Original subtitles

1

Today we're going to be talking about how to use the alternating series estimation theorem to approximate

2

the sum of the series and to find the remainder of the approximation.

3

And this particular problem we've been given the infinite sum from equals 1 to infinity of the quantity

4

negative one raised to the end minus one power times and squared divided by 10 to the power.

5

For reference here I've written the alternating series estimation theorem which we'll get to in a second.

6

We really only need this to find the remainder term R7 for now to find the approximation of the series.

7

What we need to do is list out the first several terms of this series so that we can find the sum correct

8

to 4 decimal places.

9

So if we plug n equals 1 into this series this exponent here in the negative 1 will become 1 minus 1

10

which is zero negative 1 raise to the zero power is just one plugging in here for end squared we get

11

1 squared or just 1 divided by 10 to the first power.

12

So we just have one tenth so our value here is one tenth.

13

When we plug in an equals two we'll get negative 1 race to the two minus one or negative 1 to the first

14

power which will just give us negative one negative one at times 2 squared is negative 4 divided by

15

10 squared so we get negative for over 100 and equals three.

16

If we keep going here is going to give us positive 9 over 10 to the third or one thousand.

17

And we could keep computing our terms but remember that the problem is asked us to find the sum of the

18

series correct to 4 decimal places.

19

So we're going to need to turn each of these terms into a decimal value.

20

So let's look here at one tenth our decimal value is going to be zero point 1.

21

Our decimal value for negative for over 100 will be negative zero point zero for our decimal value for

22

9.

23

Over 1000 will be zero point 0 0 9.

24

And if we keep including the fourth fifth sixth and seventh terms what we get is negative zero point

25

0 0 1 6 for the fourth term we get positive zero point zero zero zero to 5 for the fifth term we get

26

negative zero point 0 0 0 0 3 6 for the sixth term and positive zero point 0 0 0 0 0 4 9 for the seventh

27

term.

28

Now remember we're looking to approximate correct to 4 decimal places.

29

So if we basically take a series here of partial sums we start adding all of these decimals together

30

here.

31

If we take 0.1 and we subtract zero point zero for what we get here is zero point zero six.

32

That's the scale of the first two terms.

33

If we add to that point 0 0 9 some of the first three terms gives us zero point zero six nine.

34

When we subtract from that point 0 0 1 6 we get zero point zero 6 7 4 which is the sum of the first

35

four terms.

36

And then when we add to that point 0 0 0 to 5 we get zero point zero six seven six five.

37

Now we've already gotten 1 decimal place past the first four decimal places but we want to make sure

38

that if we continue adding and subtracting terms in our series from this sum that we aren't affecting

39

the fourth decimal place.

40

So we go ahead and subtract point 0 0 0 0 3 6.

41

And what we get is zero point zero.

42

6 7 6 1 4.

43

Now the reason that it's helpful to do that is because as you can see when we had this value here our

44

five in this fifth decimal place would have rounded this fourth decimal place up to seven and we would

45

have point 0 6 7 7 for our first four decimal places.

46

But now that we've subtracted our sixth term here you can see that we're not going to be rounding up

47

because the value in the fifth decimal place is a 1 leaving our first four decimal places as point 0

48

6 7 6.

49

If we add to this sum the seventh term in our series what we get is zero point zero 6 7 6 1 8 9.

50

And what we can see is that that the decimal place is still holding is a 1.

51

Even if we use the 8 to rounded up to a 2 it's not going to be rounding up our fourth decimal place.

52

So we can conclude that the terms past this point are no longer going to be affecting our fourth decimal

53

place.

54

That's as far as we had to go to make sure that our fourth decimal place would stay intact at a 6.

55

If you remember correctly this is the last term that affected our fourth decimal place.

56

So what we want to say is that the sum of the series we'll call it as sub 6 because we use the first

57

six terms to get the fourth decimal place accurately we'll say subsects is equal to approximately zero

58

point zero six seven six.

59

Now at this point we have our Some correct to four decimal places.

60

Some of our series but we don't know how accurate this approximation is as an approximation of the exact

61

sum of the series in order to find out how accurate this approximation is.

62

We need to use the alternating series estimation theorem.

63

What that tells us what the theorem tells us is that if we have a sum and we'll call that some S and

64

that sum is the sum of an alternating series or we have negative 1 to the end minus 1 Power time some

65

Series B 7 here if S is the sum of that alternating series.

66

And if that term be sub and plus 1 is always less than or equal to the piece of term and the limit as

67

end goes to infinity of the be seven series is equal to zero then our remainder term satisfies this

68

inequality here.

69

So it sounds a lot more complicated than it is.

70

All we really need to do is this.

71

Notice that this infinite sum here in the alternating series estimation theorem has this negative one

72

raised to the end minus one power multiplied by some Series B 7.

73

Well in our infinite sum we have this this value here negative one race to the minus one power which

74

makes the series alternate but it's multiplied by and squared over 10 to the power.

75

Right here it's multiplied by that.

76

So we're going to carve that away from this alternating value right here and we're going to call that

77

be saw then whatever we can carve away from this negative one race to the end minus one power is what

78

we'll call B sevens we'll say this here is going to be b c then given that we have this SB So then we

79

need to prove two things.

80

We need to prove that B sub and plus 1 is always less than or equal to be sabahan.

81

And we need to prove this limit here.

82

So let's take it one piece at a time.

83

We need to show that B.

84

And plus one or in other words whatever we get when we plug and plus one into this be value here so

85

it'll be.

86

And plus 1 squared divided by 10 to the end plus one that that is always less than or equal to be Sabun

87

which is just and squared over tend to the N.

88

How do we show this.

89

Well if we write out the first several terms of each of these series we'll be able to compare the two

90

and see if the terms in this series are always less than the corresponding terms in this series.

91

So if we plug in N equals 1 to our B sub and plus 1 series we'll get one plus one which is 2 squared

92

which is four divided by 10.

93

To the one plus one or 10 squared and we'll get 10 square there if we plug in N equals 2 We'll get two

94

plus one is three squared is 9 divided by 10 to the two plus one or 10 cubed.

95

And if we plug in an equals three we get three plus one is four squared is 16 divided by 10 to the three

96

plus one or 10 to the fourth power.

97

We need to make sure that each of these is either less than or equal to what we get when we plug in

98

1 2 and 3 to this second series here.

99

So we want to say less than or equal to we plug one into this series here we get 1 squared in the numerator

100

we get tens of the first power in the denominator.

101

If we plug in 2 we get 2 squared in our numerator which is four divided by 10 squared in our denominator.

102

If we plug in three we get three squared in the numerator which is nine divided by ten.

103

To the third in our denominator.

104

Now if we just do the division for each of these terms on our calculator what we find is that these

105

inequalities are in fact true the terms on the left are in fact always less than or equal to the terms

106

on the right.

107

So that shows us that be Saab and plus 1 will always be less than or equal to be Subhan.

108

So thats the first part of our alternating series estimation theorem.

109

The next part is proving that this limit that the limit of be Sabahan as and goes to infinity is equal

110

to zero.

111

So for that part we're just going to take the limit as and it goes to infinity of B 7.

112

Or in our case and squared divided by 10 to the power.

113

If we evaluate this at some very large number we plug in you know 1 million or 10 million on our calculator.

114

What we find is that while and squared is increasing very quickly 10 raised to the end power is increasing

115

much much faster.

116

And therefore since the denominator will be significantly larger than the numerator we'll have some

117

much smaller constant value in the numerator and some extremely large value in the denominator.

118

This is just going to tend toward zero and the limit will be zero.

119

So in fact we have proven that the limit as and goes to infinity of B-7 is equal to zero.

120

Both of those parts are true.

121

So given that we can use the alternating series estimation theorem to find this remainder term are seven.

122

Now when it comes to this remainder inequality all we really need to look at is this value here that

123

we found before remember we said that the sixth term of the series was the last term to affect our sum.

124

Well this is basically an equals six right here and so what we want to do is we want to plug 6 in for

125

and here we want to plug 6 in for and here and we want to plug 6 in for and right here.

126

So what that leaves us with is the absolute value of our subsidies is going to be equal to the absolute

127

value of S minus as subsects which is going to be less than or equal to be sub 6 plus one or in other

128

words be sub 7.

129

Now everything to the left of this less than or equal sign just says that the exact sum of the entire

130

series minus the sum of the first six terms is equal to the remainder when we're approximating the sum.

131

After six terms so we can use either one of these either the absolute value of X minus as subsects or

132

our subsects and we really just want to use our subsects So we're going to say the absolute value of

133

our subsects is less than or equal to be sub 7.

134

Now remember we have a value for B subs 7.

135

We got it in this column here that I erased when we plug 7 into our B sub and series up here.

136

And what we got for we saw then was zero point and then 5 0 0 0 0 0 0 4 9.

137

And so we just take that value for B sub 7 that we already found and we say that that's going to be

138

greater than or equal to our sub 6.

139

Now what these two things tell us together is that the some of the first six terms the approximation

140

of the some of the first six terms is zero point zero six seven six when we round to four decimal places

141

with a remainder on that approximation less than or equal 2 points 0 0 0 0 0 4 9.

142

In other words this gives us an idea of how accurate this estimation of the sun is compared to the exact

143

sum.

144

In other words if we just sum up the first six terms of the series the difference between the sum of

145

the first six terms and the sum of the entire infinite series every single term in the infinite series.

146

They're only going to be different by this value here.

147

So it goes to show you that summing up only the first few terms of an alternating series like this actually

148

gets you very very close to the exact size of the entire infinite series.

149

We're only off by this point 0 0 0 0 0 4 9 value.

150

So that's how you use the alternating series estimation theorem to find the sum correct to a certain

151

number of decimal places and get an idea of the remainder estimate.

152

In other words how accurate your sum is compared to the sum of the entire infinite series.