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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-- rather,

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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-- 2,

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just so I don't have
any funny business here

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

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And a second series,
sum from n equals 2

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to infinity of 1 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

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to work on this
question, come back,

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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

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that these might be
questions that are

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amenable to the integral test.

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One thing we can do--
you know, if you,

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if I hadn't given you that hint,
how could you figure this out?

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Well, you can look
at these integrands.

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And they 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--

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so this is the associated
function, 1 over x log x.

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This continuous
function is a function

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that we have-- you know, it
looks sort of like some things

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that we've integrated before.

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So that's one hint
for the integral test.

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Another hint for
the integral test

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is, you just don't know
very many 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

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not going to be a 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

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by 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.

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Now that we've got the
idea for the integral test,

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what do we do?

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Well, we know that this
series converges if and only

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if the associated
definite integral,

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the associated improper
definite integral converges.

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So let's do the first one first.

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What's the integral
associated with it?

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

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we-- frequently, we replace the
variable n with the variable x,

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although that doesn't
really matter.

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

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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

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and only if this integral does.

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

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is that often
there are integrals

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that are easy to compute,
while the associated series

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are hard to compute.

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So in this case,
this is an integral

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that we have tools to
know, to compute with.

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And in particular,
the tool that we have,

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is that there's a simple
little substitution that

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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

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need 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 test,

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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

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where the function explodes,
all the interesting stuff

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is whether the function is
big as it goes to infinity.

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So if you move a little
round at the bottom,

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you'll change its
numerical value,

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but you won't change whether
it 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,

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ln of x goes to infinity,
so the upper bound is also u

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equals infinity.

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And now this is
an easy integral.

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This is just ln of
u between-- well,

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ln 2, which again, really
doesn't matter, and infinity.

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And we see that at
the upper bound,

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we get ln of infinity,
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,

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and we've found 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,

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the similar-looking integrand.

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And in fact, so we
get-- so the integral

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that we want to look at
is the integral from 2

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to infinity of 1 over x
times log 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,

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du 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, minus 1

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over u goes to 0.

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So this is 0 minus, and now with
the lower bound, it's minus 1

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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

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

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the value of the integral, and
the first terms of the sum.

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I realized that I should
have said one thing

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at the beginning,
which is that we didn't

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check that the hypotheses of the
integral test are valid here.

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So remember that the
integral test only

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applies if this
function that you use

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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
the 1 over it is positive

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and decreasing.

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So it's easy to
check, in this case,

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that the conditions of
the integral test apply.

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

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if you ever want to
apply the integral test,

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that's something you should
check yourself before you

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go and apply it to anything.

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Make sure that you
really have a function

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to which it does apply.

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But then, once you have
a function to which it

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does 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,

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and an integral where
the integral converges.

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And so the corresponding sums,
the first one will diverge

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and the second
one will converge.

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I'll end there.