0:08

want to evaluate this series carefully.

So the claim here is that the sum k goes from 0

to infinity of 1 over 2 to the k is equal to 2.

But what does that even mean? Well, what it really means is this.

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It means that the sum k goes from 0 to n of one

over to two to the k is close to 2 whenever n is big.

Alright.

This sum makes sense because this is just a finite

sum it's just telling me to add one over two

to the k for the values for the values of k between 0 and n.

So this at least makes sense.

And what is close to two mean?

Well, it means as close as you want it to be to 2.

And how big does n have to be?

Well, it has to be true whenever n is larger than some fixed big

value that depends on how close you want the thing to get to 2.

Let's do the same thing, but in terms of limits.

So more precisely,

I'll define the sequence, s sub n. The sequence of partial sums.

To be the sum, k goes from 0 to n of 1 over 2 to the k.

And the claim that this series has value 2.

Just means that this sequence has limit 2. It means that the limit

as n approaches infinity of the nth partial sum is 2.

How do I evaluate that limit? Well, let's look for a pattern.

So s sub 0 lets the sum k goes from 0 to 0 of 1 over 2 of the k.

So that's just 1 over 2 to the 0, which is just 1.

S sub 1, well that's the sum k goes from 0 to 1 of 1 over 2 of the k.

That's 1 plus 1 over 2 to the first power, which is a half.

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S sub

2, well that's the sum, k goes from 0 to 2, of 1 over 2 to the k.

That's 1 plus a half, plus 1 over 2 squared, which is a fourth.

I could simplify that a little bit.

I could write that as one plus 3 4th s sub 3.

And that's the sum k goes from 0 to 3 of 1 over 2 to

the k, so that's 1 plus a half plus a 4th plus 1 over 2

to the 3rd, which is an 8th.

And I could add a half, a 4th, and 8th and get 7 8th.

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Then I could compute s sub 4, right?

And that would be 1 plus a half plus a fourth plus

an 8th plus 1 16th, and that'll be 1 plus 15 16th.

And now maybe you're beginning to the beginnings of the pattern here, right?

I've got a half, 3 quarter, 7 8th, 15 16th here.

So it's perhaps believable. I mean, this isn't a proof,

it's just a little bit of evidence but the pattern is suggesting itself.

That the n-th partial sum is 1 plus, let's, suppose

to be a number a little bit less than one.

And it's the corresponding power of two in

the denominator, and one less in the numerator.

So I could write that as 1 plus 2 to the n minus one over 2 to the n.

Now

I can simplify that a little bit.

Let's split that up as 1 plus 2 to the n over 2 to the n minus 1 over 2 to the n.

And I can write that as 2 minus 1 over 2 to the n.

I haven't proved that, but hopefully that formula seems believable.

In any case, armed with that formula, we can evaluate the limit.

Well, here we go.

I want to take the limit

of S sub n as n approaches infinity, and

now I've got that formula for the nth partial sum.

So that's the limit as n approaches infinity of 2 minus 1 over 2 to the n.

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That's a limit of a difference, which is

the difference of a limit, provided limits exist.

So this is the limit just of 2 as it

approaches infinity minus the limit of 1 over 2 to the

n as n approaches infinity.

This is the limit of a constant, which is just that constant minus,

this is the limit of a quotient, but look at what happens here.

The denominator is very, very large.

I can make the denominator as large as I like.

Right? 2 to the n can be very, very positive.

1 over a very large number is very close to 0.

So this limit is, in fact, zero.

And that means that the limit of the partial sums is just two.

That means that this series converges to two.

And I can not only see that algebraically by, by using that limit.

But I can also see it geometrically.

Geometrically, I can draw a picture like this.

Where I got a half.

A quarter of the square, an eighth of the square, a 16th

of the square, a 32nd of the square, a 64th of the square.

And all these pieces fit together, to build one unit square.

And what that's showing me, is that the sum, k goes from 1

to infinity, of 1 over 2 to the k, well, that really is 1.

So the sum, k goes from 0 to infinity, of 1 over 2 to the k Well,

that's the first term plus all the rest of these terms, so that's 1 plus 1.

That must

be 2

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