Lecture 24 : Seriesapilking/Math10560/Lectures/Lecture 24.pdf · Annette Pilkington Lecture 24 :...

Series Using limn→∞ sn to determine convergence Geometric Series geometric series not starting at n = 1 Repeating Decimals Telescoping Series. Harmonic Series. Where sum starts. Divergence Test Properties of Series

Series

So far our definition of a sum of numbers applies only to adding a finite set ofnumbers. We can extend this to a definition of a sum of an infinite set ofnumbers in much the same way as we extended our notion of the definiteintegral to an improper integral over an infinite interval.

I Example∞Xn=1

23+ . . .

We call this infinite sum a series

I Definition Given a seriesP∞

n=1 an = a1 + a2 + a3 + . . . , we let sn denoteits n th partial sum

sn = a1 + a2 + · · ·+ an.

I If the sequence {sn} is convergent and limn→∞ sn = S , then we say thatthe series

P∞n=1 an is convergent and we let

∞Xn=1

an = limn→∞

sn = S .

The number S is called the sum of the series. Otherwise the series iscalled divergent.

Annette Pilkington Lecture 24 : Series

Series

I Example∞Xn=1

23+ . . .

sn = a1 + a2 + · · ·+ an.

∞Xn=1

an = limn→∞

sn = S .

Series

I Example∞Xn=1

23+ . . .

sn = a1 + a2 + · · ·+ an.

∞Xn=1

an = limn→∞

sn = S .

Series

I Example∞Xn=1

23+ . . .

sn = a1 + a2 + · · ·+ an.

∞Xn=1

an = limn→∞

sn = S .

Using limn→∞ Sn to determine convergence/divergence

Example Find the partial sums s1, s2, s3, . . . , sn of the seriesP∞

n=112n .

Find the sum of this series. Does the series converge?

11�2n, n � 1...10

I We have s1 = 12, s2 = 1

4, s3 = 1

8. . .

I From the picture, we see that sn = 1− 12n .

n=112n = limn→∞ sn = limn→∞(1− 1

2n ) = 1.

I Therefore this series converges to S = 1.

I .... which you could have figured out from the picture :)

n=112n .

11�2n, n � 1...10

I We have s1 = 12, s2 = 1

4, s3 = 1

8. . .

n=112n = limn→∞ sn = limn→∞(1− 1

2n ) = 1.

n=112n .

11�2n, n � 1...10

I We have s1 = 12, s2 = 1

4, s3 = 1

8. . .

n=112n = limn→∞ sn = limn→∞(1− 1

2n ) = 1.

n=112n .

11�2n, n � 1...10

I We have s1 = 12, s2 = 1

4, s3 = 1

8. . .

n=112n = limn→∞ sn = limn→∞(1− 1

2n ) = 1.

n=112n .

11�2n, n � 1...10

I We have s1 = 12, s2 = 1

4, s3 = 1

8. . .

n=112n = limn→∞ sn = limn→∞(1− 1

2n ) = 1.

n=112n .

11�2n, n � 1...10

I We have s1 = 12, s2 = 1

4, s3 = 1

8. . .

n=112n = limn→∞ sn = limn→∞(1− 1

2n ) = 1.

n=112n .

11�2n, n � 1...10

I We have s1 = 12, s2 = 1

4, s3 = 1

8. . .

n=112n = limn→∞ sn = limn→∞(1− 1

2n ) = 1.

Example Recall that 1 + 2 + 3 + · · ·+ n = n(n+1)2

. Does the series

∞Xn=1

converge?

I We have the nth partial sum is sn = 1 + 2 + 3 + · · ·+ n = n(n+1)2

n=1 n = limn→∞ sn if it exists.

I limn→∞ sn = limn→∞n(n+1)

2= limx→∞

x(x+1)2

I = limx→∞x2+x

2=∞.

I Therefore this series diverges. (It does not have a finite sum)

. Does the series

∞Xn=1

converge?

2= limx→∞

x(x+1)2

I = limx→∞x2+x

2=∞.

. Does the series

∞Xn=1

converge?

2= limx→∞

x(x+1)2

I = limx→∞x2+x

2=∞.

. Does the series

∞Xn=1

converge?

2= limx→∞

x(x+1)2

I = limx→∞x2+x

2=∞.

. Does the series

∞Xn=1

converge?

2= limx→∞

x(x+1)2

I = limx→∞x2+x

2=∞.

. Does the series

∞Xn=1

converge?

2= limx→∞

x(x+1)2

I = limx→∞x2+x

2=∞.

Geometric series

The geometric series∞Xn=1

arn−1 = a + ar + ar 2 + · · ·

is convergent if |r | < 1 and its sum is

∞Xn=1

arn−1 =a

1− r|r | < 1.

If |r | ≥ 1, the geometric series is divergent.

Example Find the sum of the seriesP∞

n=1(−1)n10

4n−1 = −10 + 104− 10

16+ . . .

I We identify the values of a and r .

I a = first term = −10 in this case.

I The second term is ar , so r = term2/ term 1. Here r = 104/(−10) = −1

I Just to be sure that we are dealing with a geometric series, we check that

the n th term is arn. arn−1 = (−10) (−1)n−1

4n−1 , this is indeed the given n thterm.

I Therefore, since |r | < 1,P∞

n=1(−1)n10

4n−1 = a1−r

= −10

1− (−1)4

= −105/4

= −8.

Geometric series

arn−1 = a + ar + ar 2 + · · ·

∞Xn=1

arn−1 =a

1− r|r | < 1.

n=1(−1)n10

4n−1 = −10 + 104− 10

16+ . . .

n=1(−1)n10

4n−1 = a1−r

= −10

1− (−1)4

= −105/4

= −8.

Geometric series

arn−1 = a + ar + ar 2 + · · ·

∞Xn=1

arn−1 =a

1− r|r | < 1.

n=1(−1)n10

4n−1 = −10 + 104− 10

16+ . . .

n=1(−1)n10

4n−1 = a1−r

= −10

1− (−1)4

= −105/4

= −8.

Geometric series

arn−1 = a + ar + ar 2 + · · ·

∞Xn=1

arn−1 =a

1− r|r | < 1.

n=1(−1)n10

4n−1 = −10 + 104− 10

16+ . . .

n=1(−1)n10

4n−1 = a1−r

= −10

1− (−1)4

= −105/4

= −8.

Geometric series

arn−1 = a + ar + ar 2 + · · ·

∞Xn=1

arn−1 =a

1− r|r | < 1.

n=1(−1)n10

4n−1 = −10 + 104− 10

16+ . . .

n=1(−1)n10

4n−1 = a1−r

= −10

1− (−1)4

= −105/4

= −8.

Geometric series

arn−1 = a + ar + ar 2 + · · ·

∞Xn=1

arn−1 =a

1− r|r | < 1.

n=1(−1)n10

4n−1 = −10 + 104− 10

16+ . . .

n=1(−1)n10

4n−1 = a1−r

= −10

1− (−1)4

= −105/4

= −8.

Geometric series, another example

∞Xn=1

arn−1 = a + ar + ar 2 + · · ·

∞Xn=1

arn−1 =a

1− r|r | < 1.

n=123n = 2

27+ . . .

I a = first term = 2/3 in this case.

I The second term is ar , so r = term2/ term 1. Here r = 29/ 2

I Just to be sure that we are dealing with a geometric series, we check thatthe n th term is arn. arn−1 = (2/3) 1

3n−1 = 23n ,as required.

n=123n = a

1−r= 2/3

1− (1)3

= 2/32/3

∞Xn=1

arn−1 = a + ar + ar 2 + · · ·

∞Xn=1

arn−1 =a

1− r|r | < 1.

n=123n = 2

27+ . . .

n=123n = a

1−r= 2/3

1− (1)3

= 2/32/3

∞Xn=1

arn−1 = a + ar + ar 2 + · · ·

∞Xn=1

arn−1 =a

1− r|r | < 1.

n=123n = 2

27+ . . .

n=123n = a

1−r= 2/3

1− (1)3

= 2/32/3

∞Xn=1

arn−1 = a + ar + ar 2 + · · ·

∞Xn=1

arn−1 =a

1− r|r | < 1.

n=123n = 2

27+ . . .

n=123n = a

1−r= 2/3

1− (1)3

= 2/32/3

∞Xn=1

arn−1 = a + ar + ar 2 + · · ·

∞Xn=1

arn−1 =a

1− r|r | < 1.

n=123n = 2

27+ . . .

n=123n = a

1−r= 2/3

1− (1)3

= 2/32/3

∞Xn=1

arn−1 = a + ar + ar 2 + · · ·

∞Xn=1

arn−1 =a

1− r|r | < 1.

n=123n = 2

27+ . . .

n=123n = a

1−r= 2/3

1− (1)3

= 2/32/3

geometric series not starting at n = 1

Example Find the sum of the series

∞Xn=4

2n−1

I Note that this sequence starts at n = 4, so the formula for the sum doesnot apply as it stands.

I We can use two approaches, use the formula and subtract the missingterms or expand the series and rewrite it.

I Approach 1:P∞n=4

2n−1

3n =P∞

n=12n−1

3n −ˆ

+ 232 + 22

˜= 1/3

1−2/3−ˆ

32+6+22

˜= 1− 19

I Approach 2: rewrite the formula so that the sum starts at 1.

n=42n−1

3n = 23

34 + 24

35 + 25

36 + · · · = a + ar + ar 2 + · · · =P∞

n=1 arn−1

I a = term 1 = 23

34 , ar = term 2 = 24

35 . Therefore r = 24

35 /23

34 = 23.

I we check that 23

34 + 24

35 + 25

36 + · · · =P∞

n=1 arn−1 (true).

I Since |r | = 23< 1, we see that 23

34 + 24

35 + 25

36 + · · · = 23

34 /(1− 23) = 23

33 = 827.

∞Xn=4

2n−1

3n =P∞

n=12n−1

3n −ˆ

+ 232 + 22

˜= 1/3

1−2/3−ˆ

32+6+22

˜= 1− 19

n=42n−1

3n = 23

34 + 24

35 + 25

36 + · · · = a + ar + ar 2 + · · · =P∞

n=1 arn−1

I a = term 1 = 23

34 , ar = term 2 = 24

35 /23

34 = 23.

I we check that 23

34 + 24

35 + 25

36 + · · · =P∞

n=1 arn−1 (true).

34 + 24

35 + 25

36 + · · · = 23

34 /(1− 23) = 23

33 = 827.

∞Xn=4

2n−1

3n =P∞

n=12n−1

3n −ˆ

+ 232 + 22

˜= 1/3

1−2/3−ˆ

32+6+22

˜= 1− 19

n=42n−1

3n = 23

34 + 24

35 + 25

36 + · · · = a + ar + ar 2 + · · · =P∞

n=1 arn−1

I a = term 1 = 23

34 , ar = term 2 = 24

35 /23

34 = 23.

I we check that 23

34 + 24

35 + 25

36 + · · · =P∞

n=1 arn−1 (true).

34 + 24

35 + 25

36 + · · · = 23

34 /(1− 23) = 23

33 = 827.

∞Xn=4

2n−1

3n =P∞

n=12n−1

3n −ˆ

+ 232 + 22

˜= 1/3

1−2/3−ˆ

32+6+22

˜= 1− 19

n=42n−1

3n = 23

34 + 24

35 + 25

36 + · · · = a + ar + ar 2 + · · · =P∞

n=1 arn−1

I a = term 1 = 23

34 , ar = term 2 = 24

35 /23

34 = 23.

I we check that 23

34 + 24

35 + 25

36 + · · · =P∞

n=1 arn−1 (true).

34 + 24

35 + 25

36 + · · · = 23

34 /(1− 23) = 23

33 = 827.

∞Xn=4

2n−1

3n =P∞

n=12n−1

3n −ˆ

+ 232 + 22

˜= 1/3

1−2/3−ˆ

32+6+22

˜= 1− 19

n=42n−1

3n = 23

34 + 24

35 + 25

36 + · · · = a + ar + ar 2 + · · · =P∞

n=1 arn−1

I a = term 1 = 23

34 , ar = term 2 = 24

35 /23

34 = 23.

I we check that 23

34 + 24

35 + 25

36 + · · · =P∞

n=1 arn−1 (true).

34 + 24

35 + 25

36 + · · · = 23

34 /(1− 23) = 23

33 = 827.

∞Xn=4

2n−1

3n =P∞

n=12n−1

3n −ˆ

+ 232 + 22

˜= 1/3

1−2/3−ˆ

32+6+22

˜= 1− 19

n=42n−1

3n = 23

34 + 24

35 + 25

36 + · · · = a + ar + ar 2 + · · · =P∞

n=1 arn−1

I a = term 1 = 23

34 , ar = term 2 = 24

35 /23

34 = 23.

I we check that 23

34 + 24

35 + 25

36 + · · · =P∞

n=1 arn−1 (true).

34 + 24

35 + 25

36 + · · · = 23

34 /(1− 23) = 23

33 = 827.

∞Xn=4

2n−1

3n =P∞

n=12n−1

3n −ˆ

+ 232 + 22

˜= 1/3

1−2/3−ˆ

32+6+22

˜= 1− 19

n=42n−1

3n = 23

34 + 24

35 + 25

36 + · · · = a + ar + ar 2 + · · · =P∞

n=1 arn−1

I a = term 1 = 23

34 , ar = term 2 = 24

35 /23

34 = 23.

I we check that 23

34 + 24

35 + 25

36 + · · · =P∞

n=1 arn−1 (true).

34 + 24

35 + 25

36 + · · · = 23

34 /(1− 23) = 23

33 = 827.

∞Xn=4

2n−1

3n =P∞

n=12n−1

3n −ˆ

+ 232 + 22

˜= 1/3

1−2/3−ˆ

32+6+22

˜= 1− 19

n=42n−1

3n = 23

34 + 24

35 + 25

36 + · · · = a + ar + ar 2 + · · · =P∞

n=1 arn−1

I a = term 1 = 23

34 , ar = term 2 = 24

35 /23

34 = 23.

I we check that 23

34 + 24

35 + 25

36 + · · · =P∞

n=1 arn−1 (true).

34 + 24

35 + 25

36 + · · · = 23

34 /(1− 23) = 23

33 = 827.

∞Xn=4

2n−1

3n =P∞

n=12n−1

3n −ˆ

+ 232 + 22

˜= 1/3

1−2/3−ˆ

32+6+22

˜= 1− 19

n=42n−1

3n = 23

34 + 24

35 + 25

36 + · · · = a + ar + ar 2 + · · · =P∞

n=1 arn−1

I a = term 1 = 23

34 , ar = term 2 = 24

35 /23

34 = 23.

I we check that 23

34 + 24

35 + 25

36 + · · · =P∞

n=1 arn−1 (true).

34 + 24

35 + 25

36 + · · · = 23

34 /(1− 23) = 23

33 = 827.

Repeating Decimals

Example Write the number 0.66666666 · · · = 0.6̄ as a fraction.

I 0.666666666 · · · = 610

+ 6100

+ 61000

+ . . .

I = 610

+ 6102 + 6

103 + . . . (a geometric series).

I a = 610

and r = 6102 /

I we check 610

+ 6102 + 6

103 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore 0.666666666 · · · = 610

+ 6102 + 6

103 + · · · = 6/101−1/10

= 6/9 = 2/3

I as suspected :)

Repeating Decimals

I 0.666666666 · · · = 610

+ 6100

+ 61000

+ . . .

I = 610

+ 6102 + 6

I a = 610

and r = 6102 /

I we check 610

+ 6102 + 6

103 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore 0.666666666 · · · = 610

+ 6102 + 6

103 + · · · = 6/101−1/10

= 6/9 = 2/3

I as suspected :)

Repeating Decimals

I 0.666666666 · · · = 610

+ 6100

+ 61000

+ . . .

I = 610

+ 6102 + 6

I a = 610

and r = 6102 /

I we check 610

+ 6102 + 6

103 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore 0.666666666 · · · = 610

+ 6102 + 6

103 + · · · = 6/101−1/10

= 6/9 = 2/3

I as suspected :)

Repeating Decimals

I 0.666666666 · · · = 610

+ 6100

+ 61000

+ . . .

I = 610

+ 6102 + 6

I a = 610

and r = 6102 /

I we check 610

+ 6102 + 6

103 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore 0.666666666 · · · = 610

+ 6102 + 6

103 + · · · = 6/101−1/10

= 6/9 = 2/3

I as suspected :)

Repeating Decimals

I 0.666666666 · · · = 610

+ 6100

+ 61000

+ . . .

I = 610

+ 6102 + 6

I a = 610

and r = 6102 /

I we check 610

+ 6102 + 6

103 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore 0.666666666 · · · = 610

+ 6102 + 6

103 + · · · = 6/101−1/10

= 6/9 = 2/3

I as suspected :)

Repeating Decimals

I 0.666666666 · · · = 610

+ 6100

+ 61000

+ . . .

I = 610

+ 6102 + 6

I a = 610

and r = 6102 /

I we check 610

+ 6102 + 6

103 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore 0.666666666 · · · = 610

+ 6102 + 6

103 + · · · = 6/101−1/10

= 6/9 = 2/3

I as suspected :)

Repeating Decimals

I 0.666666666 · · · = 610

+ 6100

+ 61000

+ . . .

I = 610

+ 6102 + 6

I a = 610

and r = 6102 /

I we check 610

+ 6102 + 6

103 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore 0.666666666 · · · = 610

+ 6102 + 6

103 + · · · = 6/101−1/10

= 6/9 = 2/3

I as suspected :)

Repeating Decimals

Example Write the number 1.521212121 · · · = 1.52̄1 as a fraction.

I 1.521212121 · · · = 1.5 + 21103 + 21

105 + 21107 + . . .

I 21103 + 21

105 + 21107 + . . . is a geometric series.

I a = 21103 and r = 21

105 /21103 = 1

I we double check 21103 + 21

105 + 21107 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore1.521212121 · · · = 1.5+ 21

103 + 21105 + 21

107 +· · · = 1.5+ 21/103

1−1/102 = 3/2+21/990

I = 3/2 + 7/330 = 1004/660 = 251/165

Repeating Decimals

I 1.521212121 · · · = 1.5 + 21103 + 21

105 + 21107 + . . .

I 21103 + 21

I a = 21103 and r = 21

105 /21103 = 1

105 + 21107 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore1.521212121 · · · = 1.5+ 21

103 + 21105 + 21

107 +· · · = 1.5+ 21/103

1−1/102 = 3/2+21/990

I = 3/2 + 7/330 = 1004/660 = 251/165

Repeating Decimals

I 1.521212121 · · · = 1.5 + 21103 + 21

105 + 21107 + . . .

I 21103 + 21

I a = 21103 and r = 21

105 /21103 = 1

105 + 21107 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore1.521212121 · · · = 1.5+ 21

103 + 21105 + 21

107 +· · · = 1.5+ 21/103

1−1/102 = 3/2+21/990

I = 3/2 + 7/330 = 1004/660 = 251/165

Repeating Decimals

I 1.521212121 · · · = 1.5 + 21103 + 21

105 + 21107 + . . .

I 21103 + 21

I a = 21103 and r = 21

105 /21103 = 1

105 + 21107 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore1.521212121 · · · = 1.5+ 21

103 + 21105 + 21

107 +· · · = 1.5+ 21/103

1−1/102 = 3/2+21/990

I = 3/2 + 7/330 = 1004/660 = 251/165

Repeating Decimals

I 1.521212121 · · · = 1.5 + 21103 + 21

105 + 21107 + . . .

I 21103 + 21

I a = 21103 and r = 21

105 /21103 = 1

105 + 21107 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore1.521212121 · · · = 1.5+ 21

103 + 21105 + 21

107 +· · · = 1.5+ 21/103

1−1/102 = 3/2+21/990

I = 3/2 + 7/330 = 1004/660 = 251/165

Repeating Decimals

I 1.521212121 · · · = 1.5 + 21103 + 21

105 + 21107 + . . .

I 21103 + 21

I a = 21103 and r = 21

105 /21103 = 1

105 + 21107 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore1.521212121 · · · = 1.5+ 21

103 + 21105 + 21

107 +· · · = 1.5+ 21/103

1−1/102 = 3/2+21/990

I = 3/2 + 7/330 = 1004/660 = 251/165

Repeating Decimals

I 1.521212121 · · · = 1.5 + 21103 + 21

105 + 21107 + . . .

I 21103 + 21

I a = 21103 and r = 21

105 /21103 = 1

105 + 21107 + · · · = a + ar + ar 2 + . . . (it is)

I Therefore1.521212121 · · · = 1.5+ 21

103 + 21105 + 21

107 +· · · = 1.5+ 21/103

1−1/102 = 3/2+21/990

I = 3/2 + 7/330 = 1004/660 = 251/165

Telescoping Series.

These are series of the form similar toP

f (n)− f (n + 1). Because of the largeamount of cancellation, they are relatively easy to sum.

Example Show that the series

∞Xk=1

k2 + 7k + 12=∞X

(k + 3)− 1

(k + 4)

converges.

I S1 = 14− 1

I S2 = 14− 1

5− 1

4− 1

I S3 = 14− 1

5− 1

6− 1

4− 1

I Sn = 14− 1

5− 1

6+ · · · + 1

(n+3)− 1

(n+4)= 1

4− 1

(n+4).

(k+3)− 1

(k+4)= limn→∞ Sn = limn→∞

h14− 1

I Also check the extra example in your notes.

Telescoping Series.

∞Xk=1

k2 + 7k + 12=∞X

(k + 3)− 1

(k + 4)

converges.

I S1 = 14− 1

I S2 = 14− 1

5− 1

4− 1

I S3 = 14− 1

5− 1

6− 1

4− 1

I Sn = 14− 1

5− 1

6+ · · · + 1

(n+3)− 1

(n+4)= 1

4− 1

(n+4).

(k+3)− 1

h14− 1

Telescoping Series.

∞Xk=1

k2 + 7k + 12=∞X

(k + 3)− 1

(k + 4)

converges.

I S1 = 14− 1

I S2 = 14− 1

5− 1

4− 1

I S3 = 14− 1

5− 1

6− 1

4− 1

I Sn = 14− 1

5− 1

6+ · · · + 1

(n+3)− 1

(n+4)= 1

4− 1

(n+4).

(k+3)− 1

h14− 1

Telescoping Series.

∞Xk=1

k2 + 7k + 12=∞X

(k + 3)− 1

(k + 4)

converges.

I S1 = 14− 1

I S2 = 14− 1

5− 1

4− 1

I S3 = 14− 1

5− 1

6− 1

4− 1

I Sn = 14− 1

5− 1

6+ · · · + 1

(n+3)− 1

(n+4)= 1

4− 1

(n+4).

(k+3)− 1

h14− 1

Telescoping Series.

∞Xk=1

k2 + 7k + 12=∞X

(k + 3)− 1

(k + 4)

converges.

I S1 = 14− 1

I S2 = 14− 1

5− 1

4− 1

I S3 = 14− 1

5− 1

6− 1

4− 1

I Sn = 14− 1

5− 1

6+ · · · + 1

(n+3)− 1

(n+4)= 1

4− 1

(n+4).

(k+3)− 1

h14− 1

Telescoping Series.

∞Xk=1

k2 + 7k + 12=∞X

(k + 3)− 1

(k + 4)

converges.

I S1 = 14− 1

I S2 = 14− 1

5− 1

4− 1

I S3 = 14− 1

5− 1

6− 1

4− 1

I Sn = 14− 1

5− 1

6+ · · · + 1

(n+3)− 1

(n+4)= 1

4− 1

(n+4).

(k+3)− 1

h14− 1

Telescoping Series.

∞Xk=1

k2 + 7k + 12=∞X

(k + 3)− 1

(k + 4)

converges.

I S1 = 14− 1

I S2 = 14− 1

5− 1

4− 1

I S3 = 14− 1

5− 1

6− 1

4− 1

I Sn = 14− 1

5− 1

6+ · · · + 1

(n+3)− 1

(n+4)= 1

4− 1

(n+4).

(k+3)− 1

h14− 1

Harmonic Series.

The following series, known as the harmonic series, diverges:

∞Xk=1

ndiverges

I We can see this if we look at a subsequence of partial sums: {s2n}.I s1 = 1, s2 = 1 + 1

I s4 = 1 + 12

i> 1 + 1

I s8 = 1 + 12

i> s4 +

i> 2 + 1

I Similarly we get

s2n >n + 2

and limn→∞ sn > limn→∞n+22

=∞. Hence the harmonic series diverges.(You will see an easier proof in the next section. )

Harmonic Series.

∞Xk=1

ndiverges

I We can see this if we look at a subsequence of partial sums: {s2n}.

I s1 = 1, s2 = 1 + 12

I s4 = 1 + 12

i> 1 + 1

I s8 = 1 + 12

i> s4 +

i> 2 + 1

I Similarly we get

s2n >n + 2

Harmonic Series.

∞Xk=1

ndiverges

I s4 = 1 + 12

i> 1 + 1

I s8 = 1 + 12

i> s4 +

i> 2 + 1

I Similarly we get

s2n >n + 2

Harmonic Series.

∞Xk=1

ndiverges

I s4 = 1 + 12

i> 1 + 1

I s8 = 1 + 12

i> s4 +

i> 2 + 1

I Similarly we get

s2n >n + 2

Harmonic Series.

∞Xk=1

ndiverges

I s4 = 1 + 12

i> 1 + 1

I s8 = 1 + 12

i> s4 +

i> 2 + 1

I Similarly we get

s2n >n + 2

Harmonic Series.

∞Xk=1

ndiverges

I s4 = 1 + 12

i> 1 + 1

I s8 = 1 + 12

i> s4 +

i> 2 + 1

I Similarly we get

s2n >n + 2

Where sum starts.

Note thatconvergence or divergence is unaffected by adding or deleting a finite number of termsat the beginning of the series.

Example∞X

nis divergent

and∞X

2kis convergent.

Divergence Test

Theorem If a seriesP∞

i=1 an is convergent, then limn→∞ an = 0.

Warning The converse is not true , we may have a series where limn→∞ an = 0

and the series in divergent. For example, the harmonic series.

Proof Suppose the series∑∞

i=1 an is convergent with sum S. Sincean = sn − sn−1 and

limn→∞

sn = limn→∞

sn−1 = S

we have limn→∞ an = limn→∞ sn − limn→∞ sn−1 = S − S = 0.

This gives us a Test for Divergence:

If limn→∞ an does not exist or if limn→∞ an 6= 0, then∑∞

i=1 an is divergent.

If limn→∞ an = 0 the test is inconclusive.

Divergence Test

If limn→∞ an does not exist or if limn→∞ an 6= 0, thenP∞

Example Test the following series for divergence with the above test:∞Xn=1

n2 + 1

∞Xn=1

n2 + 1

∞Xn=1

n2 + 1

I To testP∞

n=1n2+12n2 for convergence, we check limn→∞

n2+12n2 .

I limn→∞n2+12n2 = limn→∞

1+1/n2

26= 0.

I Therefore, we can conclude thatP∞

n=1n2+12n2 diverges.

I To testP∞

n2+12n3 .

1+1/n2

2n= 0.

I In this case we can make no conclusion aboutP∞

n=1n2+12n3 .

I To testP∞

n=1n2+12n

for convergence, we check limn→∞n2+12n

I limn→∞n2+12n

= limn→∞n+1/n

2=∞ 6= 0.

n=1n2+12n

diverges.

Divergence Test

n2 + 1

∞Xn=1

n2 + 1

∞Xn=1

n2 + 1

I To testP∞

n2+12n2 .

1+1/n2

26= 0.

I To testP∞

n2+12n3 .

1+1/n2

2n= 0.

n=1n2+12n3 .

I To testP∞

n=1n2+12n

I limn→∞n2+12n

= limn→∞n+1/n

2=∞ 6= 0.

n=1n2+12n

diverges.

Divergence Test

n2 + 1

∞Xn=1

n2 + 1

∞Xn=1

n2 + 1

I To testP∞

n2+12n2 .

1+1/n2

26= 0.

I To testP∞

n2+12n3 .

1+1/n2

2n= 0.

n=1n2+12n3 .

I To testP∞

n=1n2+12n

I limn→∞n2+12n

= limn→∞n+1/n

2=∞ 6= 0.

n=1n2+12n

diverges.

Divergence Test

n2 + 1

∞Xn=1

n2 + 1

∞Xn=1

n2 + 1

I To testP∞

n2+12n2 .

1+1/n2

26= 0.

I To testP∞

n2+12n3 .

1+1/n2

2n= 0.

n=1n2+12n3 .

I To testP∞

n=1n2+12n

I limn→∞n2+12n

= limn→∞n+1/n

2=∞ 6= 0.

n=1n2+12n

diverges.

Divergence Test

n2 + 1

∞Xn=1

n2 + 1

∞Xn=1

n2 + 1

I To testP∞

n2+12n2 .

1+1/n2

26= 0.

I To testP∞

n2+12n3 .

1+1/n2

2n= 0.

n=1n2+12n3 .

I To testP∞

n=1n2+12n

I limn→∞n2+12n

= limn→∞n+1/n

2=∞ 6= 0.

n=1n2+12n

diverges.

Divergence Test

n2 + 1

∞Xn=1

n2 + 1

∞Xn=1

n2 + 1

I To testP∞

n2+12n2 .

1+1/n2

26= 0.

I To testP∞

n2+12n3 .

1+1/n2

2n= 0.

n=1n2+12n3 .

I To testP∞

n=1n2+12n

I limn→∞n2+12n

= limn→∞n+1/n

2=∞ 6= 0.

n=1n2+12n

diverges.

Divergence Test

n2 + 1

∞Xn=1

n2 + 1

∞Xn=1

n2 + 1

I To testP∞

n2+12n2 .

1+1/n2

26= 0.

I To testP∞

n2+12n3 .

1+1/n2

2n= 0.

n=1n2+12n3 .

I To testP∞

n=1n2+12n

I limn→∞n2+12n

= limn→∞n+1/n

2=∞ 6= 0.

n=1n2+12n

diverges.

Divergence Test

n2 + 1

∞Xn=1

n2 + 1

∞Xn=1

n2 + 1

I To testP∞

n2+12n2 .

1+1/n2

26= 0.

I To testP∞

n2+12n3 .

1+1/n2

2n= 0.

n=1n2+12n3 .

I To testP∞

n=1n2+12n

I limn→∞n2+12n

= limn→∞n+1/n

2=∞ 6= 0.

n=1n2+12n

diverges.

Divergence Test

n2 + 1

∞Xn=1

n2 + 1

∞Xn=1

n2 + 1

I To testP∞

n2+12n2 .

1+1/n2

26= 0.

I To testP∞

n2+12n3 .

1+1/n2

2n= 0.

n=1n2+12n3 .

I To testP∞

n=1n2+12n

I limn→∞n2+12n

= limn→∞n+1/n

2=∞ 6= 0.

n=1n2+12n

diverges.

Divergence Test

n2 + 1

∞Xn=1

n2 + 1

∞Xn=1

n2 + 1

I To testP∞

n2+12n2 .

1+1/n2

26= 0.

I To testP∞

n2+12n3 .

1+1/n2

2n= 0.

n=1n2+12n3 .

I To testP∞

n=1n2+12n

I limn→∞n2+12n

= limn→∞n+1/n

2=∞ 6= 0.

n=1n2+12n

diverges.

Properties of Series

The following properties of series follow from the corresponding laws of limits:

SupposeP

an andP

bn are convergent series, then the seriesP(an + bn),

P(an − bn) and

Pcan also converge. We haveX

can = cX

(an+bn) =X

(an−bn) =X

an−X

Example Sum the following series:

∞Xn=0

3 + 2n

πn+1.

n=03+2n

πn+1 =P∞

πn+1 +P∞

πn+1 .

πn+1 = 3π

+ 3π2 + · · · = 3/π

1−1/π= 3

(π−1)since r = 1

π< 1.

πn+1 = 1π

+ 2π2 + · · · = 1/π

1−2/π= 1

(π−2), since r = 2

π< 1.

n=03+2n

πn+1 = 3(π−1)

+ 1(π−2)

= 4π−7(π−1)(π−2)

SupposeP

an andP

P(an − bn) and

can = cX

(an+bn) =X

(an−bn) =X

an−X

∞Xn=0

3 + 2n

πn+1.

n=03+2n

πn+1 =P∞

πn+1 +P∞

πn+1 .

πn+1 = 3π

+ 3π2 + · · · = 3/π

1−1/π= 3

(π−1)since r = 1

π< 1.

πn+1 = 1π

+ 2π2 + · · · = 1/π

1−2/π= 1

(π−2), since r = 2

π< 1.

n=03+2n

πn+1 = 3(π−1)

+ 1(π−2)

= 4π−7(π−1)(π−2)

SupposeP

an andP

P(an − bn) and

can = cX

(an+bn) =X

(an−bn) =X

an−X

∞Xn=0

3 + 2n

πn+1.

n=03+2n

πn+1 =P∞

πn+1 +P∞

πn+1 .

πn+1 = 3π

+ 3π2 + · · · = 3/π

1−1/π= 3

(π−1)since r = 1

π< 1.

πn+1 = 1π

+ 2π2 + · · · = 1/π

1−2/π= 1

(π−2), since r = 2

π< 1.

n=03+2n

πn+1 = 3(π−1)

+ 1(π−2)

= 4π−7(π−1)(π−2)

SupposeP

an andP

P(an − bn) and

can = cX

(an+bn) =X

(an−bn) =X

an−X

∞Xn=0

3 + 2n

πn+1.

n=03+2n

πn+1 =P∞

πn+1 +P∞

πn+1 .

πn+1 = 3π

+ 3π2 + · · · = 3/π

1−1/π= 3

(π−1)since r = 1

π< 1.

πn+1 = 1π

+ 2π2 + · · · = 1/π

1−2/π= 1

(π−2), since r = 2

π< 1.

n=03+2n

πn+1 = 3(π−1)

+ 1(π−2)

= 4π−7(π−1)(π−2)

SupposeP

an andP

P(an − bn) and

can = cX

(an+bn) =X

(an−bn) =X

an−X

∞Xn=0

3 + 2n

πn+1.

n=03+2n

πn+1 =P∞

πn+1 +P∞

πn+1 .

πn+1 = 3π

+ 3π2 + · · · = 3/π

1−1/π= 3

(π−1)since r = 1

π< 1.

πn+1 = 1π

+ 2π2 + · · · = 1/π

1−2/π= 1

(π−2), since r = 2

π< 1.

n=03+2n

πn+1 = 3(π−1)

+ 1(π−2)

= 4π−7(π−1)(π−2)

Lecture 24 : Seriesapilking/Math10560/Lectures/Lecture 24.pdf · Annette Pilkington Lecture 24 :...

Documents

Lecture 20/21 : First and second order Linear Differential ...apilking/Math10560/Lectures/Lecture 20.pdf · Annette Pilkington Lecture 20/21 : First and second order Linear Di erential

Lecture 23 : Sequencesapilking/Math10560/Calc1Lectures/Lecture 5-… · Annette Pilkington Lecture 23 : Sequences. Continuous functionsContinuous from left and rightContinuity on

Lecture 26 : Comparison Testapilking/Math10560/Lectures/Lecture... · 2020-01-13 · Comparison Test In this section, as we did with improper integrals, we see how to compare a series

Natural Logarithm and Natural Exponentialapilking/Math10560/Lectures/Lecture... · 2020-01-13 · Natural Logarithm FunctionGraph of Natural LogarithmAlgebraic Properties of ln(x)

Left endpoint approximation - University of Notre Dameapilking/Math10560/Calc1Lectures/24... · 2018-03-17 · Approximating Area under a curve with rectangles To nd the area under

First Order Linear Di erential Equationsapilking/Math10560/Lectures/Lecture... · 2020-03-16 · First Order Linear Di erential EquationsExamplesSecond Order Linear Di erential EquationsInitial

Lecture 36: Polar Coordinatesapilking/Math10560/Lectures/Lecture 36.pdfPolar Co-ordinatesPolar to Cartesian coordinatesCartesian to Polar coordinatesExample 3Graphing Equations in

Curves de ned by Parametric equationsapilking/Math10560/Lectures/Lecture 34.pdf · Curves de ned by Parametric equationsExample 2Example 3Converting Par to CartConverting Cart to

Lecture 24

Lecture 16 : Arc Lengthapilking/Math10560/Lectures/Lecture 16.pdf · Arc Length Arc Lenth In this section, we derive a formula for the length of a curve y = f(x) on an interval [a;b]

Lecture 36: Polar Coordinates - University of Notre Dameapilking/Math10560/Lectures/Lecture 36.pdf · Annette Pilkington Lecture 36: Polar Coordinates

exp(x) = inverse of ln(x - University of Notre Dameapilking/Math10560/Lectures/Lecture 3.pdfNatural Logarithm FunctionGraph of Natural LogarithmAlgebraic Properties of ln(x) LimitsExtending

Lecture 1 : Precalculus Reviewapilking/Math10560/Calc1Lectures/1. Review Precalc .pdfLecture 1 : Precalculus Review Hyperlinks are shown in blue, download the cdf player from the Wolfram

Lecture 24

Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture

Lecture 22 : NonHomogeneous Linear Equations (Section …apilking/Math10560/Lectures/Lecture 22.pdf · homogeneous equation ay00+ by0+ cy = 0. I Since we already know how to nd y

Improper Integrals - University of Notre Dameapilking/Math10560/Lectures/Lecture 15.pdf · Improper IntegralsIn nite IntervalsArea InterpretationTheorem 1Functions with in nite discontinuitiesComparison

Lecture 16 : Arc Lengthapilking/Math10560/Lectures/Lecture 16.pdf · Annette Pilkington Lecture 16 : Arc Length. Arc Length Arc Length If f is continuous and di erentiable on the

Lecture 1 : Precalculus Revieapilking/Math10560/Calc1Lectures/1. Review Precalc .pdf · Lecture 1 : Precalculus Review ... Vertical Line Test (VLT) The graph of an equation passes

Lecture 27 :Alternating Series - University of Notre Dameapilking/Math10560/Lectures/Lecture 27.pdf · Alternating Series The integral test and the comparison test given in previous