Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

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Calculus Two: Sequences and Series

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Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

From the lesson

Series

In this second module, we introduce the second main topic of study: series. Intuitively, a "series" is what you get when you add up the terms of a sequence, in the order that they are presented. A key example is a "geometric series" like the sum of one-half, one-fourth, one-eighth, one-sixteenth, and so on.
We'll be focusing on series for the rest of the course, so if you find things confusing, there is a lot of time to catch up. Let me also warn you that the material may feel rather abstract. If you ever feel lost, let me reassure you by pointing out that the next module will present additional concrete examples.

- Jim Fowler, PhDProfessor

Mathematics

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

[SOUND].

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

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

These series have a name.

These things are called geometric series.

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

Right.

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

we've already evaluated. It converges.

And it's value is 2.

There are certain other of r, we can also

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

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

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

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

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

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

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

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

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

Right.

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

Does this series converge or diverge?

This series diverges. Because the

limit of the partial sums is infinity.

Something bad also happens when r equals negative 1.

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

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

So, r equals minus 1.

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

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

power, which is 1.

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

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

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

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

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

so on. Does this series converge or diverge?

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

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

When I add the first two terms I get 0.

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

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

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

I get 1.

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

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

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

No.

And because the sequence of partial sums

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

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

Remember, to evaluate a series,

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

Here's the partial sum.

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

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

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

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

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

partial sums?

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

I want to compute the limit of that sequence.

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

And if I can compute this limit, then

I've computed the value of the geometric series.

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

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

handle on s sub n.

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

Well, what's that equal to?

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

Right?

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

And now I can distribute.

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

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

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

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

Subtracting R times this is the same as subtracting this.

I can multiply each of the terms by r.

But now,

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

R times r to the first is r squared.

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

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

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

Now a wonderful thing happens.

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

R to the 0 survives.

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

Inside the dot dot dot, there's an r

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

And so on.

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

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

Then

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

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

So what's this equal to?

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

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

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

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

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

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

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

Okay.

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

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

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

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

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

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

Now

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

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

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

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

You can compute this limit using our limit loss.

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

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

And the limit of a numerator is the limit

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

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

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

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

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

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

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

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

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

And consequently, the original geometric series diverges.

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

Exactly.

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

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

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

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

Well, this then let's me calculate this.

If I want to calculate

the limit of 1 minus r to the n plus 1

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

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

Let's summarize the situation for geometric series.

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

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

And we know that, because we calculated the limit of

the partial sums to be 1 over 1 minus r.

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

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

So this actually

covers all of the cases

[SOUND].

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