⎨⎩

⎫⎬⎭

II. e2n2n

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ III. e2n

e2n −1⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

In each case, we just need to check the Limit at Infinity. That is, we need to check the end behavior.

I. n→∞lim 4n

3n−1 =43 ; therefore, this sequence converges.

II. n→∞lim e

2n

2n =∞ ; therefore, this one diverges.

III. n→∞lim e2n

e2n −1 =L 'H

n→∞lim e2n 2( )e2n 2( ) =1; therefore, this converges.

So, only I and III converge.

This is a typical AP question. The key is CRITICAL READING. Sometimes, the question is about Sequences, sometimes it is about Series (which we will consider later it he chapter). Sometimes the question is about convergence and sometimes it is about divergence. Ex 4 Which of the following sequences diverge?

I. 3n27n3 −1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ II. 7lnn

2n⎧⎨⎩

⎫⎬⎭

III. 4n4

2n3 −1⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

I. n→∞lim 3n2

7n3 −1 = 0 ; therefore, this sequence does not diverges.

II. n→∞lim 7lnn

2n =L 'H

n→∞lim 7 1n

⎛⎝⎜

⎞⎠⎟

2 = 0 ; therefore, this one does not diverge.

III. n→∞lim 4n4

2n3 −1=∞ ; therefore, this diverges.

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So, only III diverges. Vocabulary Infinite Sum--Defn: the sum of all the terms of a sequence. This is not possible for an arithmetic series, but might be for a geometric series. Convergent Series--Defn: an infinite series that has a total. Divergent Series--Defn: an infinite series that does not have a total. In other words,

A series is convergent if and only if ann=1

∞∑ =c .

A series is divergent if and only if ann=1

∞

∑ ≠ c .

When testing a numerical series for divergence or convergence, the two easiest tests are:

Divergence Test (nth term test): If n→∞Lim an ≠ 0→ an

n=1

∞∑ diverges.

If n→∞Lim an = 0→ no conclusion

The Divergence Test basically says that if the terms at the end are not 0, the sum will just keep getting larger and larger (i.e., diverges). While easy to use, each has a disadvantage. The Divergence Test tells you if a function has terms approaching 0. If the terms at the upper end of the series do not approach 0, the series cannot converge. But, there are many series with end-terms approaching 0 that still diverge. So half the test is inconclusive.

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OBJECTIVE Determine the convergence or divergence of a sequence. Determine the divergence of a series. Use the Alternating Series Test to check for convergence or divergence. Ex 5 What conclusion can be drawn from the Divergence Test?

I. 12nn=1

∞∑ II.

n2n n+1( )n=1

∞∑ III. 1

nn=1

∞∑

I. n→∞lim 1

2n = 0 ; so the Divergence Test is inconclusive for I. Note that

this is not enough to decide if it converges or not.

II. n→∞Lim n2n n+1( ) = n→∞Lim n2

n2 + n=1; therefore, II diverges.

III.

n→∞Lim 1n = 0 ; therefore, the Divergence Test is inconclusive for III.

I and III pass the Divergence Test

A similar test to the Divergence Test, in process, is the Alternating Series test:

The Liebnitz Alternating Series Test

If a series is Alternating Series and is an ≥ an+1 , then it is convergent if and only if

n→∞Lim an = 0 .

This means the non-alternating part of the series must be decreasing and pass the Divergence Test (that is, the sequence converges to 0).

396

Ex 6 Is the Alternating Harmonic Series convergent or divergent?

Remember that the Alternating Harmonic Series is −1( )n+1nn=1

∞∑ .

i) an =−1( )n+1n = 1n and an+1 =

−1( )n+2n+1 = 1

n+1

1n >

1n+1 , so an ≥ an+1

ii) n→∞lim −1( )n+1

n=

n→∞lim 1

n= 0

So −1( )n+1nn=1

∞∑ converges.

Ex 7 Is cos πn( )n!n=1

∞∑ convergent or divergent?

Remember that the Alternating Harmonic Series is.

i) an =cos πn( )n! = 1n! and an+1 = an =

cos π n+1( )( )n+1( )! = 1

n+1( )!

1n!>

1n+1( )! , so an ≥ an+1

ii) n→∞lim −1( )n+1

n=

n→∞lim 1

n= 0

So cos πn( )n!n=1

∞∑ converges.

397

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• 6.3 Homework Set A Determine if the sequence converges or diverges.

1. n n −1( ){ } 2. 3+ 2n2n + n2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

3. n1+ n

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ 4. 2n

3n+1⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

5.

Ln n2( )n

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ 6. n2−n{ }

7.

n!n + 2( )!

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

Determine which series might be convergent by the Divergence Test.

8.

1nn=1

∞∑ 9.

413n2 − 5

⎛

⎝⎜⎞

⎠⎟n=1

∞∑

10. 3−n 8( )n+1n=1

∞∑ 11. 2n

n=0

∞∑

12. 4n+15nn=0

∞∑

Test these series for convergence or divergence.

13.   −1( )n−1n3n=1

∞∑ 14.  

−1( )n+14n2 +1n=1

∞∑

15.   −1( )n nn +1n=1

∞∑ 16.   −1( )n n

Ln nn=1

∞∑

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17.   −1( )n−1 Ln nnn=1

∞∑ 18.  cos πn( )

n34n=1

∞∑

19.   −1( )n4n+1n=1

∞∑ 20.  cos πn( )

n54n=1

∞∑

21.  −3( )n−1nn=1

∞∑ 22.   −1( )n 2n

4n +1n=1

∞∑

6.3 Homework Set B Determine if the sequence converges or diverges.

1. 2n2 +13n− 5

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ 2. n2 +1− n{ }

3. tan−1 n2 − n3( ){ } 4. en −1− nn3

⎧⎨⎩

⎫⎬⎭

5. cosn{ } 6. 16n2 + n − 4n{ }

7. n+ 349n2 +1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ 8. n

n+1⎛⎝⎜

⎞⎠⎟n⎧

⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

Determine which series might be convergent by the Divergence Test.

9. 5 43

⎛⎝⎜

⎞⎠⎟

n

n=0

∞

∑ 10. n3e−n2n=1

∞

∑

11. 180 ⋅2−nn=0

∞

∑ 12. sinn− nn3n=0

∞

∑

13. nn2n=1

∞

∑ 14. ntan−1 4n( )n=0

∞

∑

400

6.4: The Integral Test When testing a numerical series for divergence or convergence, one of the easiest tests is:

Integral Test: ann=1

∞

∑ and xn 1

∞

∫ dx either both converge or both diverge.

The reason the Integral Test works is best explained visually. The series can be visualized as a sum of Riemann Rectangles with width = 1 and height = the term values. If the integral diverges (i.e., the sum is infinite), the rectangles can be drawn left-hand and their sum is larger than an infinite number. If the integral converges (i.e., the sum is finite), the rectangles can be drawn right-hand and their sum is smaller than a finite number.

In both situations, the function must be decreasing. While easy to use, there is a disadvantage: The Integral Test only works on functions that can be integrated. Not all functions can be integrated.

401

Ex 1 Use the Integral Test to determine which of these series converge.

I. 1nn=1

∞∑ II. 1

n2 +1n=1

∞∑ III.   1

n ln2 nn=2

∞∑

I. 1xdx

1

∞

∫ = ln x 1∞ = dne ; therefore, I diverges.

II. 1x2 +1

dx1

∞

∫ = tan−1 x1

∞= π2 −

π4 = π

4 ; therefore, II converges.

III. 1x ln2 x

dx2

∞

∫ = 1u2du

ln2

∞

∫ = −1u ln2

∞

= 0 − ln2( ) = − ln2 ; therefore, III

converges.

II and III converge.

Two notes about this example. First, I. proves that the Harmonic Series always diverges. Second, III was a series that started at n = 2 instead of n = 1. This was to avoid the problem of the first term being transfinite.

Ex 2 For what values of p does the p-series 1npn=1

∞∑ converge?

n→∞Lim 1

np = 0 for all p>0, so p must be positive.

1xpdx

1

∞

∫ = x− p+1−p +1

1

∞

. This integral will be finite as long as p>1, because this

will leave x in the denominator so the limit can go to 0. Therefore,

1npn=1

∞∑ converges for all p>1.

We will use this fact later in this chapter, when applying the Comparison Tests.

402

Ex 3 Determine whether

n1+ n2

0

∞

∑ is convergent or divergent.

x1+ x2 dx

0

∞⌠⌡⎮

= limb→∞

x1+ x2 dx

0

b⌠⌡⎮

= 12

limb→∞

duux=0

x=b

∫= 1

2limb→∞

ln 1+ x2( )0

b

= 12

limb→∞

ln 1+ b2( )− ln1⎡⎣⎢

⎤⎦⎥

= ∞

Since this integral diverges, the series

n1+ n2

0

∞

∑ diverges.

403

6.4 Homework Set A Apply the Integral Test to each of these series to determine divergence or convergence. If the series is convergent, state the improper integral value.

1.   n

n2 +1( )32n=1

∞∑ 2.   1

nLn2  nn=2

∞∑

3.   13n +1n=1

∞∑ 4.  n +1

n4n=1

∞∑

5. 1+ 18 +127 +

164 +

1125 + ...

6.  5 − 2 nn3n=1

∞∑

7.   1n2 +1n=1

∞∑ 8.   n

n2 +1n=1

∞∑

9.   15n + 2( )4n=1

∞∑ 10. ne−n2

n=1

∞∑

6.4 Homework Set B Apply the Integral Test to each of these series to determine divergence or convergence. If the series is convergent, state the improper integral value.

1. n2e−n3n=1

∞

∑ 2. 3n21+ n6n=1

∞

∑

3. 6n51+ n6n=1

∞

∑ 4. 14n + 3( )2n=0

∞

∑

5. lnnnn=2

∞

∑ 6. sinnn=1

∞

∑

7. n1+ n2n=1

∞

∑ 8. 14 − nn=1

∞

∑

404

6.5: The Direct and Limit Comparison Tests The Direct and Limit Comparison Tests are much less algebraic than the Divergence and Integral Tests. They involve comparing a series to some series that we already know converges or diverges. So what do we know? REMINDER: Factual Knowledge for Comparisons

Geometric series: 1. anrnn=0

∞

∑ converges if r <1and diverges if r ≥1

2. If anrnn=0

∞

∑ converges, it converges to a01− r

p-Series: 1. 1npn=0

∞

∑ converges if p >1and diverges if p ≤1

Note Well:

Harmonic Series: 1nn=1

∞∑ = 1+ 1

2+ 13+ 14+ ⋅ ⋅ ⋅ Diverges

Alternating Harmonic Series: 1nn=1

∞∑ = 1+ 1

2+ 13+ 14+ ⋅ ⋅ ⋅ Converges

405

The Direct Comparison Test

When comparing a given series ann=1

∞

∑ to a known series bnn=1

∞

∑ ,

1) if an ≤ bn and bnn=1

∞

∑ converges, then ann=1

∞

∑ converges.

2) if an ≥ bn and bnn=1

∞

∑ diverges, then ann=1

∞

∑ diverges.

The use of this test is very verbal. The Limit Comparison Test

When comparing a given series ann=1

∞

∑ to a known series bnn=1

∞

∑ ,

1. If n→∞lim anbn

= any positive Real number, then both converge

or both diverge.

2. If n→∞lim anbn

= 0 , then ann=1

∞

∑ converges if bnn=1

∞

∑ converges.

3. If n→∞lim anbn

= dne , then ann=1

∞

∑ diverges if bnn=1

∞

∑ diverges.

Key Idea I: What series to compare ann=1

∞

∑ to depends on where the variable is.

If n is in the base, compare to the p-Series. If n is in the exponent, compare to the Geometric Series. If n is in both the base and exponent (but notnn ), try either. Ifnn , use the Nth Root Test (section 5-4).

406

Key Idea II: Which test to use depends on the combination of the relative sizes and whether the known series converges or diverges.

bnn=1

∞

∑ converges bnn=1

∞

∑ diverges

an ≤ bn

Direct Comparison Test

Limit Comparison Test

an ≥ bn

Limit Comparison Test

Direct Comparison Test

Key Idea III: If an is rational, we want to compare to the end behavior model of the corresponding function. OBJECTIVE Use the Comparison Tests to check for convergence or divergence.

Ex 1 Does   1n3 + 2n=1

∞∑ converge?

We can make a direct comparison between   1n3 + 2n=1

∞∑ and   1

n3n=1

∞∑ .

1n3 + 2

< 1n3

because 1n3 + 2

has a larger denominator. Since   1n3n=1

∞∑

converges (it is a p-Series with p>1) and 1n3 + 2

is smaller than 1n3

, then

  1n3 + 2n=1

∞∑ converges.

Ex 2 Does   1n − 2n=3

∞∑ converge?

407

We can make a direct comparison between   1n − 2n=1

∞∑ and  1nn=1

∞∑ .

1n − 2 >

1n because 1

n − 2 has a smaller denominator. Since  1nn=1

∞∑ diverges

(it is the Harmonic Series--a p-Series with p=1) and 1n − 2 is bigger than 1

n,

then   1n − 2n=3

∞∑ diverges.

Ex 3 Does   1n2 − 4n=1

∞∑ converge?

If we can try to make a direct comparison between   1n2 − 4n=1

∞∑ and   1

n2n=1

∞∑ ,

we find that 1n2 − 4

> 1n2

(because 1n2 − 4

has a larger denominator) but

  1n2n=1

∞∑ converges (it is a p-Series with p>1). So the Direct Comparison Test

does not apply.

n→∞Lim

1n2 − 4

1n2

=n→∞Lim n2

n2 − 4=1> 0

This limit is greater than 0 and   1n2n=1

∞∑ converges. Therefore,   1

n2 − 4n=1

∞∑

converges also.

Ex 4 Does   12n −1n=1

∞∑ converge?

408

12n −1 >

12n because 1

2n −1 has a smaller denominator. But   12nn=1

∞∑

converges (it is a Geometric Series with r<1), so Direct Comparison will not work.

n→∞Lim

12n −1

12n

=n→∞Lim 2n

2n −1 =L 'H

n→∞Lim 2n ln2

2n ln2 =1

This limit is greater than 0 and   12nn=1

∞∑ converges. Therefore,   1

2n −1n=1

∞∑

converges also.

Ex 5 Does   12n + 3n=1

∞∑ converge?

We can make a direct comparison between   12n + 3n=1

∞∑ and   12nn=1

∞∑ .

12n + 3 <

12n because 1

2n + 3 has a larger denominator. Since   12nn=1

∞∑

converges and 12n + 3 is smaller than 1

2n, then   1

2n + 3n=1

∞∑ converges.

409

Ex 6 Does   1n3 + 2( )14n=1

∞∑ converge?

We can use the Limit Comparison Test between   1n3 + 2( )14n=1

∞∑ and   1

n34n=1

∞∑ .

n→∞Lim

  1

n3 + 2( )14

  1n

34

=n→∞Lim   n

34

n3 + 2( )14=1

  1n34n=1

∞∑ diverges, therefore,   1

n3 + 2( )14n=1

∞∑ diverges.

410

6.5 Homework Set A Test these series for convergence or divergence.

1.   1n2 + n +1n=1

∞∑ 2.   5

2 + 3nn=1

∞∑

3.  n +1n2n=1

∞∑ 4.   3n2nn=1

∞∑

5.  n3 +1n3 −1n=1

∞∑ 6.   n

n+1( )2nn=1

∞∑

7.  3+ cosn3nn=1

∞∑ 8.   1

1+ nn=1

∞∑

9.   n2 − 5n

n3 + n +1n=1

∞∑ 10.  sin 1nn=1

∞∑

6.5 Homework Set B Test these series for convergence or divergence.

1. sin2 nn3 +1n=1

∞

∑ 2. 1n3 −13

n=2

∞

∑ 3. 1n!n=1

∞

∑

4. 3n2 + 2n4n 2n2 + 7n −1( )n=1

∞

∑ 5. n − 5n + 2( )4n=1

∞

∑ 6. 4 + 2n1+ n3( )2n=1

∞

∑

7. 1+ 2sinnn4n=1

∞

∑ 8. nn + 4n=1

∞

∑ 9. nn + 4( )2n=1

∞

∑

10. 1n2 + 4n=1

∞

∑

411

6.6: Absolute and Conditional Convergence The four tests in the previous two sections only apply to series of positive values. There is a separate test for alternating series called the Liebnitz Alternating Series Test.

The Alternating Series Test (AST)

If a series is Alternating Series and is an ≥ an+1 , then it is convergent if and only if

n→∞Lim an = 0 .

This test was introduced in Section 6.2. Vocabulary Absolute Convergence—Defn: When an alternating series and its absolute value

are both convergent. Conditional Convergence—Defn: When an alternating series is convergent but its

absolute value are divergent. OBJECTIVE Use the Alternating Series Test to check for convergence or divergence.

Ex 1 Is cos πn( )n3n=1

∞∑ convergent or divergent?

Remember that the Alternating Harmonic Series is.

i) an =cos πn( )n3 = 1n3 and an+1 = an =

cos π n+1( )( )n+1( )3

= 1n+1( )3

1n3 >

1n+1( )3

, so an ≥ an+1

412

ii) n→∞Lim −1( )n+1

n3 =n→∞Lim 1

n3 = 0

So cos πn( )n3n=1

∞∑ converges.

Ex 2 Is cos πn( )n3n=1

∞∑ absolutely or conditionally convergent?

Ex 1 proved that cos πn( )n3n=1

∞∑ is convergent by the AST. To demonstrate

conditional vs. absolute convergence, we test the series without the alternator:

cos πn( )n3n=1

∞

∑ = 1n3n=1

∞

∑

1n3n=1

∞

∑ is a p-series with p >1 , therefore, since both the alternating and non-

alternating versions of the series converge,

cos πn( )n3n=1

∞∑ converges absolutely.

413

Ex 3 Is −1( )n+1 n+ 2( )n n+1( )n=1

∞

∑ convergent or divergent? If convergent, is it absolutely or

conditionally convergent?

i) an =−1( )n+1 n + 2( )n n +1( ) =

n + 2n2 + n

and an+1 =−1( )n+1 n +1( ) + 2( )n +1( ) n +1( ) +1( ) = n + 3

n2 + 3n + 2

n + 2n2 + n

> n + 3n2 + 3n + 2

, so an ≥ an+1

ii) n→∞Lim

−1( )n+1 n + 2( )n n +1( ) =

n→∞Lim n + 2

n2 + n= 0

So −1( )n+1 n + 2( )n n +1( )n=1

∞

∑ converges.

To test for absolute vs. conditional convergence, we need to test n + 2n n +1( )n=1

∞

∑ .

We can do the Limit Comparison Test against the end behavior model 1nn=1

∞∑ .

n→∞Lim

n + 2n n +1( )

1n

=n→∞Lim n + 2

n n +1( ) ⋅n1=

n→∞Lim n

2 + 2nn2 + n

= 1

This limit is greater than 0, therefore, both do the same thing--namely, they diverge. So,

−1( )n+1 n + 2( )n n +1( )n=1

∞

∑ is conditionally convergent.

414

Ex 4 Is −1( )n+12nn=1

∞

∑ convergent or divergent? If convergent, is it absolutely or

conditionally convergent?

i) an =−1( )n+12n = 1

2n and an+1 =

−1( )n+22n+1

= 12 ⋅2n

12n >

12 ⋅2n , so an ≥ an+1

ii) n→∞Lim

−1( )n+1

2n =n→∞Lim 1

2n = 0 So −1( )n+12nn=1

∞

∑ converges.

Furthermore, 12n

converges because it is a Geometric series with r = 12

. So,

−1( )n+12nn=1

∞

∑ converges absolutely.

415

6.6 Homework Set A Test these series for convergence or divergence. If it is convergent, determine if it has absolute or conditional convergence.

1.  −3( )n−1nn=1

∞∑ 2.   −1( )n 2n

4n +1n=1

∞∑

3.   −1( )n nn +1n=1

∞∑ 4.   −1( )n n

Ln nn=1

∞∑

5.   −1( )n−1 Ln nnn=1

∞∑ 6.  cosπn

n34n=1

∞∑

7.  −1( )n+14n2 +1n=1

∞∑ 8. −1( )n n

n

n!n=1

∞

∑

6.6 Homework Set B a) Test these series for convergence or divergence. b) If it is convergent, determine if it has absolute or conditional convergence.

1. −3( )n+124nn=1

∞

∑ 2. −1( )n n + 5( )

4nn=2

∞

∑ 3. −1( )n nn +1n=1

∞

∑

4. −1( )n 1n +1n=1

∞

∑ 5. −1( )n 1n +1n=1

∞

∑ 6. cos πn( )n!n=1

∞

∑

7. sin π

2 n⎛⎝⎜

⎞⎠⎟

nn=1

∞

∑ 8. − n6⎛⎝⎜

⎞⎠⎟

n

n=1

∞

∑ 9. −1( )n arctannn=1

∞

∑

10. −1( )n arctan πn

⎛⎝⎜

⎞⎠⎟n=1

∞

∑ 11. cos πn( )enn=1

∞

∑ 12. 2nn=1

∞

∑ sin π2 n

⎛⎝⎜

⎞⎠⎟

416

6.7: The Ratio and Nth Root Tests The Ratio Test is one of the more easily used tests and it works on most series, especially those that involve factorials and those that are a combination of geometric and p-series. Unfortunately, it is sometimes inconclusive, especially with alternating series.

Cauchy Ratio Test: If n→∞lim an+1

an <1→ it converges;

If n→∞lim an+1

an >1→ it diverges;

If n→∞lim an+1

an =1→ the test is inconclusive.

The Nth Root Test is best for series with n is both the base and the exponent.

The Nth Root Test: If

n→∞lim ann <1→ it converges;

If n→∞lim ann >1→ it diverges;

If n→∞lim ann =1→ the test is inconclusive.

OBJECTIVE Use the Ratio and Nth Root Tests to check for convergence or divergence. Since the Cauchy Ratio Test involves Limits at Infinity, It might be a good idea to review the Hierarchy of functions:

The Hierarchy of Functions

417

1. Logs grow the slowest. 2. Polynomials, Rationals and Radicals grow faster than logs and

the degree of the EBM determines which algebraic function grows fastest. For example, y = x

12 grows more slowly thany = x2 .

3. The trig inverses fall in between the algebraic functions at the value of their respective horizontal asymptotes.

4. Exponential functions grow faster than the others. (In BC Calculus, we will see the factorial function, y = n!, grows the fastest.)

5. The fastest growing function in the combination determines the end behavior, just as the highest degree term did among the algebraics.

418

Ex 1 Does   1n!n=1

∞∑ converge?

n→∞lim an+1an

=n→∞lim

1n+1( )!

1n!

=n→∞lim n!

n +1( )!

=n→∞lim n!

n +1( )n!=n→∞lim 1

n +1= 0

Since this limit is <1,   1n!n=1

∞∑ converges.

Ex 2 Does   n2nn=1

∞∑ converge?

n→∞Lim an+1an

=n→∞Lim

n +12n+1

n2n

=n→∞Lim n +1( )2n

2n+1n

=n→∞Lim n +1

2n

= 12

Since this limit is <1,   n2nn=1

∞∑ converges.

419

Ex 3 Does   n + 2n n +1( )n=1

∞∑ converge?

n→∞Lim an+1an

=n→∞Lim

n +1( ) + 2n +1( ) n +1( ) +1( )

n + 2n n +1( )

=n→∞Lim n + 3

n +1( ) n + 2( ) ⋅n n +1( )n + 2

=n→∞Lim n2 + 3n

n2 + 4n + 4=1

So the Ratio Test is inconclusive.

The easiest test would be the Limit Comparison Test:

Compare   n + 2n n +1( )n=1

∞∑ to  1nn=1

∞∑ .

limn→∞

n + 2n n +1( )1n

= limn→∞

n n + 2( )n n +1( ) =1

Since the Limit = 1,both series converge or both diverge.  1nn=1

∞∑ diverges,

therefore,

  n + 2n n +1( )n=1

∞∑ diverges also.

Ex 4 Does   1lnn n + 2( )n=1

∞∑ converge?

This series has n to an n power; therefore, the Root Test is appropriate.

420

n→∞Lim 1lnn n + 2( )n =

n→∞Lim 1ln n + 2( ) = 0

Since this limit is <1,   1lnn n + 2( )n=1

∞∑ converges.

Ex 5 Does  1+ 1

n⎛⎝⎜

⎞⎠⎟n

e−nn=1

∞∑ converge?

This series has n to an n power; therefore, the Root Test is appropriate.

n→∞Lim 1+ 1

n⎛⎝⎜

⎞⎠⎟

n

e−nn

=n→∞Lim

1+ 1n

⎛⎝⎜

⎞⎠⎟

e−1 =n→∞Lim n +1( )e

n = e >1

Since this limit is >1,  1+ 1

n⎛⎝⎜

⎞⎠⎟n

e−nn=1

∞∑ diverges.

421

6.7 Homework Set A Test these series for convergence or divergence.

1.   n2

2nn=1

∞∑ 2.   1

n nn=1

∞∑ 3.  

−3( )nn3n=1

∞∑

4.   −3( )nn!n=1

∞∑ 5.   −1( )n n

5 + nn=1

∞∑ 6.   12n( )!n=1

∞∑

7.  cosπn

3n!n=1

∞∑ 8.   10n

n +1( )42n+1n=1

∞∑

9.   n2 +12n2 +1

⎛

⎝⎜

⎞

⎠⎟

n=1

∞∑

n

10.   nn

31+3nn=1

∞∑

11. For which of the following series is the Ratio Test inconclusive. What test would you use instead?

(a)   1n3n=1

∞∑ (b)  

−3( )n−1nn=1

∞∑ (c)   n2nn=1

∞∑ (d)   n

1+ n2n=1

∞∑

6.7 Homework Set B Test these series for convergence or divergence.

1. n44nn=1

∞

∑ 2. 4nn4n=2

∞

∑ 3. e−nn!n=1

∞

∑

4. 1⋅3⋅5 ⋅7...⋅ 2n +1( )

n!n=1

∞

∑ 5. 1lnn( )nn=2

∞

∑

6. 3n −1( )nn=0

∞

∑ 7. 3n −1( )nn=0

∞

∑ 8. n!1⋅4 ⋅7 ⋅10...⋅ 3n +1( )n=1

∞

∑

422

9. 4n1+ 5n

⎛⎝⎜

⎞⎠⎟

n

n=1

∞

∑ 10. 2n ⋅n3n!n=1

∞

∑ 11. en2

n!n=1

∞

∑

12. 3n( )nn3nn=1

∞

∑

13. For which of the following series is the Ratio Test inconclusive? What test would you use instead?

(a) n2 −1n2 + 2nn=1

∞

∑ (b) 13n( )!n=1

∞

∑ (c)−1( )nn4n=2

∞

∑ (d) 12nn + 2( )33n+1n=1

∞

∑

423

6.8 General Series Summary

A series is convergent if and only if ann=1

∞∑ = c .

A series is divergent if and only if ann=1

∞

∑ ≠ c .

Previously, we investigated the seven tests that will help determine if a given series converges or not. Those tests were: 1. Divergence Test (nth term test): If

n→∞Lim an ≠ 0→ it diverges.

If n→∞Lim an = 0→ no conclusion

2. Integral Test: If f(x) is a decreasing function, then ann=1

∞

∑ and xn 1

∞

∫ dx

either both converge or both diverge.

3. Cauchy Ratio Test: If n→∞Lim an+1an

<1→ it converges;

If n→∞Lim an+1an

>1→ it diverges;

If n→∞Lim an+1an

=1→ the test is inconclusive.

4. The Alternating Series Test: An Alternating Series is convergent if and only if i) an ≥ an+1 ,

and ii) n→∞Lim an = 0 .

424

5. The Nth Root Test: If n→∞Lim ann <1→ it converges;

If n→∞Lim ann >1→ it diverges;

If n→∞Lim ann =1→ the test is inconclusive.

6. The Direct Comparison Test: When comparing a given series ann=1

∞

∑ to a

known series bnn=1

∞

∑ ,

i) if an ≤ bn and bnn=1

∞

∑ converges, then ann=1

∞

∑ converges.

ii) if an ≥ bn and bnn=1

∞

∑ diverges, then ann=1

∞

∑ diverges.

7. The Limit Comparison Test: When comparing a given series ann=1

∞

∑ to a

known series bnn=1

∞

∑ ,

i. If n→∞Lim an

bn> 0 then both converge or both diverge.

ii. If n→∞Lim an

bn= 0 , then an

n=1

∞

∑ converges if bnn=1

∞

∑ converges.

iii. If n→∞Lim an

bn=∞ , then an

n=1

∞

∑ diverges if bnn=1

∞

∑ diverges.

OBJECTIVE Find whether a given numerical series converges or diverges.

425

426

6.8 Homework 1) Which of the following sequences converge?

I.

3n5

7n4 −1⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ II.

cosnπ n

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ III.

n!e2n

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

(A) I only (B) II only (C) I and III only (D) II and III only (E) III only

2) Which of the following sequences diverge?

I.

lnn5n

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ II.

5cos3e

⎛

⎝⎜⎞

⎠⎟

n⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ III.

n!n+ 2( )!

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

(A) II only (B) III only (C) I and II only (D) I and III only (E) I, II, and III

3) Which of the following series diverge?

I.

9−n8n+1

n=1

∞

∑ II.

ln5cos 2πn( )n=3

∞

∑ III.

2n3n+1

2e( )nn=2

∞

∑

(A) I only (B) III only (C) I and II only (D) II and III only (E) I, II, and III

4) What are all values of k for which the infinite series

k5

⎛

⎝⎜⎞

⎠⎟n=1

∞∑

n

converges?

(A)

k < 5 (B)

k ≤ 5 (C)

k <1

(D) k ≤1 (E) k = 0

427

5) If f x( ) = 1− sin x( )2⎛

⎝⎜⎞⎠⎟n=0

∞∑

n

, then f π

6⎛

⎝⎜⎞

⎠⎟=

(A) 2 (B)

43

(C)

12

(D)

32

(E) divergent

6) Which of the following series converge?

I.

2n+ 3n2 + 3n+ 6n=1

∞

∑ II.

n+1en

n=1

∞

∑ III.

2n1+ n4

n=1

∞

∑

(A) I only (B) III only (C) I and III only (D) II and III only (E) I and II only

7) Which of the following series are divergent?

I.

−1( )n3n+1n=0

∞∑ II.

en

n+5( )!n=0

∞∑ III.

n!3n

n=1

∞

∑

(A) I only (B) II only (C) III only (D) I and II only (E) II and III only

8) Which of the following series are conditionally convergent?

I.

−1( )nn nn=0

∞

∑ II.

en

n+5( )!n=0

∞

∑ III.

cos πn( )nn=1

∞

∑

(A) I only (B) II only (C) III only (D) I and II only (E) I and III only

428

9) Which of the following series are divergent?

I.

−1( )n 13n+1n=0

∞

∑ II.

n+ 2( )!n!n=0

∞

∑ III.

n!3n

n=1

∞

∑

(A) I and III only (B) II only (C) III only (D) I and II only (E) II and III only

10) What are all values of p for which the infinite series

5n1+ p

n=1

∞

∑ diverges?

(A) p > −2 (B) p ≥ −2 (C) p < −2 (D) p ≤ −2 (E) All real values of p

429

Numerical Series Test Calculator Allowed 1.

I.

Ln nLn 2n

⎧⎨⎩

⎫⎬⎭

II. an = n n−1( ) III.

−1( )n n3

33 + 2n2 +1

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

(A) I only (B) III only (C) I and II only (D) I and III only (E) I, II, and III

2. Which of the following Series diverge?

I.

5−n6n−1

n=1

∞

∑ II.

3n5

7n4 −1n=1

∞

∑ III.

15n

n=1

∞

∑

(A) I only (B) III only (C) I and II only (D) II and III only (E) I, II, and III

3. Which of the following are series divergent?

I.

en

n+5( )!n=1

∞

∑ II.

ln5cos 2πn( )n=3

∞

∑ III.

2n3n+1

2e( )nn=2

∞

∑

(A) I only (B) III only (C) I and III only (D) II and III only (E) I, II, and III

430

4. What are all values of p for which the infinite series

5n1+ p

n=1

∞

∑

converges?

(A) p > −2 (B) p ≥ −2 (C) p < −2 (D) p ≤ −2 (E) All real values of p

5. Which of the following are series convergent?

I.

5n3 + 3n=1

∞

∑ II.

14+ n2

n=1

∞

∑ III.

3n+ 7( )n=2

∞

∑

(A) I only (B) III only (C) I and II only (D) I and III only (E) II and III only

6. Which of the following series diverge?

I.

−1( )n 12n+1n=1

∞

∑ II.

−1( )n cosn3n

n=1

∞

∑ III.

−1( )n+1 1nn=1

∞

∑

(A) I only (B) II only (C) I and II only (D) I and III only (E) I, II and III

431

7. If

ann=1

∞

∑ converges, then which of the following is true?

I.

ann=1

∞

∑ diverges

II.

ann=1

∞

∑ converges absolutely

III.

4ann=1

∞

∑ converges

(A) I only (B) II only (C) III only (D) II and III only (E) I, II and III

8. Which of the following series are conditionally convergent?

I.

−1( )nLn n( )n=2

∞

∑ II.

−1( )ntan−1 n( )n=1

∞

∑ III.

2n( )nn2n

n=1

∞

∑

(A) I only (B) I and III only (C) II and III only (D) I and III only (E) I, II, and III

9.

sin π6

⎛

⎝⎜⎞

⎠⎟

n

n=1

∞

∑ =

(A) 1 (B) 2 (C) 1− 3

2 (D)

32

1− 32

(E) divergent

10. Which of the following three tests will establish that the series

n2n5 +1n=1

∞

∑ converges?

432

I. Direct Comparison Test with

n−52

n=1

∞

∑

II. Limit Comparison Test with

n−32

n=1

∞

∑

III. Direct Comparison Test with

n−12

n=1

∞

∑

(A) I only (B) II only (C) I and II only (D) II and III only (E) I, II, and III Numerical Series Test Calculator Allowed

1. Use the Integral Test to determine if n2e−n3n=1

∞

∑ is convergent or divergent.

2. Use the Ratio Test to determine if 2n ⋅n3n!n=1

∞

∑ is convergent or divergent.

3. Determine whether the series nn + 4( )2n=1

∞

∑ converges or diverges.

6.1 Homework 1. 35 2. -382 3. -70 4. 5

5. 23613 6. 111 7. 2 is not in the arithmetic sequence.

8. 7th 9. 49, 56, 63

433

10. −10312 , −100, − 96 1

2 , − 93, − 89 12

11. 38 57 , 30 3

7 , 22 17 , 136

7 , 5 47 , − 2 5

7

12. arithmetic; d = 5 13. geometric; r = 13

14. geometric; r = 2 15. neither 16. neither 17. geometric; r = 5

−16 18. 354294 19. 2.815 x 1014

20. .1906030291 21. −32243 22. 1000

23. -6561256 24. 10 25. 7th

26. 15 and 45 27. 1105 , 1

21, 521 28. 5 3, 15, 15 3

434

6.2 Homework

1. 13760 2. 10 3. -7 4. 248

5. .243 6. 145 7. 125 8. -157.5 9. -143 10. 37 11. 43 12. 4 13. 14 14. 222 15. -14040 16. 531,440 17. -3.843 x 1017 18. n = 16, S = -131070 19. 25 20. 5.6 21. Divergent 6.3 Homework 1. Divergent 2. Convergent 3. Convergent 4. Convergent 5. Convergent 6. Convergent 7. Convergent 8. Might Converge 9. Might Converge 10. Divergent 11. Divergent 12. Might Converge 13. Convergent 14. Convergent 15. Convergent 16. Divergent 17. Convergent 18. Convergent 19. Convergent 20. Convergent 21. Divergent 22. Divergent 6.4 Homework

1. Convergent, 12

2. Convergent, 1ln2 3. Divergent

435

4. Convergent, 12

5. Convergent, 12 6. Convergent, 76

7. Convergent,π4 8. Divergent 9. Convergent, .0002

10. Convergent, 12e

6.5 Homework 1. Converges by LCT 2. Converges by LCT 3. Diverges by LCT 4. Converges by the Ratio Test 5. Diverges by the Divergence Test 6. Converges by the Ratio Test 7. Converges by DCT 8. Diverges by LCT 9. Diverges by LCT 10. Diverges by LCT 6.6 Homework 1. Diverges 2. Diverges 3. Conditionally Convergent 4. Diverges 5. Conditionally Convergent 6. Conditionally Conv. 7. Absolutely Convergent 8. Divergent 6.7 Homework 1. Divergent by Divergence Test 2. Converges by the Integral Test 3. Divergent by Ratio Test 4. Converges by Ratio Test 5. Divergent by Divergence Test 6. Converges by Ratio Test 7. Converges by Ratio Test 8. Converges by Ratio Test 9. Converges by Root Test 10. Diverges by Root Test

436

11a. Ratio Test inconclusive 11b. Ratio Test works 11c. Ratio Test works 11d. Ratio Test inconclusive 6.8 Homework 1. B 2. A 3. D 4. A 5. B 6. D 7. E 8. C 9. E 10. B