Original subtitles

1

In this video we're talking about how to use the ratio test to say whether or not a series converges

2

or diverges.

3

And in this particular problem we've been given the series the some from N equals 1 to infinity of three

4

to the power divided by and squared.

5

And as a reminder here we have the ratio test the ratio test is just one of our many convergence tests.

6

And what this one says is that if we find a value L So this is our critical value here the important

7

value when it comes to the ratio test L and L is equal to the limit is and goes to infinity of the absolute

8

value of this quotient here.

9

A seven plus one divided by a.

10

Well a sub n is the original series we've been given so we can go ahead and say here that a 7 is this

11

original series we've been given.

12

So we'll plug this value here in for a second.

13

A seven plus one is going to be whatever we get when we replace n with end plus 1 and this original

14

series will take this original series everywhere.

15

We have an N.

16

We'll put an end plus one instead.

17

And whatever value we get there we'll go in for a seven plus one.

18

So then we'll have the absolute value of this quotient of series.

19

We'll simplify it will take the limit is and goes to infinity and we'll find some value for L.

20

Once we do we can use the value of L to say whether or not the series converges because if L is less

21

than 1 by the ratio test the series converges.

22

If Ellas greater than one then it diverges.

23

And if L is equal to 1 then the ratio tests in particular is inconclusive and we'd have to try to use

24

another test or just say that we can't draw a conclusion about the convergence of the series.

25

So let's go ahead and work through this problem.

26

Again the first thing we need to do is find L.

27

So we'll go ahead and say L is equal to the limit as and goes to infinity at the absolute value of a

28

sub and plus 1.

29

So we're going to take this original series and we're going to replace and with and plus one.

30

So instead of three to the end we're going to get three to the end plus one.

31

And then in the denominator here instead of end squared we're going to get and plus 1 squared.

32

We have to remember to put that in parentheses so we square the entire quantity and plus 1.

33

So that's a seven plus one.

34

Then we're going to divide that by a seven and a 7 is just the original series so we're going to divide

35

this whole thing by a 7 which is three to the end divided by ends squared.

36

We're taking the absolute value of this whole thing.

37

Now what we realize is that in this case we have a fraction divided by a fraction.

38

And I remember when we are dividing fractions what we can do is take this fraction in the numerator

39

and leave it exactly as is who will say the limit as and goes to infinity.

40

Of this fraction in the numerator so three to the end plus 1 divided by and plus 1 quantities squared.

41

And then instead of dividing by this fraction in the denominator we can multiply by a reciprocal those

42

two things are the same.

43

So we just flip this upside down instead of 3 to the end over end squared.

44

We get and squared over three to the end and we just change the division to multiplication.

45

Now when you're doing the ratio test because this is always going to be the case when you have a fraction

46

over a fraction.

47

A lot of people just go straight to the second step and take the a seven plus one and multiply by the

48

reciprocal of a 7 and skip this first step.

49

Here you can do that if you want.

50

If you feel comfortable or you can write out the original quotient of fractions here and then flip this

51

denominator upside down find it's reciprocal and change the division to multiplication whichever one

52

you prefer.

53

But ultimately we're getting here to this second step and once we're to this point what we want to do

54

is look for terms that are similar and usually that means terms that have a like base.

55

So for example here we have three race to the end plus one the base is three.

56

Here we have three to the end.

57

The base is still three.

58

So what we can do is say that these two terms here are similar to one another.

59

Therefore we want to put them together in the scene fraction and you'll get better at recognizing terms

60

that are similar in these types of problems so what we're going to do is we're going to pair like terms.

61

So because these fractions are multiplied together we can swap numerators and denominators easily.

62

So we're just going to swap these two denominators to get like terms together so we're going to say

63

three is the end plus one.

64

We're going to put this denominator with this numerator and say three to the N and then multiply by

65

and squared over and plus one quantity squared.

66

And now you can see how these two terms are similar and these two terms are similar more so than the

67

arrangement that we had in this second step.

68

The reason that this is helpful is because when terms have like bases you can subtract the exponent

69

in the denominator from the exponent in the numerator.

70

So for example here this three to the end plus one divided by three to the N is the same as 3 to the

71

end plus 1 the exponent from the numerator minus the exponent in the denominator which is just N and

72

then we have.

73

And plus one minus Anwyl and minus N is zero.

74

Those two things cancel and we just have three to the first power three to the one which is just three.

75

So this whole fraction here this whole first fraction becomes three.

76

And because now we've eliminated ends we can pull that 3 out in front of the limit so we'll go and say

77

that Al is going to be equal to three times the limit as and goes to infinity of the absolute value

78

of square divided by and plus one quantity squared.

79

At this point now we want to go ahead and expand the denominator of this fraction.

80

So we have and plus one quantity squared.

81

That's the same thing as Plus one times and plus one right.

82

So we want to foil this out so end times and it's going to give us and squared and then we're going

83

to have plus and plus and it's going to be plus two and then one time's one plus one.

84

So we end up with this value here.

85

Now at this point I want to take the highest degree variable in the denominator and divide through both

86

the numerator and denominator by that value.

87

So the highest degree variable in the denominator is and squared.

88

That means we want to divide every term in the numerator and denominator by end squared so what we can

89

really say here is we can multiply the numerator here by one over and squared and we're going to multiply

90

the denominator by 1 over and squared like this.

91

This is the same thing because the numerator and denominator are the same.

92

This is really the same thing as multiplying by 1.

93

So we're not actually changing the value.

94

But the reason that this helps us out is because we get limit and to infinity is because here we take

95

and squared times one over and squared the squares here cancel and we're just left with one in the numerator

96

and denominator we have to take this one over and squared and multiply it by all three of these terms

97

here.

98

The end squared the two ends and the one so one over and squared times and squared the square it's cancell

99

and we get 1 2 and times 1 over and squared we would have to an over and squared this and here would

100

cancel and we'd be left with 1 and we'd have 2 over.

101

And so we'll get plus two over an and then 1 times 1 over and square it is of course just won over and

102

squared like this.

103

Now when we evaluated the limit as and goes to infinity we have just and in these denominators here

104

and then we have ended going to infinity in the denominator and just a constant in the numerator a constant

105

divided by infinity is going to become zero.

106

So what we're left with is 1 over 1 plus 0 plus 0 because both of these become zero when we evaluate

107

them at infinity.

108

So obviously here we just have one divided by one or one the absolute value of 1 is still just 1.

109

So we end up with 1 times 3 which is going to be equal to 3.

110

So now we can say that L is equal to 3.

111

This value of L here that we are trying to find from the beginning is going to be 3 so now we go over

112

here to our ratio test and in our case we can say L is equal to 3.

113

Well how does 3 relate to 1.

114

3 is greater than 1.

115

So we can say our is greater than 1 for this particular problem.

116

Since I was greater than 1 and the ratio test tells us that the series diverges.

117

We can say that this original series here three to the end over end squared is a divergent series by

118

the ratio test.