WEBVTT
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PROFESSOR: Hi, welcome
back to recitation.
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Today we are going to talk
about an infinite series
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and discuss its convergence.
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So in particular I have
this infinite series.
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The sum from n
equals 1 to infinity
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of 1 divided by the
product n times n plus 1.
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So what I'd like you
to do is to compute
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a few terms of the series,
compute a few partial sums,
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and use that to get
a sense for what
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you think the series is doing.
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Is it converging?
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Is it diverging?
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If it's converging,
can you figure out
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what value it's converging to?
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So why don't you
pause the video,
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take some time to try that out,
see what you get, come back
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and we can do it together.
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So this is a nice series.
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It has terms that
are easy to compute.
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And I've taken the liberty
of computing a few of them
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in advance, and I've
put them up over here.
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So for n from 1 to 5, the
terms that we're adding up
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are 1 over n times n plus 1.
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So that's when n
equals 1, that's
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1 over 1 times 2, which is 1/2.
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When it is 2, it's 1 over
2 times 3, which is 1/6.
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Then we've got 1/12,
1/20, 1/30, and so on.
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So those are the
things we're adding up.
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And then the partial sums,
the nth partial sums.
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Well, the first one is just
the first term, which is 1/2.
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The second one, we take the
first term and the second term
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and we add them together.
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So 1/2 plus 1/6 is 2/3.
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The third one, we take
the first three terms
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and add them together
and that gives us 3/4.
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And OK, so I computed the first
five partial sums here as well.
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Now, if you look
at this column-- so
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remember that the limit, that
the value of an infinite series
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is defined to be the
limit of its partial sums.
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So if we want to know,
what is the value
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of this infinite series
that we started with,
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does it converge,
does it diverge,
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what we have to do
to figure that out
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is we have to take
its partial sums
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and we have to
compute their limit.
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And if we-- if their
limit doesn't exist,
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then it diverges.
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If their limit does exist,
then the sum of the series
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is equal to what that
value of that limit is.
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And if you look at
these terms here,
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you'll see that they, there's
a little bit of a pattern
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here, right?
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So these, this is 1/2,
2/3, 3/4, 4/5, 5/6.
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That's a pretty nice
sequence of numbers.
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It's, you know, we could
probably guess at this point
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that the next one is going
to be 6/7 then 7/8 and so on.
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So that would be a guess.
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One way we can
actually prove this
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is, so we have this guess
that-- let me write it down.
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Guess is that S_n is
equal to n over n plus 1.
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Now if you wanted to confirm
this guess, what you'd
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have to do is you have
to just figure out
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how could you prove that.
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Well, one thing
you can do is you
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can say, well, S n
plus 1 is equal to S_n
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plus the next term, right?
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So in our case, S n
plus 1 is equal to S_n
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plus the next term, the n plus
first term, which in our case
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is 1 over n plus
1 times n plus 2.
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So, all right, so
that's not maybe obvious
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what to do with this, but
you could split this up,
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really you can split it
up by partial fractions.
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And you can write this as say
S_n plus-- so if you split this
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up by partial fractions,
what you'll get
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is that it's exactly equal
to 1 over n plus 1 minus 1
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over n plus 2.
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And from here it's
easy to see that, well,
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if S_n is equal to
n over n plus 1,
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then this will be
equal to 1 minus 1
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over n plus 2, which is
n plus 1 over n plus 2.
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And so using the process known
as mathematical induction,
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you have that it follows
for all values of n.
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So because of this nice
expression for the term,
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it's easy to see that this
pattern will continue forever.
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OK, and so that, you
know, that's just
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a sketch of how you
would prove this.
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Now once you've
proven this, it's
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easy to see that the-- once
you know that this is true,
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it's easy to see what
this limit is, right?
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As n goes to infinity,
this just approaches
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1 and that means
the series converges
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and the limit of
the series is 1.
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So here we have a nice example
of a series that converges
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and where it actually
is possible to compute
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the limit of this series.
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This isn't possible for
most, for all series
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or even for most series.
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Even ones with
fairly nice terms,
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it's often very difficult to
figure out what their limit is,
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but in this case
it's not hard to do
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and we have precisely that
the value of the series is 1.
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So I'll end there.