“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。

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From the course by The Ohio State University

微积分二: 数列与级数 (中文版)

45 ratings

The Ohio State University

45 ratings

“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。

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 ratio test is awesome. [MUSIC] What test should I apply? Well for this series the ratio test will work wonderfully. I can really tell that the ratio test is just going to be great for this. Because I've got these factorials and these powers. So I can expect a lot of cancellation to happen. Let's compute the limit of the ratio of neighboring terms. All right, a of n is n factorial over n to the n. And I'm trying to calculate the limit as n approaches infinity of a sub n plus one over a sub n.

And that's the limit as n approaches the infinity, what's a sub n plus one is going to replace this n with n + 1 that's n+1 factorial divided.. By (n+1) to the n+1th power divided by, what's a sub n? Well, that's just n factorial over n to the nth power. That can be simplified. First of all, I've got a fraction with fractions in the numerator and the denominator. So I can clean that up a bit. This is the limit. As n approaches infinity of n plus one factorial times n to the n divided by n factorial times n plus one to the n plus oneth power.

Now what can I do. I've got an (n+1) factorial in the numerator and an n factorial in the denominator. So this n factorial cancels everything except for the n + 1 term here. So this is the limit as n approaches infinity of just n + 1 times n to the n divided by, n factorial is gone now n plus one to the power n plus one. Well, now I've got n + 1 in the numerator, and the power of n plus one in the denominator. So, I can use this to change this n plus one reduced to nth power. So this is the limit. As N approaches infinity of N to the N over N+1 to the N. If you like, I can rewrite this a bit too. I could write this as the limit as N approaches infinity. And instead of separately raising the numerator and denominator to the Nth power, could write this as n over n plus one to be nth power. But how do i evaluate the limit? Well, analyze this limit if you really love l'Hopital's Rule you could just apply l'Hopital's Rule. I don't really like l'Hopital's Rule that much so instead i'm just going to recall a useful fact. In fact this might have been how you define the number e. The limit as N approaches infinity of one plus one over N, to the nth power, is E. Now how can I take this fact, and say something about this limit? Well I could combine this into a single fraction. So one plus one over N is N over N plus one over N. Which means, the limit of n plus one over n to the nth power, as n approaches infinity is e. Now this looks a whole lot like this, and indeed all I have to do is use the fact that the limit of a reciprocal is the reciprocal of the limit to conclude that the limit of n over n plus one to the nth power is in fact, one over e. What does that imply about the original series? Now, one over e is less than one, and that means that according to the ratio test the given series converges. We can do even better. Does the series n goes from one to infinity of n factorial divided by n over two to the nth power. Converge or diverge? Yes. This series converges. Let's see why. Well, here we go. Lets set an = n!/((n/2) to the nth). My claim is that the sum n goes from one to infinity of a sub n converges, and to justify this claim, I'm going to use the ratio test. So big L, which is the limit, as n approaches infinity of a sub n + 1 over a sub n. Well, in this case, what is that? That's the limit as n approaches infinity of this with n replaced by n + 1 And plus 1 factorial over and plus 1 over 2 to the n plus 1 power divided by a's of n which is n factorial over and over 2 with the n power. And it's kind of a mess, because I've got fractions in the numerator and denominator. So I can go and simplify that, can rewrite that as the limit n goes to infinity of n plus one factorial times n over two to the n divided by n factorial times n plus one over two to the n plus one power. I've got an n plus one factorial divided by n factorial. Most of those terms cancel except the n plus one. So I can rewrite that as just n plus one in the numerator. Let me simplify this a bit too, or at least expand it out. I can write this as n to the n divided by two to the n. So this is n to the n divided by two to the n, and the denominator here, or the n factorial goes away. But I can rewrite this as n plus one to the n plus one power divided by two to the n plus one. Now I can keep simplifying this. I've got a n plus one in the numerator. An n plus one to the n plus one power in the denominator. I can cancel one of those n plus ones in the denominator. So now I've just n plus one to the nth power in the denominator. In the numerator, I've still got n to the n. And the numerator I'm dividing by two to the n. So I can put that in the denominator. And the denominator, I'm dividing by two to the n + 1. So I can put that in the numerator. Now I've got two to the n + 1 divided by two to the n. Everything except for a single factor of two cancels. So what I'm left with here is n to the n /(n+1) to the n times two. But this, I can combine to be the limit n goes to infinity of (n/n+1) to the n and I still got that time two. But we already saw that the limit of just (n/n+1) to the n as n approaches infinity is one over e. So this whole limit is two over e and two over e is less than one. So by the ratio test, this series converges. What if that two became a three? Does the series n goes from one to infinity of n factorial over and divided by three to the n power. Converge or diverge? Now this series doesn't converge. Here's the argument that we used to show that the sum of n factorial over n over two to the n converges. Now we switched that two for a three. We should just figure out how this argument needs to be changed. So let's replace this two here with a three. And now the claim is that, that series doesn't converge anymore but that it diverges. Again, I should be applying the ratio test here. So I'm looking at the limit of the ratio of subsequent terms. But this has some two's that I swapped out for three's. Here I've got some two's that I need to swap out for three's, and here I've got some two's that I need to swap out for three's. Here's some more two's that need to be replaced with three's, and here we got three to the n + 1 over three to the n. So instead of multiplying by two. I'm now multiplying by three. This two becomes a three, and here's the worst part. This two becomes a three, and three over e is not less than one. Three over e is bigger than one. And because L is bigger than one, the ratio test says that this series diverges. Let me leave you with a question. Does the series n goes from one to infinity of n factorial divided by n over e to the nth power converge or diverge? [SOUND]

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