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PROFESSOR: Hi.
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Last time, we had discussed
series and sequences.
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00:00:37,360 --> 00:00:41,010
Today, we're going to turn our
attention to a rather special
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00:00:41,010 --> 00:00:45,520
situation, a situation in which
every term in our series
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00:00:45,520 --> 00:00:46,720
is positive.
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For this reason, I have entitled
today's lecture
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00:00:49,630 --> 00:00:51,160
'Positive Series'.
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00:00:51,160 --> 00:00:54,760
Before we can do full justice
to positive series, however,
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00:00:54,760 --> 00:00:58,770
there are a few topics that we
must discuss as preliminaries.
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00:00:58,770 --> 00:01:02,710
The first of these is simply
the process of ordering.
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00:01:02,710 --> 00:01:05,290
Now this is a rather
strange situation.
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00:01:05,290 --> 00:01:07,550
Because in the finite
case, it turns out
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00:01:07,550 --> 00:01:08,670
to be rather trivial.
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00:01:08,670 --> 00:01:12,550
By way of illustration, let's
suppose the set 'S' consists
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00:01:12,550 --> 00:01:16,860
of the numbers 11,
8, 9, 7, and 10.
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00:01:16,860 --> 00:01:18,510
Now there are many ways
in which we could
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00:01:18,510 --> 00:01:19,610
order the set 'S'.
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00:01:19,610 --> 00:01:22,530
We can arrange them so that any
one of these five elements
27
00:01:22,530 --> 00:01:24,340
comes first, any one
of the remaining
28
00:01:24,340 --> 00:01:25,790
four second, et cetera.
29
00:01:25,790 --> 00:01:28,270
But suppose that we want
to arrange these
30
00:01:28,270 --> 00:01:29,940
according to size.
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00:01:29,940 --> 00:01:33,990
Observe that there is a rather
straightforward, shall we say,
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00:01:33,990 --> 00:01:37,240
binary technique, whereby we
can order these elements.
33
00:01:37,240 --> 00:01:40,090
By binary, I mean, let's look
at these two at the time.
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00:01:40,090 --> 00:01:43,020
We look at 11 and 8, and we
throw away the larger of the
35
00:01:43,020 --> 00:01:44,220
two, which is 11.
36
00:01:44,220 --> 00:01:47,020
Then we compare 8 with
9, throwaway 9
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00:01:47,020 --> 00:01:48,180
because that's bigger.
38
00:01:48,180 --> 00:01:51,520
Compare 8 and 7, throw away 8.
39
00:01:51,520 --> 00:01:55,240
Compare 7 and 10,
throw away 10.
40
00:01:55,240 --> 00:01:59,685
The survivor, being 7, is
therefore the least member of
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00:01:59,685 --> 00:02:00,790
our collection.
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00:02:00,790 --> 00:02:04,810
In a similar way, we can delete
7 and start looking for
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00:02:04,810 --> 00:02:06,770
the next number of our set.
44
00:02:06,770 --> 00:02:10,449
And in this way, eventually
order the elements of 'S'
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00:02:10,449 --> 00:02:11,390
according to size--
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00:02:11,390 --> 00:02:14,080
7, 8, 9, 10, 11.
47
00:02:14,080 --> 00:02:14,690
OK.
48
00:02:14,690 --> 00:02:20,690
Clearly, 7 is the smallest
element of 'S' and 11 is the
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00:02:20,690 --> 00:02:22,860
largest element of S.
50
00:02:22,860 --> 00:02:27,210
And in terms of nomenclature, we
say that 7 is a lower bound
51
00:02:27,210 --> 00:02:30,840
for 'S', 11 is an upper
bound for 'S'.
52
00:02:30,840 --> 00:02:34,070
You see, in terms of a picture,
all we're saying is
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00:02:34,070 --> 00:02:37,160
that when the elements of 'S'
are ordered according to size,
54
00:02:37,160 --> 00:02:41,220
7 is the furthest to
the left, 11 is the
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00:02:41,220 --> 00:02:43,040
furthest to the right.
56
00:02:43,040 --> 00:02:46,490
We could even talk more about
the nomenclature by saying--
57
00:02:46,490 --> 00:02:48,210
and this sounds like
a tongue twister.
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00:02:48,210 --> 00:02:51,290
This is why I had you read this
material first in the
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00:02:51,290 --> 00:02:53,660
last assignment and the
supplementary notes, so that
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00:02:53,660 --> 00:02:55,990
part of this will at least
seem like a review.
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00:02:55,990 --> 00:02:59,740
Observe that 7 is called the
greatest lower bound for 'S',
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00:02:59,740 --> 00:03:04,570
simply because any number larger
than 7 cannot be a
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00:03:04,570 --> 00:03:08,350
lower bound for 'S', simply
because 7 would be smaller
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00:03:08,350 --> 00:03:09,880
than that particular number.
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00:03:09,880 --> 00:03:13,690
And in a similar way, 11 is
called the least upper bound
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00:03:13,690 --> 00:03:17,650
for 'S', because any number
smaller than 11 would be
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00:03:17,650 --> 00:03:20,520
exceeded by 11, and hence
could not be an
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00:03:20,520 --> 00:03:22,260
upper bound for 'S'.
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00:03:22,260 --> 00:03:24,110
Now the interesting
point is this.
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00:03:24,110 --> 00:03:27,000
Hopefully, at this stage of the
game, you listened to what
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00:03:27,000 --> 00:03:30,810
I've had to say, and you say,
my golly, this is trivial.
72
00:03:30,810 --> 00:03:32,720
And the answer is,
it is trivial.
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00:03:32,720 --> 00:03:36,500
But remember what our main
concern was in our last
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00:03:36,500 --> 00:03:40,460
lecture when we introduced the
concept of infinite series and
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00:03:40,460 --> 00:03:41,490
sequences--
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00:03:41,490 --> 00:03:45,740
that many things that were
trivial for the finite case
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00:03:45,740 --> 00:03:49,800
became rather serious dilemmas
for the infinite case.
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00:03:49,800 --> 00:03:52,960
In other words, my claim is
that these results are far
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00:03:52,960 --> 00:03:55,260
more subtle for infinite sets.
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00:03:55,260 --> 00:03:57,160
And I think the best
way to do this is
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00:03:57,160 --> 00:03:59,000
by means of an example.
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00:03:59,000 --> 00:04:03,560
See now, let 'S' be the set of
numbers where the n-th number
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00:04:03,560 --> 00:04:05,860
is 'n' over 'n + 1'.
84
00:04:05,860 --> 00:04:08,480
In other words, the first member
of 'S' will be 1/2, the
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00:04:08,480 --> 00:04:13,300
second member, 2/3, the third
member, 3/4, et cetera.
86
00:04:13,300 --> 00:04:16,050
Now, let's take look to see
what the least upper
87
00:04:16,050 --> 00:04:17,350
bound for 'S' is.
88
00:04:17,350 --> 00:04:20,500
And sparing you the details, I
think you can observe at this
89
00:04:20,500 --> 00:04:24,070
stage of the game, especially
based on the homework of the
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00:04:24,070 --> 00:04:30,390
last unit, that the limit of
the sequence 'S' is 1.
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00:04:30,390 --> 00:04:34,220
In fact, 1 is the smallest
number which exceeds every
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00:04:34,220 --> 00:04:35,690
member in this collection.
93
00:04:35,690 --> 00:04:38,740
In other words, 1 is the least
upper bound for 'S'.
94
00:04:38,740 --> 00:04:41,945
But observe the interesting
case that here--
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00:04:41,945 --> 00:04:43,820
and by the way, notice the
abbreviation that we use in
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00:04:43,820 --> 00:04:46,700
our notes. 'LUB', least
upper bound.
97
00:04:46,700 --> 00:04:48,890
'GLB', greatest lower bound.
98
00:04:48,890 --> 00:04:51,730
But 1 is the least upper
bound for 'S'.
99
00:04:51,730 --> 00:04:55,740
Yet the fact remains that the
least upper bound is not a
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00:04:55,740 --> 00:04:57,160
member of 'S' itself.
101
00:04:57,160 --> 00:05:00,960
In other words, there is no
number 'n' such that 'n' over
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00:05:00,960 --> 00:05:03,470
'n + 1' is equal to 1.
103
00:05:03,470 --> 00:05:06,320
You see, notice that as these
numbers increase, they get
104
00:05:06,320 --> 00:05:09,620
bounded by 1, but 1 is
not a member of 'S'.
105
00:05:09,620 --> 00:05:12,380
In other words, here's an
example where the least upper
106
00:05:12,380 --> 00:05:16,110
bound of a set does not have
to be a member of the set.
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00:05:16,110 --> 00:05:18,740
And a companion to this would
be an example where the
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00:05:18,740 --> 00:05:21,950
greatest lower bound is not
a member of the set.
109
00:05:21,950 --> 00:05:26,030
And to this end, simply let
the n-th member of 'S'
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00:05:26,030 --> 00:05:29,220
arranged by sequence be '1/n'.
111
00:05:29,220 --> 00:05:31,000
The n-th member is '1/n'.
112
00:05:31,000 --> 00:05:33,020
Therefore, 'S' is what?
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00:05:33,020 --> 00:05:37,590
The set consisting of 1,
1/2, 1/3, et cetera.
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00:05:37,590 --> 00:05:41,550
Observe that as the terms go
further and further out, they
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00:05:41,550 --> 00:05:44,650
get arbitrarily close
to 0 in size--
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00:05:44,650 --> 00:05:47,310
every one of these terms, '1/n',
no matter how big 'n'
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00:05:47,310 --> 00:05:48,890
is is greater than 0.
118
00:05:48,890 --> 00:05:52,150
In other words, then, observe
that 0 will be the greatest
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00:05:52,150 --> 00:05:56,350
lower bound for 'S', but 0
is not a member of 'S'.
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00:05:56,350 --> 00:05:59,690
In other words, it seems that
things which are quite trivial
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00:05:59,690 --> 00:06:02,550
for finite collections have
certain degrees of
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00:06:02,550 --> 00:06:04,870
sophistication for infinite
collections.
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00:06:04,870 --> 00:06:08,440
So what we're going to do now
is to establish a few basic
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00:06:08,440 --> 00:06:09,780
definitions.
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00:06:09,780 --> 00:06:11,380
And we'll do it in a
rather formal way.
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00:06:11,380 --> 00:06:13,790
Our first definition
is the following.
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00:06:13,790 --> 00:06:17,000
Given the set of numbers, 'S',
'M' is called an upper bound
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00:06:17,000 --> 00:06:21,020
for 'S', If 'M' is greater than
or equal to 'x' for all
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00:06:21,020 --> 00:06:22,410
'x' in 'S'.
130
00:06:22,410 --> 00:06:26,230
In other words, if 'M' exceeds
every member of 'S', 'M' is
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00:06:26,230 --> 00:06:29,920
called an upper bound for 'S'.
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00:06:29,920 --> 00:06:32,590
As I say, these these
definitions are very
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00:06:32,590 --> 00:06:34,140
straightforward.
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00:06:34,140 --> 00:06:35,430
The companion--
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00:06:35,430 --> 00:06:37,380
well, let's go one
step further.
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00:06:37,380 --> 00:06:39,900
By the way, observe that I
have in the interest of
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00:06:39,900 --> 00:06:43,100
brevity left out the
corresponding definitions for
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00:06:43,100 --> 00:06:43,740
lower bounds.
139
00:06:43,740 --> 00:06:45,680
But they are quite analogous.
140
00:06:45,680 --> 00:06:49,760
In other words, a lower bound
for 'S' would be a number that
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00:06:49,760 --> 00:06:53,080
was less than or equal to
each member in 'S'.
142
00:06:53,080 --> 00:06:56,830
At any rate, then, 'M' is called
a least upper bound for
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00:06:56,830 --> 00:07:00,580
'S' if first of all, 'M' is
an upper bound for 'S'.
144
00:07:00,580 --> 00:07:05,010
And secondly, if 'L' is less
than 'M', 'L' is not an upper
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00:07:05,010 --> 00:07:06,090
bound for 'S'.
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00:07:06,090 --> 00:07:08,260
In other words, least upper
bound means what?
147
00:07:08,260 --> 00:07:12,190
Anything smaller cannot
be an upper bound.
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00:07:12,190 --> 00:07:15,540
Notice in terms of our previous
remark that the least
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00:07:15,540 --> 00:07:19,630
upper bound need not
belong to 'x'.
150
00:07:19,630 --> 00:07:21,860
The companion to this
would be what?
151
00:07:21,860 --> 00:07:25,620
A greatest lower bound would be
a number which is a lower
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00:07:25,620 --> 00:07:28,790
bound such that anything
greater could
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00:07:28,790 --> 00:07:30,470
not be a lower bound.
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00:07:30,470 --> 00:07:34,950
Again, all of this is easier to
see in terms of a picture.
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00:07:34,950 --> 00:07:39,220
Visualize 'S' as being this
interval with little 'm' and
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00:07:39,220 --> 00:07:41,910
capital 'M' being the endpoints
of the interval.
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00:07:41,910 --> 00:07:45,280
Observe that capital 'M' is
the least upper bound.
158
00:07:45,280 --> 00:07:47,900
Little 'm' is the greatest
lower bound.
159
00:07:47,900 --> 00:07:52,470
Anything smaller than little
'm' will be lower bound.
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00:07:52,470 --> 00:07:56,690
Anything greater than capital
'M' will be an upper bound.
161
00:07:56,690 --> 00:08:00,770
And notice that nothing in the
set 'S' itself can be either
162
00:08:00,770 --> 00:08:03,140
an upper bound or
a lower bound.
163
00:08:03,140 --> 00:08:08,210
Because anything inside 'S'
appears to the right of little
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00:08:08,210 --> 00:08:11,230
'm' and to the left
of capital 'M'.
165
00:08:11,230 --> 00:08:14,220
And one final definition
as a preliminary.
166
00:08:14,220 --> 00:08:18,410
A set, 'S', is called bounded if
it has both an upper and a
167
00:08:18,410 --> 00:08:19,790
lower bound.
168
00:08:19,790 --> 00:08:23,080
And the key property that
we have to keep track of
169
00:08:23,080 --> 00:08:24,110
throughout this--
170
00:08:24,110 --> 00:08:25,210
and we won't prove this.
171
00:08:25,210 --> 00:08:27,760
In other words, in more
rigorous advanced math
172
00:08:27,760 --> 00:08:30,730
courses, this is proven
as a theorem.
173
00:08:30,730 --> 00:08:33,739
For our purposes, it seems
self-evident enough so that
174
00:08:33,739 --> 00:08:35,100
we're willing to accept it.
175
00:08:35,100 --> 00:08:37,830
And so rather than to belabor
the point, let us just accept
176
00:08:37,830 --> 00:08:41,870
as a key property that every
bounded set of numbers has a
177
00:08:41,870 --> 00:08:44,360
greatest lower bound and
at least upper bound.
178
00:08:44,360 --> 00:08:48,140
In other words, if a set is
bounded, we can find a
179
00:08:48,140 --> 00:08:52,680
smallest upper bound and
a largest lower bound.
180
00:08:52,680 --> 00:08:55,390
And this completes the
first portion of
181
00:08:55,390 --> 00:08:57,550
our preliminary material.
182
00:08:57,550 --> 00:09:00,050
The next portion of our
preliminary material before
183
00:09:00,050 --> 00:09:04,690
studying positive series
involves what we mean by a
184
00:09:04,690 --> 00:09:09,010
monotonic non-decreasing
sequence.
185
00:09:09,010 --> 00:09:12,580
A sequence is called monotonic
non-decreasing--
186
00:09:12,580 --> 00:09:16,160
and if you don't frighten at
these words, it's almost
187
00:09:16,160 --> 00:09:18,100
self-evident what this
thing means.
188
00:09:18,100 --> 00:09:21,120
It means that no term can
be smaller than the
189
00:09:21,120 --> 00:09:22,580
one that came before.
190
00:09:22,580 --> 00:09:26,370
In other words, the n-th term
is less than or equal to the
191
00:09:26,370 --> 00:09:29,070
'n plus first term'
for each 'n'.
192
00:09:29,070 --> 00:09:32,070
And wording that more
explicitly, it says what?
193
00:09:32,070 --> 00:09:35,010
'a sub 1' is less than or equal
to 'a sub 2' is less
194
00:09:35,010 --> 00:09:37,380
than or equal to 'a
sub 3', et cetera.
195
00:09:37,380 --> 00:09:40,030
And I hope that it's clear by
this time that it's not
196
00:09:40,030 --> 00:09:42,560
self-evident that a sequence
has to have this.
197
00:09:42,560 --> 00:09:45,510
Remember the subscripts simply
tell you the order in which
198
00:09:45,510 --> 00:09:46,610
the terms appear.
199
00:09:46,610 --> 00:09:49,850
It has no bearing on the
size of the term.
200
00:09:49,850 --> 00:09:52,860
For example, in an arbitrary
sequence, recall that the
201
00:09:52,860 --> 00:09:56,100
second term can be smaller in
magnitude than the first term.
202
00:09:56,100 --> 00:10:00,540
However, if the terms are
non-decreasing sequentially
203
00:10:00,540 --> 00:10:03,350
this way, the sequence
is called monotonic
204
00:10:03,350 --> 00:10:04,560
non-decreasing.
205
00:10:04,560 --> 00:10:07,260
And the problem that comes up
is, or the important question
206
00:10:07,260 --> 00:10:11,130
that comes up is, what's so
important about monotonic
207
00:10:11,130 --> 00:10:13,000
non-decreasing sequences?
208
00:10:13,000 --> 00:10:15,730
And the answer is that for
such sequences, two
209
00:10:15,730 --> 00:10:17,530
possibilities exist.
210
00:10:17,530 --> 00:10:21,600
In other words, either the terms
can keep getting larger
211
00:10:21,600 --> 00:10:23,695
and larger without bound--
212
00:10:23,695 --> 00:10:24,850
see?
213
00:10:24,850 --> 00:10:26,940
In other words, that the
sequences 'a sub n' has no
214
00:10:26,940 --> 00:10:28,160
upper bound.
215
00:10:28,160 --> 00:10:31,140
In which case, we say that the
limit of 'a sub n' as 'n'
216
00:10:31,140 --> 00:10:34,070
approaches infinity
is infinity.
217
00:10:34,070 --> 00:10:36,320
And for example, the
ordinary counting
218
00:10:36,320 --> 00:10:37,950
sequence has this property.
219
00:10:37,950 --> 00:10:42,400
See, 1, 2, 3, 4, 5, et cetera,
is a monotonic
220
00:10:42,400 --> 00:10:43,565
non-decreasing sequence.
221
00:10:43,565 --> 00:10:45,930
In fact, it's monotonic
increasing.
222
00:10:45,930 --> 00:10:49,000
Every member of the sequence is
greater than the one that
223
00:10:49,000 --> 00:10:49,980
came before.
224
00:10:49,980 --> 00:10:52,440
But as you go further and
further out, the terms
225
00:10:52,440 --> 00:10:54,750
increase without upper bound.
226
00:10:54,750 --> 00:10:56,000
OK?
227
00:10:56,000 --> 00:11:00,170
Now what is the other
possibility for a monotonic
228
00:11:00,170 --> 00:11:01,980
non-decreasing sequence?
229
00:11:01,980 --> 00:11:06,450
After all, the opposite, or the
only other possibility, is
230
00:11:06,450 --> 00:11:09,520
that if the 'a sub n'-- if it's
false, that the sequence
231
00:11:09,520 --> 00:11:12,460
doesn't have an upper bound,
then of course it must have an
232
00:11:12,460 --> 00:11:12,960
upper bound.
233
00:11:12,960 --> 00:11:14,350
That's the second case.
234
00:11:14,350 --> 00:11:16,950
In other words, suppose the
sequence has an upper bound.
235
00:11:16,950 --> 00:11:19,710
Then the interesting point
is that the limit of this
236
00:11:19,710 --> 00:11:21,790
sequence exists.
237
00:11:21,790 --> 00:11:25,980
And not only does it exist, but
the limit itself is the
238
00:11:25,980 --> 00:11:28,500
least upper bound
of the sequence.
239
00:11:28,500 --> 00:11:32,060
In other words, in the case
where the sequence is
240
00:11:32,060 --> 00:11:35,680
non-decreasing, if
it's bounded--
241
00:11:35,680 --> 00:11:36,900
if it's bounded--
242
00:11:36,900 --> 00:11:39,180
the least upper bound
will be the limit.
243
00:11:39,180 --> 00:11:41,150
Now you see, here's where we
use that key property.
244
00:11:41,150 --> 00:11:45,020
Namely, if a sequence is
bounded, it must have a least
245
00:11:45,020 --> 00:11:45,990
upper bound.
246
00:11:45,990 --> 00:11:48,650
Let's call that least
upper bound 'L'.
247
00:11:48,650 --> 00:11:50,780
And by the way, you'll notice
I wrote the word proof in
248
00:11:50,780 --> 00:11:51,910
quotation marks.
249
00:11:51,910 --> 00:11:54,920
It's simply to indicate that I
prefer to give you a geometric
250
00:11:54,920 --> 00:11:57,630
proof here rather than
an analytic one.
251
00:11:57,630 --> 00:12:00,060
But the analytic proof
follows word for
252
00:12:00,060 --> 00:12:02,780
word from this picture.
253
00:12:02,780 --> 00:12:05,530
In other words, it just
translates in the usual way.
254
00:12:05,530 --> 00:12:08,750
And we'll drill on this with
the exercises and the notes
255
00:12:08,750 --> 00:12:09,770
and the textbook.
256
00:12:09,770 --> 00:12:11,690
But at any rate, the
idea is this.
257
00:12:11,690 --> 00:12:14,960
To prove that 'L' is a limit,
what must I do?
258
00:12:14,960 --> 00:12:18,700
I must show that if I choose
any epsilon greater than 0,
259
00:12:18,700 --> 00:12:22,860
that all the terms beyond a
certain one lie between 'L
260
00:12:22,860 --> 00:12:25,920
minus epsilon' and
'L plus epsilon'.
261
00:12:25,920 --> 00:12:27,320
Now here's the way this works.
262
00:12:27,320 --> 00:12:29,360
Let's see if we can just read
the diagram and get this thing
263
00:12:29,360 --> 00:12:30,390
very quickly.
264
00:12:30,390 --> 00:12:35,060
First of all, at least one term
in my sequence has to be
265
00:12:35,060 --> 00:12:37,590
between 'L minus epsilon'
and 'L'.
266
00:12:37,590 --> 00:12:39,730
And the reason for that
is simply this.
267
00:12:39,730 --> 00:12:41,800
Remember 'L' is a least
upper bound.
268
00:12:41,800 --> 00:12:44,890
Because 'L' is a least upper
bound, 'L minus epsilon'
269
00:12:44,890 --> 00:12:46,580
cannot be an upper bound.
270
00:12:46,580 --> 00:12:50,080
Now if no term can get beyond
'L minus epsilon', then
271
00:12:50,080 --> 00:12:53,320
certainly 'L minus epsilon'
would be an upper bound.
272
00:12:53,320 --> 00:12:55,920
The fact that 'L minus epsilon'
isn't an upper bound,
273
00:12:55,920 --> 00:13:00,770
therefore, means that at least
one 'a sub n', say 'a sub
274
00:13:00,770 --> 00:13:04,220
capital N', is in this
interval here.
275
00:13:04,220 --> 00:13:09,340
Notice also that because 'L' is
an upper bound, no 'a sub
276
00:13:09,340 --> 00:13:11,320
n' can get beyond here.
277
00:13:11,320 --> 00:13:13,840
In other words, no 'a
sub n' lies between
278
00:13:13,840 --> 00:13:15,700
'L' and 'L plus epsilon'.
279
00:13:15,700 --> 00:13:18,560
Because in that particular case,
if that were to happen,
280
00:13:18,560 --> 00:13:20,960
'L' couldn't be an
upper bound.
281
00:13:20,960 --> 00:13:22,720
OK, so far, so good.
282
00:13:22,720 --> 00:13:24,980
That follows just from
the definition of
283
00:13:24,980 --> 00:13:26,370
the least upper bound.
284
00:13:26,370 --> 00:13:30,320
Now, we use the fact that the
sequence is non-decreasing.
285
00:13:30,320 --> 00:13:31,660
And that says what?
286
00:13:31,660 --> 00:13:36,165
That if little 'n' is greater
than capital 'N', 'a sub
287
00:13:36,165 --> 00:13:39,430
little n' is greater than or
equal to 'a sub capital N'.
288
00:13:39,430 --> 00:13:42,360
In other words, what this means
is if you list the 'a
289
00:13:42,360 --> 00:13:46,740
sub n's along this line, as 'n'
increases, the terms move
290
00:13:46,740 --> 00:13:49,500
progressively from
left to right.
291
00:13:49,500 --> 00:13:56,680
In other words, notice then that
if 'n' is greater than or
292
00:13:56,680 --> 00:14:04,190
equal to capital 'N', 'L minus
epsilon' is less than 'a sub
293
00:14:04,190 --> 00:14:07,220
capital N', which in turn is
less than or equal to 'a sub
294
00:14:07,220 --> 00:14:10,220
little n', which in turn is
less than or equal to 'L'.
295
00:14:10,220 --> 00:14:12,900
And that's just another
geometric way of saying that
296
00:14:12,900 --> 00:14:16,050
'a sub n' is within
epsilon of 'L'.
297
00:14:16,050 --> 00:14:19,630
And that's exactly the
definition that the 'a sub n's
298
00:14:19,630 --> 00:14:21,250
converge to the limit 'L'.
299
00:14:21,250 --> 00:14:24,520
Now I went over this rather
quickly simply because this is
300
00:14:24,520 --> 00:14:25,840
done in the textbook.
301
00:14:25,840 --> 00:14:28,500
And all I'm trying to do with
this lecture is to give you a
302
00:14:28,500 --> 00:14:31,040
quick overview of
what's going on.
303
00:14:31,040 --> 00:14:33,970
Well, you see, we're now roughly
15 minutes into our
304
00:14:33,970 --> 00:14:37,460
lecture and we haven't come
to the main topic yet.
305
00:14:37,460 --> 00:14:40,410
My claim is that the main
topic works very, very
306
00:14:40,410 --> 00:14:44,140
smoothly once we understand
these preliminaries.
307
00:14:44,140 --> 00:14:47,600
The main topic, you see, is
called positive series.
308
00:14:47,600 --> 00:14:49,850
And the definition of a positive
series is just as the
309
00:14:49,850 --> 00:14:51,230
name implies.
310
00:14:51,230 --> 00:14:55,890
If each term in the series is
at least as great as 0-- in
311
00:14:55,890 --> 00:14:58,470
other words, if each term
is non-negative--
312
00:14:58,470 --> 00:15:01,080
then the series is
called positive.
313
00:15:01,080 --> 00:15:04,010
Now why are positive
series important?
314
00:15:04,010 --> 00:15:07,320
And why do they tie in with
our previous discussion?
315
00:15:07,320 --> 00:15:10,290
Well, let's answer the
last question.
316
00:15:10,290 --> 00:15:13,140
The reason they tie in with
our previous discussion is
317
00:15:13,140 --> 00:15:17,990
that we have already seen that
by the sum of a series, we
318
00:15:17,990 --> 00:15:20,820
mean the limit of the sequence
of partial sums.
319
00:15:20,820 --> 00:15:24,150
To go from one partial sum to
the next, you add on the next
320
00:15:24,150 --> 00:15:25,290
term in the series.
321
00:15:25,290 --> 00:15:28,430
If each of the 'a sub n's
is positive, or at least
322
00:15:28,430 --> 00:15:32,590
non-negative, notice then that
the sequence of partial sums
323
00:15:32,590 --> 00:15:34,860
is monotonic non-decreasing.
324
00:15:34,860 --> 00:15:35,670
Why?
325
00:15:35,670 --> 00:15:38,790
Because to go from the n-th
partial sums to the 'n plus
326
00:15:38,790 --> 00:15:43,460
first' partial sum, you add
on 'a sub 'n plus 1''.
327
00:15:43,460 --> 00:15:47,560
And since 'a sub 'n plus 1''
is at least as big as 0, it
328
00:15:47,560 --> 00:15:51,040
means that the 'n plus first'
partial sum can be no smaller
329
00:15:51,040 --> 00:15:52,650
than the n-th partial sum.
330
00:15:52,650 --> 00:15:57,860
In other words, therefore, if
the series summation 'n' goes
331
00:15:57,860 --> 00:16:01,780
from 1 to infinity, 'a sub n' is
a positive series, it must
332
00:16:01,780 --> 00:16:05,900
either diverge to infinity, or
else it converges to the limit
333
00:16:05,900 --> 00:16:09,400
'L', where 'L' is the least
upper bound for the sequence
334
00:16:09,400 --> 00:16:11,140
of partial sums.
335
00:16:11,140 --> 00:16:11,660
OK?
336
00:16:11,660 --> 00:16:15,470
Now qualitatively, that's
the end of story.
337
00:16:15,470 --> 00:16:16,820
In other words, we
now know what?
338
00:16:16,820 --> 00:16:20,380
For a positive series, it either
diverges to infinity,
339
00:16:20,380 --> 00:16:24,570
or else it converges
to a sum, a limit.
340
00:16:24,570 --> 00:16:25,820
And that limit is what?
341
00:16:25,820 --> 00:16:29,460
The least upper bound for the
sequence of partial sums.
342
00:16:29,460 --> 00:16:32,610
The problem is that
quantitatively, we would like
343
00:16:32,610 --> 00:16:36,590
to have some criteria for
determining whether a positive
344
00:16:36,590 --> 00:16:38,610
series falls into the
first category
345
00:16:38,610 --> 00:16:40,190
or the second category.
346
00:16:40,190 --> 00:16:42,950
Notice, how can we tell whether
the series converges
347
00:16:42,950 --> 00:16:44,900
or whether it diverges?
348
00:16:44,900 --> 00:16:47,670
And you see, the reading
material in the text for this
349
00:16:47,670 --> 00:16:52,360
assignment focuses attention
on three major tests.
350
00:16:52,360 --> 00:16:55,060
And these are the ones I'd like
to go over with you once
351
00:16:55,060 --> 00:16:57,080
lightly, so to speak.
352
00:16:57,080 --> 00:17:01,300
The first test is called
the comparison test.
353
00:17:01,300 --> 00:17:03,140
And it almost sounds
self-evident.
354
00:17:03,140 --> 00:17:05,869
I just want to outline
how these proofs go.
355
00:17:05,869 --> 00:17:08,869
Because I think that once you
hear these things spoken, as
356
00:17:08,869 --> 00:17:12,780
you read the material, the
formal proofs will fit into a
357
00:17:12,780 --> 00:17:16,010
pattern much more nicely than if
you haven't heard the stuff
358
00:17:16,010 --> 00:17:18,060
said out loud, you see.
359
00:17:18,060 --> 00:17:19,079
The idea is this.
360
00:17:19,079 --> 00:17:22,270
Let's suppose that we know that
summation 'n' goes from 1
361
00:17:22,270 --> 00:17:27,150
to infinity, 'C sub n', is a
convergent positive series.
362
00:17:27,150 --> 00:17:29,070
In other words, all of
these are positive,
363
00:17:29,070 --> 00:17:30,920
and the series converges.
364
00:17:30,920 --> 00:17:33,700
Suppose I now have another
sequence of numbers, 'u sub
365
00:17:33,700 --> 00:17:37,210
n', where each 'u sub
n' is positive--
366
00:17:37,210 --> 00:17:39,800
see, it's at least as big as 0--
but can be no bigger than
367
00:17:39,800 --> 00:17:42,170
'C sub n' for each 'n'.
368
00:17:42,170 --> 00:17:45,660
Then the statement is that the
series formed by adding up the
369
00:17:45,660 --> 00:17:48,080
'u sub n's must also converge.
370
00:17:48,080 --> 00:17:50,810
Notice what you're saying is,
here's a bunch of positive
371
00:17:50,810 --> 00:17:53,920
terms that can't
get too large.
372
00:17:53,920 --> 00:17:56,360
And these terms in magnitude
are less
373
00:17:56,360 --> 00:17:57,560
than or equal to these.
374
00:17:57,560 --> 00:17:59,760
Therefore, what you're saying is
that this sum can't get too
375
00:17:59,760 --> 00:18:01,030
large either.
376
00:18:01,030 --> 00:18:03,770
And if you want to verbalize
that so that it becomes more
377
00:18:03,770 --> 00:18:06,080
formal, the idea is this.
378
00:18:06,080 --> 00:18:11,700
Let 'T sub n' denote the n-th
partial sums of the series
379
00:18:11,700 --> 00:18:12,980
involving the 'C sub n's.
380
00:18:12,980 --> 00:18:17,220
In other words, let 'T sub n'
be 'C sub 1' plus et cetera,
381
00:18:17,220 --> 00:18:19,020
up to 'C sub n'.
382
00:18:19,020 --> 00:18:22,780
And let 'S sub n' be the n-th
term in the sequence of
383
00:18:22,780 --> 00:18:25,470
partial sums of the series
involving the 'u's.
384
00:18:25,470 --> 00:18:29,010
In other words, let 'S sub
n' be 'u1' plus et cetera
385
00:18:29,010 --> 00:18:31,080
up to 'u sub n'.
386
00:18:31,080 --> 00:18:32,630
Now the idea is-- lookit.
387
00:18:32,630 --> 00:18:35,870
We know that each of the 'u's
in magnitude is smaller than
388
00:18:35,870 --> 00:18:37,120
each of the 'C's.
389
00:18:37,120 --> 00:18:40,350
Consequently, the sum of all
the 'u's must be no greater
390
00:18:40,350 --> 00:18:42,520
than the sum of all the 'C's.
391
00:18:42,520 --> 00:18:46,460
In other words, n-th partial
sum, 'S sub n', is less than
392
00:18:46,460 --> 00:18:50,800
or equal to the partial sum,
'T sub n' for each 'n'.
393
00:18:50,800 --> 00:18:56,230
Now, by our previous result,
knowing that these series
394
00:18:56,230 --> 00:18:59,990
summation 'C sub n' converges,
it converges to its least
395
00:18:59,990 --> 00:19:02,140
upper bound of partial sums.
396
00:19:02,140 --> 00:19:05,450
In other words, the limit of
'T sub n' as 'n' approaches
397
00:19:05,450 --> 00:19:08,970
infinity is some number 'T',
where 'T' is the least upper
398
00:19:08,970 --> 00:19:12,680
bound of the sequence of partial
sums of the series.
399
00:19:12,680 --> 00:19:16,520
Well in particular, then, since
each 'S sub n' is less
400
00:19:16,520 --> 00:19:20,570
than or equal to 'T sub n', 'S
sub n' itself certainly cannot
401
00:19:20,570 --> 00:19:23,030
exceed the least upper
bound, namely 'T'.
402
00:19:23,030 --> 00:19:25,130
In other words, 'S sub n'
can be no bigger than
403
00:19:25,130 --> 00:19:26,760
'T' for each 'n'.
404
00:19:26,760 --> 00:19:28,700
Well, what does this mean?
405
00:19:28,700 --> 00:19:31,350
It means, then, that
the sequence 'S sub
406
00:19:31,350 --> 00:19:32,900
n' itself is bounded.
407
00:19:32,900 --> 00:19:34,990
Well, it's bounded.
408
00:19:34,990 --> 00:19:38,130
It's a monotonic non-decreasing
sequence.
409
00:19:38,130 --> 00:19:40,820
Therefore, its limit exists.
410
00:19:40,820 --> 00:19:43,840
Not only does it exist, but it's
the least upper bound of
411
00:19:43,840 --> 00:19:45,430
the sequence of partial sums.
412
00:19:45,430 --> 00:19:47,660
In other words, this proof
is given in the text.
413
00:19:47,660 --> 00:19:51,700
All I want you to see is that
step by step, what this proof
414
00:19:51,700 --> 00:19:55,780
really does is it compares
magnitudes of terms.
415
00:19:55,780 --> 00:19:59,430
In other words, if one batch of
terms can't get lodged, if
416
00:19:59,430 --> 00:20:01,220
term by term, everything--
417
00:20:01,220 --> 00:20:05,310
another sequence is less than
these terms, then that second
418
00:20:05,310 --> 00:20:07,830
sum can't get too
large either.
419
00:20:07,830 --> 00:20:10,080
And this is just a formalization
of that.
420
00:20:10,080 --> 00:20:13,590
There are a few notes that we
should make first of all.
421
00:20:13,590 --> 00:20:18,930
Namely, the condition that 'u
sub n' be between 0 and 'C sub
422
00:20:18,930 --> 00:20:23,980
n' for all 'n' can be weakened
to cover the case where this
423
00:20:23,980 --> 00:20:27,290
is true only beyond
a certain point.
424
00:20:27,290 --> 00:20:29,380
Look, let me show you
what I mean by this.
425
00:20:29,380 --> 00:20:36,030
Let's suppose I look at the
sequence 1 plus 1/2 plus 1/3--
426
00:20:36,030 --> 00:20:37,640
let's do at a different one.
427
00:20:37,640 --> 00:20:43,410
1 plus 1/2 plus 1/4 plus 1/8
plus 1/16, et cetera.
428
00:20:43,410 --> 00:20:45,880
In other words, the geometric
series each of
429
00:20:45,880 --> 00:20:47,810
whose terms is 1/2.
430
00:20:47,810 --> 00:20:50,950
This series we know
converges OK.
431
00:20:50,950 --> 00:20:53,700
Now what the comparison test
says is, suppose you have
432
00:20:53,700 --> 00:20:54,950
something like this--
433
00:20:54,950 --> 00:21:05,400
1 plus 1/3 plus 1/5 plus 1/9
plus 1/17, et cetera.
434
00:21:05,400 --> 00:21:07,980
See, notice that each of these
terms is less than the
435
00:21:07,980 --> 00:21:10,050
corresponding term here.
436
00:21:10,050 --> 00:21:13,380
Consequently, since these
terms add up to a finite
437
00:21:13,380 --> 00:21:16,210
amount, these terms
here must also add
438
00:21:16,210 --> 00:21:17,600
up to a finite amount.
439
00:21:17,600 --> 00:21:20,950
But suppose for the sake of
argument I decide to replace
440
00:21:20,950 --> 00:21:25,120
the first term here by
10 to the sixth.
441
00:21:25,120 --> 00:21:26,530
I'll make this a million.
442
00:21:26,530 --> 00:21:29,720
Now notice that this sum is
going to become much larger.
443
00:21:29,720 --> 00:21:33,680
But the point is when I change
this from a 1 to 10 to the
444
00:21:33,680 --> 00:21:37,230
sixth, even though I made the
sum larger, I didn't change
445
00:21:37,230 --> 00:21:38,620
the finiteness of it.
446
00:21:38,620 --> 00:21:41,800
In other words, if I replace the
first four terms here by
447
00:21:41,800 --> 00:21:45,150
fantastically large numbers and
then keep the rest of the
448
00:21:45,150 --> 00:21:49,470
series intact, sure, I've made
to sum very, very large.
449
00:21:49,470 --> 00:21:52,600
But I've only change it by a
finite amount, which will in
450
00:21:52,600 --> 00:21:54,840
effect, not change
the convergence.
451
00:21:54,840 --> 00:21:58,140
That's all I was saying over
here, that the comparison test
452
00:21:58,140 --> 00:21:59,550
really hinges on what?
453
00:21:59,550 --> 00:22:02,360
Beyond a certain term, you
could stop making the
454
00:22:02,360 --> 00:22:03,800
comparison.
455
00:22:03,800 --> 00:22:06,640
And the second observation is
the converse to what we're
456
00:22:06,640 --> 00:22:07,730
talking about.
457
00:22:07,730 --> 00:22:12,590
Namely, if we know that 'u sub
n' is at least as big as 'd
458
00:22:12,590 --> 00:22:16,730
sub n' for each 'n', where the
series summation 'n' goes from
459
00:22:16,730 --> 00:22:21,990
1 to infinity, 'd sub n' is a
positive divergent series,
460
00:22:21,990 --> 00:22:27,360
then this series, summation 'u
sub n', must also diverge
461
00:22:27,360 --> 00:22:31,510
since its convergence would
imply the convergence of this.
462
00:22:31,510 --> 00:22:35,980
In other words, notice that
by the comparison test, if
463
00:22:35,980 --> 00:22:41,290
summation 'u sub n' converged,
the 'd n's, being less than
464
00:22:41,290 --> 00:22:44,230
the 'u n's would mean that
summation 'dn' end would have
465
00:22:44,230 --> 00:22:45,220
to converge also.
466
00:22:45,220 --> 00:22:47,990
At any rate, these are
the two portions of
467
00:22:47,990 --> 00:22:49,760
the comparison test.
468
00:22:49,760 --> 00:22:52,620
And this is what goes into
the comparison test.
469
00:22:52,620 --> 00:22:54,540
Now the interesting thing,
or the draw back to the
470
00:22:54,540 --> 00:22:56,620
comparison test is
simply this--
471
00:22:56,620 --> 00:23:00,760
that 99 times out of 100, if
you can find a series to
472
00:23:00,760 --> 00:23:04,500
compare a given series with, you
probably would have known
473
00:23:04,500 --> 00:23:06,730
whether the given series
converged or diverged in the
474
00:23:06,730 --> 00:23:07,870
first place.
475
00:23:07,870 --> 00:23:10,050
In other words, somehow or
other, to find the right
476
00:23:10,050 --> 00:23:13,570
series to compare something with
is a rather subtle thing
477
00:23:13,570 --> 00:23:16,440
if you didn't already know the
right answer to the problem.
478
00:23:16,440 --> 00:23:19,370
Well, at any rate, what I'm
trying to drive at is that the
479
00:23:19,370 --> 00:23:25,170
comparison test has as one of
its features a proof for a
480
00:23:25,170 --> 00:23:28,525
more interesting test-- a test
that's far less intuitive--
481
00:23:28,525 --> 00:23:30,980
called the 'ratio test'.
482
00:23:30,980 --> 00:23:32,970
The ratio test says
the following.
483
00:23:32,970 --> 00:23:35,600
Let's suppose again you're
given a positive series.
484
00:23:35,600 --> 00:23:39,610
What you do now is form a
sequence whereby each term in
485
00:23:39,610 --> 00:23:42,200
the sequence is the
ratio between two
486
00:23:42,200 --> 00:23:44,190
terms in the series.
487
00:23:44,190 --> 00:23:48,150
In other words, what I do is I
form the sequence by taking
488
00:23:48,150 --> 00:23:51,050
the second term divided by the
first term, the third term
489
00:23:51,050 --> 00:23:53,910
divided by the second term, the
fourth term divided by the
490
00:23:53,910 --> 00:23:54,590
third term.
491
00:23:54,590 --> 00:23:58,100
Now let's call that general
term 'u sub n'.
492
00:23:58,100 --> 00:24:00,360
This seems a little bit
abstract for you.
493
00:24:00,360 --> 00:24:03,230
Let's look at a more
tangible example.
494
00:24:03,230 --> 00:24:06,380
Suppose I take the series
summation 'n' goes from 1 to
495
00:24:06,380 --> 00:24:10,210
infinity, '10 to the n'
over 'n factorial'.
496
00:24:10,210 --> 00:24:14,620
Notice that in this case, the
n-th term is '10 to the n'
497
00:24:14,620 --> 00:24:16,340
over 'n factorial'.
498
00:24:16,340 --> 00:24:20,910
The 'n plus first' term is '10
to the 'n plus 1'' over ''n
499
00:24:20,910 --> 00:24:22,770
plus 1' factorial'.
500
00:24:22,770 --> 00:24:26,830
So 'u sub n' is the ratio
of the 'n plus first'
501
00:24:26,830 --> 00:24:28,850
term to the n-th term.
502
00:24:28,850 --> 00:24:32,120
In other words, '10 to the 'n
plus 1'' over ''n plus 1'
503
00:24:32,120 --> 00:24:36,700
factorial' divided by '10 to
the n' over 'n factorial'.
504
00:24:36,700 --> 00:24:40,100
'10 to the 'n plus 1'' divided
by '10 to the n' is simply 10.
505
00:24:40,100 --> 00:24:43,740
And ''n plus 1' factorial'
divided by 'n factorial' is
506
00:24:43,740 --> 00:24:45,560
simply 'n plus 1'.
507
00:24:45,560 --> 00:24:48,220
In other words, observe the
structure of the factorials
508
00:24:48,220 --> 00:24:51,900
that you get from the n-th to
the 'n plus first' simply by
509
00:24:51,900 --> 00:24:54,200
multiplying by 'n plus 1'.
510
00:24:54,200 --> 00:24:56,350
Again, the computational
details will be
511
00:24:56,350 --> 00:24:58,870
left for the exercises.
512
00:24:58,870 --> 00:25:01,570
At any rate, then, in this
particular case, 'u sub n'
513
00:25:01,570 --> 00:25:03,950
would be '10 over 'n plus 1''.
514
00:25:03,950 --> 00:25:06,220
Now here's what the
ratio test says.
515
00:25:06,220 --> 00:25:09,480
Assuming that the limit 'u sub
n' as 'n' approaches infinity
516
00:25:09,480 --> 00:25:12,220
exists, call it 'rho'.
517
00:25:12,220 --> 00:25:17,660
Then, the series converges if
rho less than 1 and diverges
518
00:25:17,660 --> 00:25:19,490
if rho is greater than 1.
519
00:25:19,490 --> 00:25:22,750
And the test fails
if rho equals 1.
520
00:25:22,750 --> 00:25:26,830
In other words, if the terms get
progressively smaller, so
521
00:25:26,830 --> 00:25:30,040
that the ratio and the limit
stays less than 1, then the
522
00:25:30,040 --> 00:25:31,450
series converges.
523
00:25:31,450 --> 00:25:34,590
If on the other hand, the ratio
in the limit exceeds 1,
524
00:25:34,590 --> 00:25:37,170
that means the terms are getting
big fast enough so
525
00:25:37,170 --> 00:25:39,020
that the series diverges.
526
00:25:39,020 --> 00:25:42,430
Let me point out an important
observation here.
527
00:25:42,430 --> 00:25:46,310
Notice the difference between
the limit equaling 1 and each
528
00:25:46,310 --> 00:25:48,490
term of the sequence
being less than 1.
529
00:25:48,490 --> 00:25:51,030
In other words, notice that even
if 'u sub n' is less than
530
00:25:51,030 --> 00:25:55,970
1 for every 'n', rho
may still equal 1.
531
00:25:55,970 --> 00:25:59,350
For example, look at the terms
'n' over 'n plus 1''.
532
00:25:59,350 --> 00:26:02,980
For each 'n', 'n' over 'n
plus 1' is less than 1.
533
00:26:02,980 --> 00:26:05,810
Yet the limit as 'n' approaches
infinity is exactly
534
00:26:05,810 --> 00:26:07,360
equal to 1.
535
00:26:07,360 --> 00:26:10,810
Now again, the formal proof of
this is given in the book.
536
00:26:10,810 --> 00:26:13,160
But I thought if I just take
a few minutes to show you
537
00:26:13,160 --> 00:26:16,150
geometrically what's happening
here, you'll get a better
538
00:26:16,150 --> 00:26:19,060
picture to understand what's
happening in the text.
539
00:26:19,060 --> 00:26:22,170
Let's prove this in the case
that rho is less than 1.
540
00:26:22,170 --> 00:26:25,310
Pictorially, if rho is less than
1, that means there's a
541
00:26:25,310 --> 00:26:27,310
space between rho and 1.
542
00:26:27,310 --> 00:26:30,600
That means I can choose an
epsilon such that rho plus
543
00:26:30,600 --> 00:26:34,660
epsilon, which I'll call it 'r',
is a positive number, but
544
00:26:34,660 --> 00:26:36,500
still less than 1.
545
00:26:36,500 --> 00:26:39,370
Now, by definition of rho
being the limit of the
546
00:26:39,370 --> 00:26:43,470
sequence 'a sub 'n plus 1'' over
'a sub n', that means I
547
00:26:43,470 --> 00:26:47,350
can find the capital 'N' for
this given epsilon, such that
548
00:26:47,350 --> 00:26:51,820
whenever 'n' is greater than
capital 'N', that 'a sub 'n
549
00:26:51,820 --> 00:26:56,040
plus 1' over 'a sub n' is less
than rho plus epsilon.
550
00:26:56,040 --> 00:26:59,180
In other words, is less than
'r', where 'r' is some fixed
551
00:26:59,180 --> 00:27:01,380
number now less than 1.
552
00:27:01,380 --> 00:27:05,450
Now let me apply this to
successive values of 'n'.
553
00:27:05,450 --> 00:27:09,150
In other words, taking
'n' to be capital--
554
00:27:09,150 --> 00:27:13,010
see, looking at this thing here,
taking 'n' to be capital
555
00:27:13,010 --> 00:27:18,030
'N', I have 'a' over capital 'N
plus 1', 'a sub capital 'N
556
00:27:18,030 --> 00:27:20,990
plus 1'' over 'a sub N'
is less than 'r'.
557
00:27:20,990 --> 00:27:24,960
In other words, 'a sub capital
'N plus 1'' is less than 'r'
558
00:27:24,960 --> 00:27:26,500
times 'a sub N'.
559
00:27:26,500 --> 00:27:31,720
Similarly, 'a sub capital 'N
plus 2'' over 'a sub capital
560
00:27:31,720 --> 00:27:34,240
'N plus 1'' is also
less than 'r'.
561
00:27:34,240 --> 00:27:38,190
In other words, 'a sub capital
'N plus 2'' is less than 'r'
562
00:27:38,190 --> 00:27:41,920
times 'a sub capital
'N plus 1''.
563
00:27:41,920 --> 00:27:45,890
But 'a sub 'N plus 1'' in
turn is less than 'r'
564
00:27:45,890 --> 00:27:47,300
times 'a sub N'.
565
00:27:47,300 --> 00:27:51,290
Putting this in here, 'a sub
'N plus 2' is less than 'r
566
00:27:51,290 --> 00:27:53,240
squared' times 'a sub N'..
567
00:27:53,240 --> 00:27:57,170
At any rate, if I now sum these
inequalities, you see
568
00:27:57,170 --> 00:27:58,450
what I get is what?
569
00:27:58,450 --> 00:28:01,430
The limit as we go from 'n plus
1' to infinity-- in other
570
00:28:01,430 --> 00:28:03,880
words, the tail end of
this particular sum
571
00:28:03,880 --> 00:28:05,340
is less than what?
572
00:28:05,340 --> 00:28:08,310
Well, I can factor out the
'a sub N' over here.
573
00:28:08,310 --> 00:28:10,680
And what's left inside
is what?
574
00:28:10,680 --> 00:28:14,780
'r' plus 'r squared' plus
'r cubed', et cetera.
575
00:28:14,780 --> 00:28:19,550
But this particular series is a
convergent geometric series.
576
00:28:19,550 --> 00:28:22,720
In other words, this must
converge because this
577
00:28:22,720 --> 00:28:25,870
converges, and this is less
than, term by term,
578
00:28:25,870 --> 00:28:27,280
the terms over here.
579
00:28:27,280 --> 00:28:30,910
In other words, notice that the
proof of the ratio test
580
00:28:30,910 --> 00:28:34,690
hinges on knowing two things,
the comparison test and the
581
00:28:34,690 --> 00:28:37,110
convergence of a geometric
series.
582
00:28:37,110 --> 00:28:40,490
Again, the reason I go through
this as rapidly as I do is
583
00:28:40,490 --> 00:28:44,360
that every detail is done
magnificently in the textbook.
584
00:28:44,360 --> 00:28:47,640
All I want you to see here is
the overview of how these
585
00:28:47,640 --> 00:28:49,730
tests come about.
586
00:28:49,730 --> 00:28:54,060
Finally, we have another
powerful test called the
587
00:28:54,060 --> 00:28:55,360
'integral test'.
588
00:28:55,360 --> 00:28:58,740
And the integral test
essentially equates positive
589
00:28:58,740 --> 00:29:01,480
series with improper
integrals.
590
00:29:01,480 --> 00:29:05,500
By the way, I have presented the
material, so to speak, in
591
00:29:05,500 --> 00:29:08,640
the order of appearance
in the textbook.
592
00:29:08,640 --> 00:29:12,940
You see, the comparison test,
ratio test, and integral test
593
00:29:12,940 --> 00:29:15,690
are given in that order
in the text.
594
00:29:15,690 --> 00:29:17,760
And so, I kept the
same order here.
595
00:29:17,760 --> 00:29:20,470
However, it's interesting to
point out that the integral
596
00:29:20,470 --> 00:29:24,450
test, in a way, is a companion
of the comparison test.
597
00:29:24,450 --> 00:29:26,600
And let me show you first
of all what the
598
00:29:26,600 --> 00:29:28,130
integral test says.
599
00:29:28,130 --> 00:29:31,570
It says, and I've written out
the formal statements here.
600
00:29:31,570 --> 00:29:34,250
I'll show you pictorially what
this means in a moment.
601
00:29:34,250 --> 00:29:37,040
But it says, suppose there is
a decreasing continuous
602
00:29:37,040 --> 00:29:39,410
function, 'f of x'.
603
00:29:39,410 --> 00:29:40,760
Notice the word continuous
in here.
604
00:29:40,760 --> 00:29:43,610
That guarantees that we can
integrate 'f of x'.
605
00:29:43,610 --> 00:29:48,360
Such that 'f' evaluated at each
integral value of 'x',
606
00:29:48,360 --> 00:29:52,240
say 'x' equals 'n', is 'u sub
n', where 'u sub n' is the
607
00:29:52,240 --> 00:29:54,860
n-th term of the positive
series 'u1'
608
00:29:54,860 --> 00:29:56,650
plus 'u2', et cetera.
609
00:29:56,650 --> 00:29:59,830
Then, what the integral test
says is that the series
610
00:29:59,830 --> 00:30:03,760
summation 'n' goes from 1 to
infinity 'u sub n', and the
611
00:30:03,760 --> 00:30:08,220
integral 1 to infinity, ''f
of x' dx', either converge
612
00:30:08,220 --> 00:30:11,140
together or diverge together.
613
00:30:11,140 --> 00:30:13,420
In other words, we can test
a particular series for
614
00:30:13,420 --> 00:30:16,110
convergence by knowing
whether a particular
615
00:30:16,110 --> 00:30:18,200
improper integral converges.
616
00:30:18,200 --> 00:30:20,390
And to show you what this
thing means pictorially,
617
00:30:20,390 --> 00:30:21,860
simply observe this.
618
00:30:21,860 --> 00:30:26,220
See, what we're saying is,
suppose that when you plot the
619
00:30:26,220 --> 00:30:27,800
terms of the series--
620
00:30:27,800 --> 00:30:31,300
so, 'u1', 'u2', 'u3', 'u4'--
621
00:30:31,300 --> 00:30:34,800
that these happen to be the
integral values of a
622
00:30:34,800 --> 00:30:37,770
continuous curve, 'y' equals
'f of x', which not only
623
00:30:37,770 --> 00:30:41,060
passes through these points, but
is a continuous decreasing
624
00:30:41,060 --> 00:30:43,100
function as this happens.
625
00:30:43,100 --> 00:30:47,300
Then what the statement is is
that the sum of these lengths
626
00:30:47,300 --> 00:30:50,200
converges if and only
if the area under
627
00:30:50,200 --> 00:30:52,610
the curve is finite.
628
00:30:52,610 --> 00:30:54,200
And the proof go something
like this.
629
00:30:54,200 --> 00:30:56,530
It's a rather ingenious
type of thing.
630
00:30:56,530 --> 00:31:00,470
You see, notice that if this
height is 'u sub 1', and the
631
00:31:00,470 --> 00:31:04,070
base of the rectangle is 1,
notice that numerically, the
632
00:31:04,070 --> 00:31:06,600
area of the rectangle--
it's quite in general.
633
00:31:06,600 --> 00:31:11,000
If the base of a rectangle is 1,
numerically the area of the
634
00:31:11,000 --> 00:31:14,080
rectangle equals the height,
because the area is the base
635
00:31:14,080 --> 00:31:14,930
times the height.
636
00:31:14,930 --> 00:31:17,620
If the base is 1, the area
equals the height.
637
00:31:17,620 --> 00:31:19,510
So the idea is simply this.
638
00:31:19,510 --> 00:31:20,370
Lookit.
639
00:31:20,370 --> 00:31:23,060
Suppose, for example,
that we look at this
640
00:31:23,060 --> 00:31:24,630
diagram over here.
641
00:31:24,630 --> 00:31:28,490
Notice that in this diagram,
if we stop at n, the area
642
00:31:28,490 --> 00:31:32,500
under this curve is the integral
from 1 to 'n', or 1
643
00:31:32,500 --> 00:31:35,350
to 'n plus 1', because of how
these lines are drawn.
644
00:31:35,350 --> 00:31:40,320
See, notice that the first
height stops at the number 2.
645
00:31:40,320 --> 00:31:43,850
The second base stops at
number 3, et cetera.
646
00:31:43,850 --> 00:31:46,950
The idea is this, though, that
the area under the curve in
647
00:31:46,950 --> 00:31:51,130
general, from 1 to 'n plus 1',
is integral from 1 to 'n plus
648
00:31:51,130 --> 00:31:53,060
1', ''f of x' dx'.
649
00:31:53,060 --> 00:31:58,320
On the other hand, the area of
the rectangles are what?
650
00:31:58,320 --> 00:32:02,970
'u1' plus 'u2' plus
'u3', up to 'u n'.
651
00:32:02,970 --> 00:32:07,190
In other words, for any given
'n', 'u1' up to u n' is
652
00:32:07,190 --> 00:32:10,110
greater than or equal to this
particular integral.
653
00:32:10,110 --> 00:32:14,290
Consequently, taking the
limit as 'n' goes to
654
00:32:14,290 --> 00:32:16,220
infinity, we get what?
655
00:32:16,220 --> 00:32:20,990
The summation 'u n' is greater
than or equal to integral from
656
00:32:20,990 --> 00:32:23,380
one to infinity ''f of x' dx'.
657
00:32:23,380 --> 00:32:26,530
Consequently, if this integral
diverges, meaning that this
658
00:32:26,530 --> 00:32:30,030
gets very large, the fact that
this can be no less than this
659
00:32:30,030 --> 00:32:33,150
means that this too
must also diverge.
660
00:32:33,150 --> 00:32:37,340
Correspondingly, if we now do
the same problem, but draw the
661
00:32:37,340 --> 00:32:41,980
thing slightly differently,
notice that now in this
662
00:32:41,980 --> 00:32:47,770
particular picture, the area
under the curve is integral
663
00:32:47,770 --> 00:32:50,780
from 1 to 'n', ''f of x' dx'.
664
00:32:50,780 --> 00:32:53,400
On the other hand, the area of
the rectangles are what?
665
00:32:53,400 --> 00:32:56,840
It's 'u2' plus 'u3'
up to 'u n'.
666
00:32:56,840 --> 00:33:01,220
But now you see the rectangles
are inscribed under the curve.
667
00:33:01,220 --> 00:33:05,220
Consequently, 'u2' plus et
cetera, up to 'u n', is less
668
00:33:05,220 --> 00:33:06,860
than this integral.
669
00:33:06,860 --> 00:33:10,767
Therefore, if I add 'u1' onto
both sides, the sum 'u1', et
670
00:33:10,767 --> 00:33:14,920
cetera, up to 'u n', is less
than 'u1' plus integral 1 to
671
00:33:14,920 --> 00:33:17,090
'n', ''f of x' dx'.
672
00:33:17,090 --> 00:33:20,680
If I now take the limit as n
goes to infinity, you see this
673
00:33:20,680 --> 00:33:23,920
becomes summation un.
674
00:33:23,920 --> 00:33:26,330
n goes from 1 to infinity.
675
00:33:26,330 --> 00:33:32,880
This becomes integral from 1
to infinity, ''f of x' dx'.
676
00:33:32,880 --> 00:33:37,120
And therefore, if this now
converges, the sum on the
677
00:33:37,120 --> 00:33:38,630
right is finite.
678
00:33:38,630 --> 00:33:41,780
Since the sum on the left cannot
exceed the sum on the
679
00:33:41,780 --> 00:33:45,240
right, the sum on the left
must also be finite.
680
00:33:45,240 --> 00:33:48,960
Consequently, the convergence
of the integral implies the
681
00:33:48,960 --> 00:33:50,930
convergence of the series.
682
00:33:50,930 --> 00:33:54,320
Again, I apologize for doing
this this rapidly.
683
00:33:54,320 --> 00:33:57,780
All I wanted you to do was
to get an idea of what's
684
00:33:57,780 --> 00:33:58,510
happening here.
685
00:33:58,510 --> 00:34:01,760
Because as I say, the book is
magnificent in this section.
686
00:34:01,760 --> 00:34:03,960
The proofs are very well
self-contained.
687
00:34:03,960 --> 00:34:07,790
At any rate, this gives us three
rather powerful tests,
688
00:34:07,790 --> 00:34:11,190
which I will drill you on in
the exercises for testing
689
00:34:11,190 --> 00:34:13,350
convergence of positive
series.
690
00:34:13,350 --> 00:34:16,590
What we're going to do next time
is to come to grips with
691
00:34:16,590 --> 00:34:17,989
a much more serious problem.
692
00:34:17,989 --> 00:34:22,969
And that is, what do you do if
the series isn't positive?
693
00:34:22,969 --> 00:34:24,380
But that's another story.
694
00:34:24,380 --> 00:34:26,360
And so until next
time, good bye.
695
00:34:26,360 --> 00:34:29,170
696
00:34:29,170 --> 00:34:32,370
Funding for the publication of
this video was provided by the
697
00:34:32,370 --> 00:34:36,420
Gabriella and Paul Rosenbaum
Foundation.
698
00:34:36,420 --> 00:34:40,590
Help OCW continue to provide
free and open access to MIT
699
00:34:40,590 --> 00:34:44,800
courses by making a donation
at ocw.mit.edu/donate.
700
00:34:44,800 --> 00:34:49,535