“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

p series.

[SOUND] When I say p series,

this is what I mean.

I mean the series, sum n=1 to infinity, of 1/n to the p.

For some fixed real number p.

The question is then for which p does the series converge?

The claim is the following.

This series converges if p is bigger than one,

and it diverges if p is less than or equal to one.

We can check this using coshen condensation.

Well, let's just remember what coshen condensation says.

It says to analyse the convergence of a given series,

it's enough to look at the convergence of a condense series.

In this case, I'm trying to analyze the convergence, this P series.

And here's the condense series.

How did I get this?

Well this is 2 to the n times the 2 to the nth term of the original series.

Okay, so now all I've got to do is just determine whether this condensed series

converges or diverges.

How am I going to do that?

Well this series can be rewritten.

This series is the same, its just 1 over 2 to the n to the p minus 1 power.

Alright, look, I've got a 2 to the n in the numerator, and

a power of 2 to the n in the denominator.

So I can rewrite that power in the denominator as just the p minus 1 power.

Now this could also rewrite I could rewrite this as the sum and

goes from zero to infinity of one over two to the P-1 to the Nth power.

Why is this an improvement?

Well what kind of series is this?

This series is a just a Geometric series with common ratio 1 over 2 to the P-1.

And by condensation, if this series converges or diverges, so

too does the original p series.

So I've reduced the entire question down to just knowing whether or

not this geometric series converges or diverges.

And I know how to analyze that.

If p is bigger than 1, then this common ratio is less than 1 and

in that case, the series converges.

And if P is less than or equal to 1, then this common ratio is bigger than or

equal to 1.

And in that case the series diverges.

If you've already seen integrals you could also do this by using the integral test.

So the integral test reduces the question of this series as convergence

to the question about this integral.

In this case I'm trying to determine whether or not this series converges.

And that happens if and

only if the integral from 1 to infinity of 1 over x to p dx.

Has a finite value if this interval converges.

This is true in this case because this function f of x equals one over x to the p

is positive and decreasing on the interval that I care about.

Okay, so all I've got to do now is just analyze this interval and

if this interval has a finite value,

the series converges If this interval ends up being infinity, this series diverges.

So let's see if I can compute this integral, so

I am trying to integrate from 1 to infinity, 1 over the P dx.

And by definition this is the limit of

as n approaches infinity of the integral from 1 to big N.

Of 1 over x to the p dx.

And now instead of writing 1 over x to the p dx, we're going to write x to the -p.

So this is the limit, n goes to infinity.

The integral from 1 to big N of x to the -p dx.

Now I want to evaluate this so

I really want to assume that P isn't 1, because I'm going to write down

a derivative of this and different things happen if P is 1.

But we already handled a case where P is 1.

Anyhow that's just the harmonic series.

Okay, so let's keep going assuming that P is not 1.

Then I can evaluate this definite integral by finding an anti-derivative of x to

the minus p, right, via the fundamental theorem of calculus.

So, this is the limit n goes to infinity.

What's an anti-derivative of x to the minus p?

X to the minus p plus 1 over minus p plus 1.

All right, [INAUDIBLE].

Add 1.

To that exponent and I divide by that same quantity.

I'm evaluating this from 1 to N.

So now I just got to plug in N and plug in 1 and take the difference.

So this is the limit big N goes to infinity

of N to the- P + 1 over- P + 1

Minus what I get when I plug in 1 which is 1 over -p + 1.

I could combine these two fractions, this is the limit n goes to infinity

of N to the -p + 1 over -p + 1.

Okay. Now the whole story

is being told by this limit.

This limit has a finite value, then so does this integral.

And that means that the series converges.

How do I check that?

The story is being told by this power, here.

Right. N to the p plus 1.

What happens.

So, if minus p plus 1 is positive, then end to a positive power.

This is going to be very large and

this limit then be infinity that means that the series will diverge.

So then, the series diverges.

If, on the other hand, -p + 1 is negative, then I'm taking a big number.

But I'm raising it to a negative power.

It's going to be ending up close to 0.

And in that case, this integral will have a finite value.

And that means that this series converges.

So then converges.

And may be it's a little bit complicated to think about what this two

conditions are.

Instead of saying minus p plus one is positive,

I could instead just say p is less than one.

And instead of saying minus p plus one is negative,

I could just say that p is bigger than one.

So now we've shown the whole story, right?

If p < 1 or equal to 1, the series diverges.

And if p > 1, then the p-series converges.

Even when the p-series converges,

the actual values that we're getting can be quite mysterious.

For example, the sum of 1/n squared, n goes from 1 to infinity,

it's pi squared / 6.

The sum of the reciprocals of the fourth powers, the sum of one over into

the fourth, n goes from one to infinite, it's pi to the fourth over ninety, the sum

of the reciprocals of the sixth powers, the sum n goes from one to infinite of one

over n to the sixth it's pi to the sixth over nine hundred and forty five.

So this is 2, 4, 6.

What about odd numbers?

Right, what is the sum of 1 over n cubed, n goes to infinity.

Well hundreds of years ago Euler calculated an approximation, right,

he calculated some of the decimal digits of this number and then not so

long ago Apery in 1978.

Showed that this was an irrational number.

But as far as we know it's not a rational multiple of pi cubed.

I think that story's pretty cool.

Right?

I mean, in the 1730s, Euler approximates some number.

And over 200 years later, Apery comes along and

says that number that you were approximating is irrational.

To be doing mathematics means that you're part of a community

that will out live you.

You're joining into a conversation with people from hundreds of years ago.

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