0:08

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.

So I'll write a sub 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 infinity.

What's a sub n plus one?

I just gotta replace these n's with n plus one.

That's n plus one factorial divided by n plus one to the n plus oneth 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 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.

2:37

And if you like, I can just rewrite this a bit too or analyze this limit.

If you really love l'HĂ´pital's rule, you could just apply l'HĂ´pital's rule.

I don't really like l'HĂ´pital's rule that much.

So instead, I'm just going to recall a useful fact.

In fact, this might've 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?

3:03

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.

And 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.

Let's set a sub n equal to n factorial over N over two to the n.

And my claim

is that the sum, n goes from one to infinity of a, sub n converges.

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 plus one, 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 plus one.

It's n plus one factorial over n plus one, over two to the n plus one power.

Divided by a sub n which is N factorial over N over two to the nth power.

This is kind of a mess because I've got fractions, a numerator, and a denominator.

So I can 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 an N factorial.

Most of those terms cancel except for the N plus one.

So, I can rewrite that as just N plus one on the numerator.

Let me simplify this a bit too, or at least let's expand it out.

I can write this as n to the

n, divided by two to the n.

So it's n to the n, divided by two to the n.

And the denominator here, well the n factorial goes away but I can rewrite this

as n plus n to the n plus one power, divided by two to the n plus one.

Now I can keep simplifying this.

I've got an an 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 demoninator.

So now I've just got n plus one to the nth power in the denominator.

And in the numerator, I've still got n to the n.

6:09

And the numerator I'm dividing by two to the n.

So I can put that in the denominator.

And in the denominator, I am diving by the two to

the n plus one so I can put that in the numerator.

I've got two to the n plus one 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, over N plus one to the N,

times two. But this, I can combine to be the limit.

N goes to infinity of N over N plus one to the N.

And that's to get that times two.

But we already started that, the limit of this N over N

plus one 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 n divided

by three to the nth 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 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

am looking at the limit of the ratio of subsequent terms.

But this has some twos that I swapped out for threes.

Here I've got some twos that I need to swap out for threes.

And here I got some twos.

That I need to swap out for threes.

And there's some more twos that need to be replaced with threes.

And here we got three to the n plus one 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 big 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?

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