“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 harmonic series again. >> [SOUND]. [MUSIC] >> You've already seen that the harmonic series diverges. So we've shown that the harmonic series diverges by grouping together the terms, and then showing that each of the groups is bigger than one half. So summing the harmonic series is even worse than summing 1 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 and so on. We can also check this with the integral test. So let's set An = 1/n. And I'm trying to analyze the convergence of the series n goes from 1 to infinity of An, that's just the harmonic series. And to help me, I'll look at the function f(x) = 1/x. And note that An = f(n). I want to make sure that f is positive and decreasing. Right, so I need the function f to be positive, And decreasing on

the interval that I care about, which is the interval from 1 to infinity. Well I know that f is positive, because if x is at least 1, even if x is at least 0, right, f(x), which is 1 over x, is positive. I mean, the reciprocal of a positive number is positive. I need to check that f is decreasing. And what that means is that if a > b, then I need that f(a) < f(b) on the interval I'm thinking about here. Well that's true, because if I've got a and b, say they're both positive numbers even, Then how does 1/a compare to 1/b if a is bigger than b? Well if I've got a larger positive number, but I take its reciprocal, now it's smaller. So this is telling me that f is a decreasing function. Now I'll check that the integral diverges. Well let's compute the integral from 1 to infinity of f(x) dx. By definition, this integral is the limit as big N approaches infinity of the integral from 1 to big N of f(x) dx. Now in this case, f is just the 1 over function. It's the function that sends x to 1/x. So this is the limit as N approaches infinity of the integral from 1 to N of 1/x dx. Now, to compute this definite integral, I'm going to use the fundamental theorem of calculus. And recall that an anti-derivative for the 1 over x function is the natural logarithm. So by the fundamental theorem of calculus, this is the limit as n approaches infinity. The NI derivative which is a natural log, at the one end point minus the NI derivative natural log at the other end point, but the log of 1 is zero. So this is just asking what's the limit, as N approaches infinity of the natural log of N. What happens when you take the natural log of an enormous number? Well, admittedly, it's less enormous, but it's still as large as you'd like. By choosing big N big enough, I can make log N as large as you'd like. Which is to say that the limit of log N as N approaches infinity is infinity. So this integral diverges. So the same is true of the series. Right so by the integral test.

The sum, n goes form 1 to infinity, of 1 / n. The harmonic series diverges as well. So now we've got two proofs that the harmonic series diverges.

Anytime in mathematics that you've got two different proofs of the same theorem, it's worth thinking about how those proofs relate. In this case which proof do you think is easier? In a certain sense the integral test is easier. It requires less creativity to just integrate 1 / x than to do that complicated grouping.

And yet the grouping argument predates integration. The grouping argument is from the 1300s. And integration, right, wasn't really around until Newton, say.

So yes, although the integral test seems like the easier way to show the harmonic series diverges, the integral test is dependent on the huge edifice of calculus. It's requiring more technology. And the grouping argument, while somehow requiring more creativity doesn't require so much theory to be built up. [SOUND]

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