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Hi.

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Welcome back to recitation.

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In lecture, you've been learning
about computing,

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rather not computing, but
determining whether series

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converge or diverge, and
different tests for that.

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In particular, you've learned
the integral test.

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So here are a couple of series
that you haven't seen before.

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The sum from n equals
2 to infinity.

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2, just so I don't have any
funny business here, of

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dividing by 0, of 1 over
n times log of n.

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And a second series, sum from
n equals 2 to infinity of 1

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over n times log of n
quantity squared.

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So the question is, do these
series converge or diverge?

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So why you pause the video, take
some time to work on this

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question, come back, and we
can work on it together.

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Welcome back.

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Before you left, I gave you a
little hint that these might

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be questions that are amenable
to the integral test. One

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thing we can do, you know, if
you, if I hadn't given you

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that hint, how could you
figure this out?

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Well, you can look at
these integrands.

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And that there don't really
look a lot like anything

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you've seen before.

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But the associated functions,
right, so this is the

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associated function,
1 over x log x.

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This continuous function is
a function that we have--

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you know, it looks sort of like
some things that we've

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integrated before.

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So that's one hint for the
integral test. Another hint

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for the integral test is, you
just don't know very many

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tests right now.

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So it's kind of a small
selection of options.

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Another thing is that, you know,
it's not going to be a

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nice series.

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It's not going to have a
nice numerical value.

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This log n thing is
behaving badly.

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You're not going to be able to
compute values, exact values,

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of the partial sums, any nicer
than they looked just by

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writing down that sum.

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So OK.

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So after we've got the idea
of the integrals--

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that's how we get the idea for
the integral test. Now that

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we've got the idea for the
integral test, what do we do?

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Well, we know that this series
converges if and only if the

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associated definite integral,
the associated improper

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definite integral converges.

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So let's do the first one first.
What's the integral

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associated with it?

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Well, we take the integrand,
and, you know, we frequently,

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we replace the variable n with
the variable x, although that

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doesn't really matter.

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And so what we do, is we look
at the integral of this

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integrand over the
same region.

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So in this case, from
2 to infinity.

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And so what we know is that this
sum converges if and only

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if this integral does.

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So then, the reason this is a
nice thing to do, is that

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often there are integrals that
are easy to compute, while the

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associated series are
hard to compute.

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So in this case, this is an
integral that we have tools to

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know, to compute with.

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And in particular, the tool that
we have, is that there's

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a simple little substitution
that will work on this series.

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So that's the substitution,
u equals log x.

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So we make the substitution
u equals ln of x.

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Then du is equal to
1 over x times dx.

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So this is the integral of--
so 1 over x dx, that's du.

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And so it's 1 over u du.

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This is a definite integral,
so I also need

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to change my bounds.

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So when x is 2, u is ln of 2,
although the lower bound

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doesn't really matter very much
when we do the integral

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test, because you know, if you
change it a little bit, that's

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not going to change
the, you know--

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as long as you don't move it
across a place where the

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function explodes, all the
interesting stuff is whether

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the function is big as
it goes to infinity.

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So if you move a little round
at the bottom, you'll change

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its numerical value, but you
won't change whether it

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converges or diverges.

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But in any case, ln of 2.

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And then when x goes to
infinity, ln of x goes to

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infinity, so the upper bound
is also u equals infinity.

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And now this is an
easy integral.

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This is just ln of u between--

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well, ln 2, which again, really

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doesn't matter, and infinity.

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And we see that at the upper
bound, we get ln of infinity,

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which is infinity.

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So this thing is infinity.

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So our original series
diverges.

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OK.

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So we've applied the integral
test here, and we've found

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that our series diverges.

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What about this second one?

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Well, here, we can again apply
the integral test, the

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similar-looking integrand.

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And in fact, so we get, so the
integral that we want to look

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at is the integral from 2 to
infinity of 1 over x times log

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of x squared dx.

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And the same substitution
is going to work here.

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So we're going to use the
substitution u equals ln x, du

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equals 1 over x times dx.

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And the bounds are going to be
the same. ln 2 to infinity.

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OK.

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But what happens when we
make this substitution?

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Well, the 1 over x dx
is still the du.

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But then here, this time, we
have 1 over u squared, right?

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Because we've got an ln of
x squared, and u is ln x.

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So OK.

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So we get 1 over u squared.

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So again, this is easy
to integrate.

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So this is 1 over u squared, so
that's going to be minus 1

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over u when we integrate it
between ln 2 and infinity.

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OK.

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So we take the two values here,
as u goes to infinity,

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minus 1 over u goes to 0.

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So this is 0 minus, and now
with the lower bound, it's

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minus 1 over ln 2, and
this is just ln 2.

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So this integral converges to a
nice finite value, ln of 2,

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so that means the sum
converges as well.

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All right?

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And in fact, if you go back and
look at the lecture video,

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you'll see that you can actually
bound the value of

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the sum in terms of this value,
ln of 2, the value of

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the integral, and the first
terms of the sum.

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I realized that I should have
said one thing at the

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beginning, which is that we
didn't check that the

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hypotheses of the integral
test are valid here.

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So remember that the integral
test only applies if this

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function that you use is a
decreasing positive function

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on the interval in question.

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And I didn't actually check
those conditions.

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They're easy to see in
this case, right?

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Because for n bigger than 2, n
is positive and increasing,

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and log n is positive and
increasing, so the product is

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positive and increasing, so
there the 1 over it is

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positive and decreasing.

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So it's easy to check, in this
case, that that the conditions

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of the integral test apply.

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In the, when you're doing this
out in the real world, if you

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ever want to apply the
integral test, that's

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something you should check
yourself before you go and

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apply it to anything.

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Make sure that you really
have a function to

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which it does apply.

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But then, once you have a
function to which it does

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apply, especially if it's a
nice, easy to integrate

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function like this, you
can easily apply it.

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And in this case, we applied
it both directions.

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We saw an integral where the
integral diverges, and an

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integral where the integral
converges.

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And so the corresponding sums,
the first one will diverge and

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the second one will converge.

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I'll end there.

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