“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

级数

在这第二个模块中，我们将介绍第二个主要学习课题：级数。直观地说，将数列的项按照它们的顺序依次加起来就会得到“级数”。一个主要示例是“几何级数”，如二分之一、四分之一、八分之一、十六分之一，以此类推的和。在本课程的剩余部分我们将重点学习级数，因此如果你在有些地方感到疑惑，将会有大量时间来弄清楚。另外我还要提醒你，这个课题可能会令人感到相当抽象。如果你曾经为此困惑，我保证下一个模块提供的实例会让你感到豁然开朗。

- Jim Fowler, PhDProfessor

Mathematics

Lets sum the reciprocals of squares. [SOUND] There's a few different approaches to trying to determine weather or not this series converges or diverges. So here's the series I'm interested in. The sum n goes from 1 to infinity of 1 over n squared. And that series converges if and only if the series n starts at 2 to infinity of 1 over n squared converges. Whether or not I include the n = 1 term doesn't affect the convergence of the series at all.

Well, note that 0 is less than or equal to 1 over n squared is less than or equal to 1 over n squared- n, at least as long as n is 2 or more. I don't want to plug in n equals 1 here, because then I'd be divide it by 0. But this is true because this denominator is smaller than this denominator so this fraction is bigger than this fraction. And what is that mean about the series? Well I can do a comparison test then. So by comparison test the sum of 1 over n squared minus n, n goes from 2 to infinity, converges implies that the smaller sum,

the sum n goes from 2 to infinity of 1 over n squared converges. So, now I want to analyze the sum of 1 over n squared minus n. Well it's going to turn out that this series telescopes. Let's see how that goes. So the series that I'm interested in analyzing is the sum n goes from 2 to infinity of 1/n squared- n, all right? I'd like to evaluate that series. Now how can I do that? The trick is to notice the following, what's 1 over n- 1- 1 over n?

I can put that over a common denominator. So this is n over n times n- 1- n- 1 over n times a- 1. And now that I have got another common denominator, I can do the subtraction. This is n- n- 1, over and times n- 1. Well n- n- 1, that's just 1. And the denominator is n squared- n.

N squared- n. So evaluating this series is really the same as evaluating 1/n-1- 1/n, the sum of these. And before plunging into the infinity on top, let's just do this from 2 to some value big N.

So what happens when I plug in two2 I get 1 over 2 minus 1 is 1 minus 1 over 2. Then I plug in n equals 3. I get one over 3 minus 1 which is 2 minus 1 over 3.

Then I plug in n equals 4, and I get 1 over 3 minus 1 over 4. And then I keep on going until I plug in n and I get 1 over big N minus 1 minus 1 over n.

But lots and lots of stuff cancels. This is exactly what I mean by telescoping. Right? What cancels, this minus a half and this a half, this minus a third and this a third, this minus a fourth will cancel something in there. Everything else in the middle will die. They'll be a 1 over N- 1 term with a negative sign in front of it which will cancel this. The only thing that survives is this initial term, 1/1 and this last term -1/N. So the sum of this little n from 2 to big N is this. Now how do I evaluate then the infinite series? Well I just take a limit as N goes to infinity. So the limit as N goes to infinity of the sum of n=2 to N of (1/n-1- 1/n) = the limit As N goes to infinity of 1 over 1 minus 1 over N. Well, as N goes to infinity, 1 over N, this term is going to 0. So it's 1 over 1 minus something very close to 0. This limit is 1. And that means that this original series not only converges, but I know its value Its value with 1. Why is that significant? Well knowing that this series converges, that means that this series converges. And if this series, little n from 2 to infinity of 1 over n squared converges that means that the original series that I'm interested in, the sum of the reciprocals of the squares, that converges. And that's great. But that's not the only way to determine if this series converges. So yeah, I want to play quotie condensation but to what series? I'm still working on this series, the sum of 1 over n squared, n from 1 to infinity. But instead of looking at that series, I'm going to write it like this. The sum of just the ans, were an is 1 over n squared. Now I want to note what about this sequence [INAUDIBLE]. It's a sequence all of whose terms are positive and it's a decreasing sequence. So it's the sort of thing that I'm allowed to apply Koshi condensation to. And what does Koshi condensation say? Well it says that this series converges if and only if the condensed series converges. If and only if the sum of 2 to the n times the 2 to the nth term, n goes from 0 to infinity, converges. So what's the condensed series in this case? I'm just asking what's this series in this case? Well that's the sum, n goes from 0 to infinity, of 2 to the n, times, what's the 2 to then nth term of the sequence? It's 1 over, instead of n, I'm going to write 2 to the n squared.

But I can simplify this. This is something times 1 over the same thing squared. Well that's the sum n goes from 0 to infinity just 1 / 2 to the n. And is that series converge or diverge? Well that series is just a geometric series. All right, so this series, we've already seen that this series converges and, in fact, we know its valley, its valley was 2. And consequently, because this condense series converges, so too must the original series converge. The sum of 1 over n squared is then goes from 1 to infinity, converges So we've seen that this series converges by comparing to a telescoping series that we know converges. And by using Cauchy condensation. And there's other ways. We could have also used the integral test. So we're seeing lots of different methods to prove the same result, the sum of 1 over n squared, n goes from 1 to infinity converges. And usually we’re happy with that. Usually we’re happy that a series convert his or diverges.

But in this case we can ask the more refined question, to what does this convergence series converge? So numerically the sum of the first two terms 1 over 1 squared was 1 over 2 squared it's five quarters and we can keep on going. These are some of the first three terms, some of the first four terms, five terms, six terms, seven terms, eight terms, nine terms. A little bit over 1.5. We could try something a lot more terms too, right? Here is some of the first 100 terms about 1.63 Sum the first 200 terms first 300, 400. If we add up the first thousand terms all right so 1 over 1 + 1 over 2 squared + 1 over 3 squared + 1 over a thousand squared.

And we're getting just above 1.64. The exact answer is really pretty surprising. Numerically, right, we're getting about 1.64 after adding up the first thousand terms.

Pi squared over 6 is about 1.64, and it turns out that the sum of the reciprocals of the squares. Is pi squared over 6. These series evaluate the pi squared over 6 is a shocking result. To really leave you wondering why, why is something like that true? Unfortunately, we have to wonder just a little bit longer. [SOUND]

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