WEBVTT

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So what we're going
to attempt to do

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is evaluate this
sum right over here,

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evaluate what this
series is, negative 2

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over n plus 1 times n plus
2, starting at n equals 2,

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all the way to infinity.

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And if we wanted to see what
this looks like, it starts at n

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equals 2.

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So when n equals 2, this
is negative 2 over 2

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plus 1, which is 3, times
2 plus 2, which is 4.

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Then when n is equal to 3,
this is negative 2 over 3

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plus 1, which is 4, times
3 plus 2, which is 5.

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And it just keeps going
like that, negative 2

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over 5 times 6.

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And it just keeps
going on and on and on.

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And now, it looks pretty clear
that each successive term

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is getting smaller.

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And it's getting
smaller reasonably fast.

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So it's reasonable to assume
that even though you have

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an infinite number
of terms, it actually

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might give you a finite value.

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But it doesn't jump out
at me, at least the way

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that I've looked
at it right now,

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as to what this sum
would be, or how

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to actually figure out that sum.

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So what I want you to do
now is pause this video.

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And I'm going to give you a hint
about how to think about this.

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Try to dig up your memories
of partial fraction expansion,

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or partial fraction
decomposition,

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to turn this expression into
the sum of two fractions.

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And that might help us think
about what this sum is.

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So I'm assuming you've
given a go at it.

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So let's try to
manipulate this thing.

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And let's see if we can rewrite
this as a sum of two fractions.

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So this is negative
2 over-- and I'm

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going to do this in two
different colors-- n plus 1

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times n plus 2.

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And we remember from our
partial fraction expansion

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that we can rewrite this as
the sum of two fractions,

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as A over n plus 1
plus B over n plus 2.

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And why is this reasonable?

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Well, if you're
adding two fractions,

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you want to find a
common denominator, which

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would be a multiple of
the two denominators.

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This is clearly a multiple of
both of these denominators.

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And we learned in partial
fraction expansion

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that whatever we have up here,
especially because the degree

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here is lower than the degree
down here, whatever we have

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up here is going to be a degree
lower than what we have here.

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So this is a first-degree
term in terms of n,

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so these are going to be
constant terms up here.

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So let's figure out
what A and B are.

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So if we perform
the addition-- well,

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let's just rewrite
both of these with

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the same common denominator.

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So let's rewrite
A over n plus 1,

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but let's multiply the numerator
and denominator by n plus 2.

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So we multiply the numerator
by n plus 2 and the denominator

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by n plus 2.

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I haven't changed the value
of this first fraction.

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Similarly, let's do the same
thing with B over n plus 2.

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Multiply the numerator and the
denominator by n plus 1, so

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n plus 1 over n plus 1.

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Once again, I haven't change
the value of this fraction.

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But by doing this, I now
have a common denominator,

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and I can add.

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So this is going to be equal
to n plus 1 times n plus 2 is

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our denominator.

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And then our numerator--
let me expand it out.

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This is going to be,
if I distribute the A,

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it is An plus 2A.

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So let me write
that, An plus 2A.

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And then let's distribute
this B, plus Bn plus B.

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Now, what I want to do
is I want to rewrite this

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so I have all of the n terms.

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So for example, An plus Bn--
I could factor an n out.

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And I could rewrite that
as A plus B times n, those

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two terms right over there.

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And then these two
terms, the 2A plus B,

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I could just write it
like this, plus 2A plus B.

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And, of course, all of that is
over n plus 1 times n plus 2.

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So how do we solve for A and B?

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Well, the realization
is that this thing

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must be equal to negative 2.

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These two things must
be equal to each other.

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Remember, we're making
the claim that this,

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which is the same thing
as this, is equal to this.

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That's the whole reason
why we started doing this.

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So we're making the
claim that these two

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things are equivalent.

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We're making this claim.

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So everything in the numerator
must be equal to negative 2.

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So how do we do that?

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It looks like we have
two unknowns here.

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To figure out two unknowns, we
normally need two equations.

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Well, the realization
here is, look,

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we have an n term on
the left-hand side here.

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We have no n term here.

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So you literally could view
this, instead of just viewing

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this as negative
2, you could view

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this as negative 2 plus
0n, plus 0 times n.

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That's not "on."

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That's 0-- let me write
it this way-- 0 times n.

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So when you look at
it this way, it's

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clear that A plus B is
the coefficient on n.

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That must be equal to 0.

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A plus B must be equal to 0.

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And this is kind of
bread-and-butter partial

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fraction expansion.

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We have other videos on that
if you need to review that.

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And the constant part, 2A plus
B, is equal to negative 2.

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And so now we have two
equations in two unknowns.

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And we could solve it a
bunch of different ways.

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But one interesting way is
let's multiply the top equation

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by negative 1.

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So then this becomes negative
A minus B is equal to-- well,

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negative 1 times 0 is still 0.

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Now we can add these
two things together.

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And we are left with 2A minus
A is A, plus B minus B-- well,

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those cancel out.

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A is equal to negative 2.

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And if A is equal to
negative 2, A plus B is 0,

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B must be equal to 2.

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Negative 2 plus 2 is equal to 0.

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We solved for A. And then I
substituted it back up here.

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So now we can rewrite all
of this right over here.

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We can rewrite it as
the sum-- and actually,

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let me do a little bit instead.

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Let me just write it as
a finite sum as opposed

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to an infinite sum.

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And then we can just take the
limit as we go to infinity.

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So let me rewrite it like this.

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So this is the sum from n
equals 2-- instead to infinity,

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I'll just say to capital
N. And then later, we

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could take the limit as this
goes to infinity of-- well,

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instead of writing this, I can
write this right over here.

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So A is negative 2.

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So it's negative
2 over n plus 1.

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And then B is 2,
plus B over n plus 2.

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So once again, I've just
expressed this as a finite sum.

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Later, we can take the limit
as capital N approaches

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infinity to figure out
what this thing is.

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Oh, sorry, and B-- let
me not write B anymore.

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We now know that B
is 2 over n plus 2.

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Now, how does this actually
go about helping us?

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Well, let's do what
we did up here.

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Let's actually write out what
this is going to be equal to.

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This is going to be
equal to-- when n is 2,

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this is negative 2/3, so
it's negative 2/3, plus 2/4.

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So that's n equals-- let me
do it down here, because I'm

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about to run out of real estate.

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That is when n is equal to 2.

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Now, what about when
n is equal to 3?

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When n is equal to 3, this
is going to be negative 2/4

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plus 2/5.

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What about when n is equal to 4?

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I think you might see a pattern
that's starting to form.

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Let's do one more.

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When n is equal to
4, well, then, this

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is going to be
negative 2/5-- let

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me do that same blue color--
negative 2/5 plus 2/6.

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And we're just
going to keep going.

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Let me scroll down to
get some space-- we're

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going to keep going all the
way until the N-th term.

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So plus dot dot dot plus
our capital N-th term,

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which is going to be negative
2 over capital N plus 1 plus 2

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over capital N plus 2.

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So I think you might
see the pattern here.

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Notice, from our first when
n equals 2, we got the 2/4.

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But then when n equals 3,
you had the negative 2/4.

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That cancels with that.

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When n equals 3, you had 2/5.

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Then that cancels when n
equals 4 with the negative 2/5.

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So the second term cancels
with-- the second part,

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I guess, for each
n, for each index,

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cancels out with the first
part for the next index.

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And so that's just going to
keep happening all the way

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until n is equal to capital N.

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And so this is going to
cancel out with the one right

00:10:02.300 --> 00:10:03.110
before it.

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And all we're going
to be left with

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is this term and this
term right over here.

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So let's rewrite that.

00:10:15.950 --> 00:10:19.090
So we get-- let's
get more space here.

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This thing can be rewritten
as the sum from lowercase n

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equals 2 to capital
N of negative 2

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over n plus 1 plus
2 over n plus 2

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is equal to-- well, everything
else in the middle canceled

00:10:39.440 --> 00:10:39.940
out.

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We're just left
with negative 2/3

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plus 2 over capital N plus 2.

00:10:50.460 --> 00:10:53.110
So this was a huge
simplification right over here.

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And remember, our original sum
that we wanted to calculate,

00:10:57.000 --> 00:11:00.510
that just has a limit as
capital N goes to infinity.

00:11:00.510 --> 00:11:04.620
So let's just take the limit
as capital N goes to infinity.

00:11:04.620 --> 00:11:06.170
So let me write it this way.

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Well, actually, let
me write it this way.

00:11:07.878 --> 00:11:10.810
The limit-- so we can
write it this way.

00:11:10.810 --> 00:11:15.350
The limit as
capital N approaches

00:11:15.350 --> 00:11:20.030
infinity is going to be equal
to the limit as capital N

00:11:20.030 --> 00:11:22.239
approaches infinity
of-- well, we just

00:11:22.239 --> 00:11:23.280
figured out what this is.

00:11:23.280 --> 00:11:33.240
This is negative 2/3 plus
2 over capital N plus 2.

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Well, as n goes to
infinity, this negative 2/3

00:11:36.180 --> 00:11:37.600
doesn't get impacted at all.

00:11:37.600 --> 00:11:40.375
This term right over here, 2
over an ever larger number,

00:11:40.375 --> 00:11:42.000
over an infinitely
large number-- well,

00:11:42.000 --> 00:11:43.520
that's going to go to 0.

00:11:43.520 --> 00:11:47.990
And we're going to be
left with negative 2/3.

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And we're done.

00:11:48.750 --> 00:11:54.740
We were able to figure out the
sum of this infinite series.

00:11:54.740 --> 00:11:58.460
So this thing right over here
is equal to negative 2/3.

00:11:58.460 --> 00:12:01.320
And this type of series is
called a telescoping series--

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telescoping, I should say.

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This is a telescoping series.

00:12:09.270 --> 00:12:12.130
And a telescoping series
is a general term.

00:12:12.130 --> 00:12:14.500
So if you were to
take its partial sums,

00:12:14.500 --> 00:12:18.430
it has this pattern right over
here, where, in each term,

00:12:18.430 --> 00:12:20.020
you're starting to
cancel things out.

00:12:20.020 --> 00:12:23.270
So what you're left with
is just a fixed number

00:12:23.270 --> 00:12:25.520
of terms at the end.

00:12:25.520 --> 00:12:27.130
But either way,
this was a pretty--

00:12:27.130 --> 00:12:29.421
it's a little bit hairy, but
it was a pretty satisfying

00:12:29.421 --> 00:12:30.640
problem.