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Let's add up the terms in a geometric
progression in general.

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[SOUND].

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Pick a value for r.
Here's a series I want to evaluate.

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The sum k goes from zero to infinity of
sum number r raised to the kth power.

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These series have a name.

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These things are called geometric series.

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We've already seen this when r equals 1
half.

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Right.

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So if I plug in r equals 1 half and look
at this geometric series, this series

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we've already evaluated.
It converges.

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And it's value is 2.

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There are certain other of r, we can also

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figure out what's happening with this
series right away.

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Instead of considering 1 half, let's think
back to this

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original geometric series, and let's plug
in 1 for r.

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If I plug in 1 for r, does this geometric
series converge or diverge?

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Well, it's not too hard to think about.

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Let's start writing down what this series
looks like.

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This series would be the zeroth term would
be 1 to the zero, which is 1.

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The k equals 1 term would be 1 to the 1,
which is 1.

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The k equal 2 term would be 1 squared,
which is 1.

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Right.

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This series is just 1 plus, 1 plus, 1 plus
1 and so on.

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Does this series converge or diverge?

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This series diverges.
Because the

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limit of the partial sums is infinity.

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Something bad also happens when r equals
negative 1.

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So yeah, instead of looking at r equals 1,
let's plug in

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r equals minus 1 And see what sort of
terrible thing happens.

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So, r equals minus 1.

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Well, then I could start writing down the
first few terms of this series as well.

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When I plug in k equals 0, I get minus 1
to the zeroth

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power, which is 1.

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When I plug in k equals 1, I get minus 1
to the first power, which is minus 1.

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Then I plug in k equals 2, and I get minus
1 squared, which is plus 1.

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Then I plug in k equals 3, and I get minus
1 cubed, which is minus 1.

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So what this series looks like, is 1 minus
1 plus

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1 minus 1, plus 1 minus 1, plus 1 minus 1,
and

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so on.
Does this series converge or diverge?

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More precisely, what's the limit of the
partial sums?

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Well, when I add just the first term I get
1.

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When I add the first two terms I get 0.

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When I add the first three terms together,
I get 1.

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When I add the first four terms together,
I get 0.

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And this pattern continues.
When I add the first five terms together,

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I get 1.

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When I add the first six terms together, I
get 0.

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The partial sums are flip-flopping between
1 and 0.

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And does the sequence 1, 0, 1, 0, 1, 0,
does that sequence converge?

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No.

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And because the sequence of partial sums

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doesn't converge, the original series
doesn't convert either.

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Let's think about what happens for some
other values of r.

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Remember, to evaluate a series,

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I want to take a limit of the sequence of
partial sums, and symbols.

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Here's the partial sum.

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S sub n is the nth partial sum of this
geometric series, and it's

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the sum of terms from k equals 0 all the
way up to n.

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So here's the k equals 0 term, here's the
k equals 1 term.

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Dot dot dot, hides all the other terms,
and then here's the k equals n term.

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And the real question is, what's the limit
of these

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partial sums?

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If I take the limit of this as n goes to
infinity, that's the value of this series.

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I want to compute the limit of that
sequence.

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So, I really want to compute the limit of
this, the nth partial sum.

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And if I can compute this limit, then

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I've computed the value of the geometric
series.

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Well, I've been calling this nth partial
sum s sub n.

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So to compute the limit of s sub n, first
thing to do is to get a better

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handle on s sub n.

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And the trick for doing that is to
multiply s sub n by 1 minus r.

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Well, what's that equal to?

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Well, that's 1 minus r times, here's an
expression for s sub n.

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Right?

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The nth partial sum is just the sum of the
terms, k from 0 up to k equals n.

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And now I can distribute.

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So here I've just written down 1 minus r
times that quantity,

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I haven't changed anything.
But by distributing, I can

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rewrite this as 1 times s sub n minus r
times s sub n.

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Now, I can do a little bit more
manipulation here.

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Subtracting R times this is the same as
subtracting this.

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I can multiply each of the terms by r.

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But now,

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each of these terms also can be rewritten.
R times r to the 0 is r to the first.

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R times r to the first is r squared.

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And so on, until I get to r times r to the
n, which is really r to the n plus 1.

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So, instead of writing it this way, I
could rewrite r times sub n as minus

88
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r to the first plus r squared and it's
like it's r to the n plus 1.

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Now a wonderful thing happens.

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Lots of these terms cancel terms up here.
Well, how so?

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R to the 0 survives.

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But r to the first is killed by this
subtract r to the first here.

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Inside the dot dot dot, there's an r

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squared, and that's killed by this r
squared term.

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And so on.

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All of the terms up here, except for r to
the

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zero, all of these terms are killed by
something down here.

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Then

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I've also got this extra r to the n plus

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1 term that doesn't have anything up here
to kill.

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So what's this equal to?

102
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Well, this ends up being equal to r to the
0, which survives.

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All of the middle stuff dies, and this
last r to

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the n plus 1 term survives and I've
written it down here.

105
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But r to the 0 can be written as just 1 So

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I could rewrite this as 1 minus r to the n
plus 1.

107
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All of this is to say that 1 minus r times
this nth partial sum is 1

108
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minus r to the n plus 1.
Let's divide by 1 minus r.

109
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Okay.

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So we've got 1 minus r times sub n is 1
minus r to the n plus 1.

111
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And assuming that r is not 1.
Well, I can divide both sides

112
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by 1 minus r and I get that 1 minus r
times s sub n over 1

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minus r is equal to 1 minus r to the n
plus 1 over 1 minus r.

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Now I can cancel the 1 minus r's, and I've
got a formula for the nth partial sum.

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The nth partial sum is 1 minus r to the n
plus 1 over 1 minus r.

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Now

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I'll take the limit.
So, the limit of s sub n.

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Well, here's a formula for s sub n.
So, the limit of sub n is just the limit

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as n goes to infinity of 1 minus r to the
n plus 1 over 1 minus r.

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That's the limit we have to calculate.
When does that limit exist?

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You can compute this limit using our limit
loss.

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This is the limit of a quotient, so it's
the quotient of the limits.

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The limit of the denominator is the limit
of a constant.

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And the limit of a numerator is the limit

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of a difference, which is the difference
of the limits.

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All that is to say, that, using the limit
laws, this ends up being 1 minus

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the limit as n approaches infinity of r to
the n plus 1 over 1 minus r.

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Now, how do I calculate this limit?
What's the limit

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of r to the n plus 1 as n approaches
infinity?

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Well, the situation is that if r is bigger
than 1, or r is less than minus 1,

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then the limit of r to the n plus 1 as it
approaches infinity isn't a finite number.

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And consequently, when this happens, when
r is bigger than 1 or

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less than minus 1, then the limit of the
partial sums diverges.

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And consequently, the original geometric
series diverges.

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But if r is between minus 1 and 1, then
we're good.

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Exactly.

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In that case, if r is between minus 1 and
1, then the limit

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of r to the n plus 1 as n approaches
infinity is equal to 0.

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If I take a number between minus 1 and 1
and raise

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it to an enormous power, that number gets
very close to 0.

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Well, this then let's me calculate this.

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If I want to calculate

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the limit of 1 minus r to the n plus 1

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over 1 minus r, that then is 1 over 1
minus r.

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Because the limit of r to the n plus 1 is
0.

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Let's summarize the situation for
geometric series.

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So the geometric series, the sum k goes
from zero to infinity of r to the

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k, is equal to 1 over 1 minus r if r is
between minus 1 and 1.

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And we know that, because we calculated
the limit of

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the partial sums to be 1 over 1 minus r.

151
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In the case where r equal 1 or r equals
minus 1, we already saw it diverged.

152
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And if r is bigger than 1 or r is less
than minus 1, then it also diverges.

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So this actually

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covers all of the cases

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[SOUND].