“微积分二：数列与级数”将介绍数列、无穷级数、收敛判别法和泰勒级数。本课程不仅仅满足于得到答案，而且要做到知其然，并知其所以然。 注意：此课程的注册将在2018年3月30日结束。如果您在该日期之前注册，您将可以在2018年9月之前访问该课程。

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From the course by The Ohio State University

微积分二: 数列与级数 (中文版)

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“微积分二：数列与级数”将介绍数列、无穷级数、收敛判别法和泰勒级数。本课程不仅仅满足于得到答案，而且要做到知其然，并知其所以然。 注意：此课程的注册将在2018年3月30日结束。如果您在该日期之前注册，您将可以在2018年9月之前访问该课程。

From the lesson

幂级数

在第五个模块中，我们学习幂级数。截至目前为止，我们一次讲解了一种级数；对于幂级数，我们将讲解整个系列取决于参数 x 的级数。它们类似于多项式，因此易于处理。而且，我们关注的许多函数，如 e^x，也可表示为幂级数，因此幂级数将轻松的多项式环境带入棘手的函数域，如 e^x。

- Jim Fowler, PhDProfessor

Mathematics

The most interesting series in the world.

[MUSIC]

Power series don't always converge.

But when they do, they usually converge absolutely.

Well, here's the theorem.

Suppose that this power series converges.

When I plug in x knot, x sub-zero.

Well then, that same series converges absolutely for

any value of x between -x knot and x knot.

Well let's prove the theorem.

Well the assumption is that the power series converges when I plug in x knot.

And a consequence of convergence Is that the limit of the nth term,

which in this case is a sub n times x naught to the nth power,

well that limit must be 0 because the series converges when I plug in x naught.

But a convergent sequence is a bounded sequence.

What that means is that they're Is an m

so that for all n

this is no bigger than m in absolute value.

So we'll write that down.

A sub n x naught to the nth power is less than or

equal to m Now pick an x.

I want to pick that x to be between minus x0 and x0.

So I'll just write that here.

I'll pick some value of x that's in the interval

between minus absolute value of x0 and x0.

So this just means x is an element of this interval

Between minus the absolute value of x naught and the absolute value of x naught.

I'm writing it in this funny way just because x naught might be negative.

My goal is to show that with that value of x, the power series converges absolutely.

Now watch this.

The absolute value of a sub n times x to the nth power.

Well, that's just equal to the absolute value of a sub n times x naught to the nth

power times the absolute value of x to the n over x naught to the n.

I mean this equality is just because x naught to the n divided by x naught to

the n, you get an x to the n here and these are just equal.

But, what else do I know?

I know that this part here,

a sub n times x naught to the nth power, is bounded by big M.

So this is less than or equal to big M, and

I could rewrite this as the absolute value of x/x naught to the nth power.

But that helps me make a comparison.

Well how so?

Lets say r equal to the absolute value of x over x.

And then lets think about x series, the sum n goes from zero to infinity

of n times r to the n, exactly we are term here, but am just summing them all up.

and that series converges.

Well, since x was chosen to be between negative absolute value of x naught and

absolute value of x naught, this quantity r, the ratio between x and x naught,

an absolute value, that is less than 1.

And consequently, this just geometric series, well it converges.

So what about the original series?

So by the direct comparison test, what do I know?

I know that the sum n goes from zero to infinity of the absolute value of

a sub n times x to the n converges.

Original power series converges absolutely.

That value of x.

Remember back to what the original statement of the theorem was.

I'm just assuming convergence at a point in that I'm deducing

something much stronger, absolute convergence on a whole interval.

There's an important corollary of this theorem.

So consider this power series, then there is an R, so that,

that series converges absolutely, for any value of x,

between -R and R, and it diverges whenever the absolute value of x is bigger than R.

Meaning whenever x is bigger than R, or x is smaller than -R.

I'm not seeing anything about what happens when x is equal to R or

when x is equal to negative R.

But at least on this interval,

I'm getting absolute convergence that red R has a name.

This r is called the radius of convergence.

[SOUND]

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