“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

Powers versus factorials.

[NOISE] We've seen the convergence of some

series that involve both powers and factorials.

We've seen that this series, the sum n goes from 1 to infinity of n

factorial over n/2 to the nth power Converges.

But this series, the sum [INAUDIBLE] goes from 1 to infinity of n factorial over

n over 3 to the nth power, diverges.

Well, this says something about the relative sizes of n factorial, and

n over 2 to the nth power.

Now, since this is a convergent series,

the limit of the nth term must be equal to 0.

So held at the limit as n approaches infinity of n factorial over n over 2 to

the nth power is 0.

That's also the case that I've looked at the limit over here.

The limit of n!

over n over 3 to the nth power as n approaches infinity is infinity

not just because the series diverges but

because the way in which we show the series diverge by applying the ratio test.

For 2 is too small.

In that case the denominator ends up over powering the numerator.

And 3 is too big.

In that case the numerator ends up over powering the denominator.

These two forces are balanced.

When 2 and 3 are placed by e.

So if I look at this sequence, the sequence a sub n = n factorial over

n/e to the nth power in that case the limit as n approaches infinity

of the ratio between the n+1st and the nth term = 1.

So the limit of the ratio between subsequent terms is 1.

And that means if we were to look at the series and

try to apply the ratio test, the ratio test would be silent in this case.

But still, I think this is a really interesting sequence to consider.

Let me just warn you that these

do not imply that the limit of a sub n as n approaches infinity is 1.

That's not even true!

All right, but what then I'm trying to get at here, right?

What we've seen is that n factorial must be

way smaller than (n/2) to the nth power.

Because this limit was = 0.

We've also seen that n factorial must be a ton bigger than (n/3) to the nth power.

And the question now is, if n factorial is way smaller than n over 2,

and n factorial's way bigger than n over 3,

how then does n factorial really compare to n over e to the n?

We can analyze the sequence a sub n a little bit more,

if we're willing to use an integral.

Okay, so let's evaluate log(n factorial) or at least approximate it.

Log(n factorial) = log(1 * 2 * 3 all the way up to n).

But that's log of a product which is the sum of the logs.

So this is the sum k goes from 1 to n I'm just log of k.

Now I could approximate this sum with an interval.

This is approximately the interval from 1 to n, of log x, dx.

I've got to figure out what is this definite interval.

I need to know n i derivative for log x, but it turns out the derivative

of x log x minus x is equal to log x.

So there is an n i derivative of log x.

And I can use that n i derivative to evaluate this integral.

So this integral is, let me write down the n i derivative, x Log X

minus X evaluated at at n and 1, and take the difference.

So this is n log n minus n, that's what I get when I plug in n,

minus what I get when I plug in 1, which is 1 times

log 1 minus 1 Well this is n log n minus l.

Log 1 is 0 minus negative 1 plus 1.

This is approximately n log n minus n.

Now we can simplify this.

Okay so I've got log of n factorial.

Is approximately n log n minus n.

Let me rewrite this side, instead of n log n, I'll write that as log of n to the n.

And instead of minus n, I'll write minus log, e to the n.

And now this is a difference of logs,

which is the log of the ratio, n to the n, over e to the n.

Which I could also write as log of n over e to the nth power.

So now I've got log of n!

is approximately log of n over e to the nth power, so

then you'd hope that I just conclude that n!

is approximately (n/e)n.

So how good is this approximation?

Well, we can look at some numerical evidence.

I mean here I've calculated 10 factorial.

It's about 3 times 10 to the 6th.

And here I've got (10/e)10 and it's not so far off.

Try it with some bigger numbers as well.

Here is 25!, it's about 1.55 times 10 to the 25th.

Here's (25/e) to the 25th power.

Well it's 1.23, about, times 10 to the 24th.

And that's not so terrible.

This is usually called Stirling's approximation.

A better approximation is this.

That n factorial is approximately n over e to the nth power times this factor,

the square root of two pi n.

And this usually goes by the name Stirling's approximation.

[SOUND]

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