“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

审敛法

在第三个模块中，我们学习用于确定级数是否收敛的各种审敛法：特别地，我们将说明比值审敛法、根值审敛法和积分审敛法。

- Jim Fowler, PhDProfessor

Mathematics

The root test.

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Here's another convergence test.

So here is the root test.

You want to analyse a series with positive terms.

You're supposed to calculate this limit, call it big L.

Big L is the limit as n goes to infinity of the nth root of the nth term.

If big L is less than 1 then the series, the sum of these events converges.

If big L is bigger than 1 then the series diverges.

And if big L is equal to 1, well than the root test is inconclusive.

We have to try something else.

You might think that there are some reasons to really love the root test.

Here is an example.

Let's look at the sum, n goes from 1 to infinity of 1 over n to the nth power.

I don't know whether that converges or diverges [LAUGH].

I'm going to use against my better judgement the root test.

So what am I supposed to do?

Well this is a series all of whose terms are positive, so I'll calculate big L.

The limit as n approaches infinity of the nth root of the nth term.

Well this is the limit as n approaches infinity.

What's the nth root of the nth power?

Well that's just n.

So this is the limit of 1/n as n approaches infinity.

That is 0 and 0 is less than 1.

So by the root test,

the series converges.

But in that case, I could have just used comparison test.

So here is the thing to notice, one over

n to the nth power is smaller than one over n squared.

Maybe the only confusing case is when n is 1 and which case these are equal but

then when n is 2 is 1 over 2 squared is less 1 over 2 squared,

when n equals 3 this is 1 over 3 cube just way less than 1 over 3 squared.

Well, 1 over n to the n, as I've already pointed out, is positive.

And I know that the sum n goes from 1 to infinity, of 1 over n squared, converges.

That's a p series of p equals 2.

So, by just the comparison test, I've got

a series now which is term Y's less than a convergent series,

so by the comparison test, the sum 1 over N to the N and

goes from 1 to infinity converges as well.

And that's when I'm not so impressed with the root test.

Here's what happens, right?

People go out in the world and they're given series problems and

sometimes those series involves something to the nth power.

And people see with something to the nth power,

I'm going to apply the root test because then I'll get rid of the nth power.

And yeah, that's true.

But you could also have applied the ratio test.

And the ratio test is good, not just when you've got powers,

but also when you've got factorials floating around.

That's not to say that the root test is just entirely useless.

It's just not as useful I think as people make it out to be.

All right, the ratio test is often easier to apply and

it's more likely to cause some useful cancellation.

We're going to see in the future some more instances of where the root test does

come in handy.

But for the time being I think your first

inclination should be to reach for the ratio test.

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