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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From the course by The Ohio State University

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

Power Series

In this fifth module, we study power series. Up until now, we had been considering series one at a time; with power series, we are considering a whole family of series which depend on a parameter x. They are like polynomials, so they are easy to work with. And yet, lots of functions we care about, like e^x, can be represented as power series, so power series bring the relaxed atmosphere of polynomials to the trickier realm of functions like e^x.

- Jim Fowler, PhDProfessor

Mathematics

Half open intervals.

[MUSIC]

Well, as an example, let's consider this series, the sum, n

goes from 1 to infinity, of x to the n, over n.

Well it's a power series, so what's its interval of convergence?

Well fore we address that question, we ask the easier question.

What's its radius of convergence? Well, apply the ratio test.

So, I'm going to think about where this series converges

absolutely, for which values of x does it converge absolutely.

So I'm wondering when does this converge, and the

ratio test tells me to look at this limit.

The limit, as n approaches infinity of the n plus first term, which is x to

the n plus 1 over n plus 1, divided by the nth term which is x to the n over n.

And we're looking at this with absolute

value bars and asking about absolute convergence.

So the ratio test asks me to consider for

which values of x is this limit less than 1?

Well I can simplify this limit somewhat.

All right, this is the limit as n approaches infinity.

I've got x to the n plus one over x to the n here.

Those cancel and just leave me with an x.

And here I've got n plus 1 in the denominator of

the numerator and an n in the denominator of the denominator,

this ends up being n over n plus 1. So now, what's this limit?

Well, the limit of n over n plus 1, as n approaches infinity, that's 1 and

this is just a constant, so this limit of the constant as far as n is concerned.

So this limit is just the absolute value of x.

So this series converges absolutely when the absolute value of x is less

than 1, and it diverges when the absolute value of x is bigger

than 1. So what is the radius of convergence?

Well, this is telling me that the radius of convergence is 1, so

I could plot the points on the number line where this series converges.

And what I know thus far, is that it converges, when the absolute value of x

is less than 1, meaning that x is between minus 1 and 1, and it diverges when

the absolute value of x is bigger than 1.

So I know it diverges out here, and it diverges down here.

What about the end points?

Well, exactly.

Alright, what happens when x is equal to 1, right?

Does the series converge or diverge at 1? What happens when x is equal to minus 1?

Does the series converge or diverge at minus 1?

All I know so far is

that it converges absolutely between minus 1 and 1 and it diverges out here, but

I haven't actually addressed the question of whether

or not this series converges at the endpoints.

We can plug in x equals 1 and then recognize the series.

Yeah, so, if we're looking at this series, the sum n goes from 1

to infinity of x to the n over n, and I plug in x

equals 1, what do I get?

Well then, this is just the sum n goes from 1

to infinity of 1 to the n, which is 1 over n.

Does this series converge or diverge?

That's the harmonic series, and the harmonic series

diverges, which means that this series, when x is equal to one, diverges.

Now, what about x

equals negative one?

Well, in that case, I'm going to plug in minus 1 here, and I'll get the sum n goes

from 1 to infinity of minus 1 to the n over n, and

that's the alternating harmonic series, and that converges, albeit conditionally.

We can summarize this.

So putting this all together, we can write the following.

We can say

that the interval of convergence is the half open interval

minus 1 to 1 but closed on the minus 1 side because this series

converges here but doesn't converge at 1.

So the interval of convergence is this half open interval.

And note

just how complicated this was.

We've got our interval of convergence, and in

the interior of that interval, we've got absolute convergence.

And it wasn't too hard to figure out the radius of that

interval, but the story became way more complicated at the end points.

We had one endpoint where the series converged and another endpoint where

the series diverged, and in general, this is how it's going to work out.

It won't be hard for you to find the radius of convergence,

but it might be really painful to analyze the story at the endpoints.

And it's possible that the series could diverge at both endpoints.

It might converge at both endpoints.

It might just converge at one endpoint.

Story at the endpoints is more complicated.

[SOUND]

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