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Let's compute the value of another series.

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[SOUND]

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[MUSIC]

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A bit of algebra will make a huge
difference here.

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1 divided k plus 1 times k is the same as
1 over k minus 1 over k plus 1.

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See why this is true, right?

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I can write these over a common
denominator of k plus 1 times k.

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That means that 1 over k is k plus 1 over
k plus 1

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times k, and 1 over k plus 1 is k over k
plus 1 times k.

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So now I've got k plus 1 minus k in the
numerator, and that's just 1.

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Why is that helpful?

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It'll take us a little while to see how
that algebra fact helps.

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But in the meantime, remember what our
goal is.

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Our goal is to evaluate this series.

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And instead of just diving straight into
that, let's evaluate a

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slightly simpler sum.

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Instead of k goes from 1 to infinity,
let's just

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do k goes from 1 to 5 of the same thing.

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Of course, this is small enough that we
can just do the calculation.

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So to calculate this, I just plug in k
equal to 1, k

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equals 2, k equals 3, k equals 4, k equals
5 into this expression.

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And add it all up.
Let's see.

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I plug in k equals 1, and I get 1 over 2
times 1.

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And I'll add that to what I get when I
plug in k equals two which is

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one over three times two.

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And I'll add that to k equals three term
which is one over four times three.

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And I'll add that to the k equals four
term which is one over five times four

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and I'll add that to the k equals five
term which is one over six times five.

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So all I've gotta do is just do this
arithmetic.

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And I can simplify this a bit.

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Instead of writing over two times one,
I'll just write a half.

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Here I've got one over three times two is
a sixth.

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Here I've got one over four times three,
that's a 12th.

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Here I've got one over five times four,
that's a 20th.

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Here I've got one over six times five.
That's a 30th.

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All I've gotta do is add up these
fractions, which I can do.

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I put them over the common denominator of
60.

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One half in 60ths, is 30 60ths.
1

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6ths in 60ths is ten 60ths, over a common
denominator of 60, this becomes

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5 60ths, common denominator of 60, this
becomes 3 60ths.

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Common denominator of 60.

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This becomes 2/60.

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And just add up the numerators, 30 plus 10
is 40 plus 5 is 45 plus 3 is 48

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plus 2 is 50.
So this entire thing is just 50/60.

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But instead of writing 50/60, I could just
right 5/6.

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Alternatively, we can make use of the
algebra fact from the beginning.

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What was the algebra fact?

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It was a fact that 1 over k plus 1 times k

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is the same as 1 over k minus 1 over k
plus 1.

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Now, why is that helpful?

51
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Well, that means, instead of running down
1 over

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k plus 1 times k, I can write down this.

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So when k equals 1, instead of writing
down this, I'd write down this.

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One over one

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minus one over two and when k equals two

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instead of writing it on this, I'd write
down this.

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One half minus a third.

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And when k equals three, instead of
writing down

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1 over 4 times 3 I'd write down this.

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A third minus a fourth.

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And when and k equals four instead of
writing down this,

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one over k equals one times k, I'm
going to write down.

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A fourth minus a fifth, and the last term,
the k equal 5 term,

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instead of 1 over 6 times 5, would be 1
over 5 minus 1 over 6.

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Why is this an improvement?

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Well, the cool thing that happens here is
that most of these terms end up canceling.

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Take a look.

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I've got a minus one half, plus one half,
minus a third, plus

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a third, minus a fourth, plus a fourth,
minus a fifth, plus a fifth.

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The only terms that survive here are this
initial 1 over

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1 term and this last negative 1 over 6
term.

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And that means this entire thing ends up
just being equal

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to 1 over 1 minus 1 over 6, which is 5
sixths.

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This trick has a name.

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We say that the series telescopes.

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And we'll call such a thing a telescoping
series.

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What about the original series with
infinitely many terms?

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This is the original series that I want to
compute the value of.

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And this is by definition equal to the
limit as n approaches infinity.

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Of the sum k goes from 1 to n of 1 over k
plus 1 times k.

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We can compute this limit.

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Well, this limit is the limit as n
approaches

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infinity of the sum.

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K goes from 1 to n and instead of writing
this.

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All right.
1 over k minus 1 over k plus 1.

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And this'll be a much better way to write
it, right?

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Cause what is this?

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This is then the limit as n approaches
infinity of what

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[UNKNOWN]

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in the first term, 1 over 1 minus 1 over
2.

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[SOUND].

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Plus the next term.
The k equals 2 term.

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Which is plus 1/2 minus 1/3.

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And I'll write dot, dot, dot, plus the
last term.

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Which is 1 over n minus 1 over n plus 1.

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That's when k equals n.
But practically all of these terms die.

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This minus one half is killed by this one
half.

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This minus a third is killed by something.
Every single term here, there's a term

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in here which kills this one over n.

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The only thing that's left is this initial
one over

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one and this final minus one over n plus
one.

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So this limit is just the limit as n goes
to infinity of one

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over one minus one over n plus one.
But what is that limit?

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Well, this is one minus one over an
enormous number.

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All right?

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And this limit

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is just 1.
Let's formulate this as a general trick.

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Pose that I want to take a sum k goes from
1 to n of some function

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evaluated at k minus the same function
evaluated at k plus 1.

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Then this is what.

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Well it's f of one minus f of two plus f
of two minus

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f of three.

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And so on until finally I get to the last
term when k equals n.

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It's f of n minus f.
Of n plus 1.

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And just like before, practically all the
terms cancel.

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This f of 2 and this f of 2 cancel.

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This negative f of 3 cancels something.
And the previous term here cancels this

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f of n, so this sum is equal to f of 1
minus f Of n plus 1.

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And taking a limit lets us say something
about the infinite series.

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Now suppose I want to calculate the sum k
goes

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from 1 to infinity of f of k minus f of k
plus 1.

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Well this infinite series is by definition
just the limit

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as n approaches infinity of the sum k goes
from one to n of

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f of k minus f of k plus one.
But I just saw how to calculate this.

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This is now the limit.
As n approaches infinity

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[INAUDIBLE]

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but what's this?
This is f of 1 minus f of n plus 1.

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But this is now a limit of a difference.

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And the limit of this first term is the
limit of a constant which is f of 1.

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So this is equal to f of 1 minus the
limit.

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As n approaches infinity of the function
f.

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Let's do another example.

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This fact again.
Let's try it.

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Let's try to evaluate the sum k goes from
1

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to infinity of k over k plus 1 factorial.

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And the trick is to re-write this as a

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telescoping sum, so I gotta cook up some
function.

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And the function I'm going to use

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will be the function f of k is one over k
factorial.

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And I'm just going to check that f of k
minus.

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F of k plus 1.

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Well what is that?
That's 1 k factorial minus 1

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over k plus 1 factorial.
But right in

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this common denominator of k plus 1
factorial,

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this is k plus 1 over k plus 1 factorial
minus 1 over.

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K plus one factorial.

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And what happens here is that I've got a k
plus one factorium

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in the denominator and a k plus one minus
one in the numerator.

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That's a k.

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So indeed, this sum is the sum k goes from
one to infinity

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of this function.
1 over k factorial minus this function

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at k plus 1.
Which is 1 over k plus 1 factorial.

153
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But now, this infinite series is just its

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value when, of this function when k equals
1.

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Which is 1 over 1 factorial minus the
limit.

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As n approaches

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infinity of this function.
I'll write one over n plus one factorial.

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But this limit is zero and that means that
this infinite series has value one and it.

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[SOUND]

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[SOUND]