1
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Let's use the Monotone Convergence
Theorem.

2
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[NOISE]

3
00:00:03,877 --> 00:00:08,776
Well here's a complicated

4
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sequence.
It's a sequence that starts with its first

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term equal to 1 and subsequent terms, a
sub n plus 1, we define in terms

6
00:00:19,412 --> 00:00:24,600
of a sub n.
Will be the square root of a sub n plus 2.

7
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This sequence is bounded above by 2.
Well let's see why.

8
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Let's suppose that a sub n is between 0
and 2.

9
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Then what do I know about the next term, a
sub n plus 1?

10
00:00:39,160 --> 00:00:40,927
Well, then a sub n plus 1 must also be

11
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non-negative, and that's just' cause of
the square root of something.

12
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But how big can a sub n plus 1 possibly
be?

13
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Well, a sub n plus 1 is the square root of
a sub n, which could be as big as

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2 plus 2.

15
00:00:55,180 --> 00:00:57,830
So a sub n plus 1 can be as big as

16
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the square root of 2 plus 2, which is
equal to 2.

17
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And this is great.
Right?

18
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This is telling me that if I know that a
particular term is stuck

19
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between 0 and 2, then the next term is
also stuck between 0 and 2.

20
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And what do I know about the first term?

21
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Well, the first term is definitely between
0 and 2.

22
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Right?
A sub 1 is between 0 and 2.

23
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And that means, that the next term must
also be between 0 and 2.

24
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And that means that the next term, a sub 3
must be between 0 and 3.

25
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And that means that the next term, which
is a sub 4 must be between 0 and 2.

26
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That means that the next term,

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which is a sub 5 must be between 0 and 2,
and so on.

28
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And what this actually is telling me, is
that no matter

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what n is, a sub n is between 0 and 2.

30
00:01:58,100 --> 00:02:01,010
And the sequence is nondecreasing.

31
00:02:01,010 --> 00:02:06,050
To show that this sequence is
nondecreasing, I'm going to show that

32
00:02:06,050 --> 00:02:09,380
a sub m is less than or equal to the next
term,

33
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a sub m plus 1.

34
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And of course, I've got a formula for the
next term.

35
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That's just the square root of a sub m
plus 2.

36
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So the question, is this true?

37
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Is a sub n less than or equal to the next
term, the square root of a sub n plus 2.

38
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And since a sub n is non-negative, I can

39
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figure out when this happens, just by
squaring both sides.

40
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So, I'm wondering,

41
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is a sub n squared, less than or equal to
a sub n plus 2?

42
00:02:41,350 --> 00:02:44,290
I'm going to subtract a sub n squared from
both sides.

43
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So I'm wondering, when is negative a sub n

44
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squared, plus a sub n, plus 2 non
negative.

45
00:02:53,210 --> 00:02:54,775
Now, I can factor this.

46
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This is asking when is 0 less than or
equal to, how does this factor.

47
00:03:00,190 --> 00:03:05,290
Got a 2 there, so I'll write a 2,1, 2
minus a sub n, 1

48
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plus a sub n, is how this quadratic
polynomial factors.

49
00:03:10,690 --> 00:03:14,880
So the question is, when is this product
non-negative?

50
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Well in order for this product to be
non-negative, either one

51
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of these is 0, or they're both positive or
they're both negative.

52
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It can't happen, that both these

53
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terms are negative.

54
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So the only possibility is that a sub n is
2, or

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a sub n is minus 1, or both of these terms
are positive.

56
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And that happens when a sub n is between
minus 1 and 2.

57
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So the upshot of this whole argument.

58
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is that as long a sub n is between minus 1
and 2, then

59
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a sub n is less than or equal to a sub n
plus 1.

60
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But I know that a sub n is between minus 1
and 2, because just

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little while ago, we saw that a sub n was
stuck between 0 and 2.

62
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And consequently, this sequence is
nondecreasing.

63
00:04:03,070 --> 00:04:04,640
What does that all mean?

64
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So get the sequence, and the terms of the
sequence are between 0 and 2, and that

65
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means that the sequence is bounded.
And the sequence is nondecreasing.

66
00:04:17,340 --> 00:04:18,800
Means the sequence is Monotone.

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So, we've got a Monotone bounded sequence,
and by the Monotone

68
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Convergence Theorem, that means that the
limit of the sequence exists.

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We can try to get some numerical evidence
to, guess the limit.

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So, the first term in this sequence, was
just defined to be 1.

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To get the next term, I use this recursive
definition.

72
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I'll add 2 and take the square

73
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root, and that gives me the a sub 2 term,

74
00:04:46,050 --> 00:04:52,190
and that's the square root of 3, which is
about 1.73.

75
00:04:52,190 --> 00:04:54,310
Now to compute the next term, what do I
do?

76
00:04:54,310 --> 00:05:00,484
Well, using this formula, I'll add 2 and
then take the square root of all

77
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that, that will give me the a sub 3 term,
and that's about 1.93.

78
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Now to compute the

79
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a sub 4 term.
Right?

80
00:05:08,955 --> 00:05:12,410
It's the same sort of process, I'll add 2.

81
00:05:12,410 --> 00:05:18,280
And then take the square root of all that,
and that'll give me the a sub 4 term.

82
00:05:18,280 --> 00:05:23,671
And that's approximately 1.98.
And if I keep on going, well

83
00:05:23,671 --> 00:05:28,624
you might believe then, that the limit of
a sub

84
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n, as n approaches infinity, is in fact,
2.

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[NOISE]