Here's another example. Consider the sum of n to the n over n

factorial. Well, you might be able to

argue, based on what you know about n to the n that this diverges.

That's certainly what I would guess.

Now lets say we needed to show that explicitly.

Well we could try the

ratio test. Lets see what we get.

A sub n is a to the n over n factorial.

A sub n plus one is n plus 1 quantity to the n plus 1.

Over quantity n plus one factorial.

When we compute this limit row, we get n plus

one to the n plus one over n plus one factorial times n factorial

over n to the n. As before, there's quite a bit of

cancellation going on with these factorials.

And we're left with, the limit, as n goes to infinity

of n plus 1 to the n plus 1 over n plus 1, times 1 over n to the n.

There's some cancelling that goes on, there.

And what

do we get? we get, after a little bit of rewriting a

limit that we've seen before. The limit is n goes to infinity of

quantity one plus one over n all to the f, that is e.

That means the ratio of the subsequent terms tends to a number.

Those bigger than 1.

But it is precisely e, that means this series diverges

as we thought. Now sometimes.

It's not clear exactly which test ought to be used.

Consider the following complicated looking series.

You've got odd numbers, and the numerator and

the product of even numbers, and the denominator.

And one over two to the n. What do we do?

Well lets try the ratio test.

In this case a sub n is the product of odd numbers up to, 2 n minus 1 divided

by the product of even numbers up two n with an extra 1 over 2 to the n.

In the denomiator.

When we substitute in n plus 1,

we have to up the index on all these terms.

But notice the amount of cancellation that happens when we compute

the ratio that an plus 1 to an.

After all of that cancellation, We're really

only left with just a couple of terms.

That would be 2n plus 1 in the numerator, and

2 times quantity 2n plus 2 in the denominator.

The leading order terms in numerator and denominator are 2n and 4n

respectively. That yields a limit of one half.

Therefore this series converges. It converges to something very nice.

You'll learn to what, soon.

Sometimes, it's not always clear

which test is going to be best. Consider the sum as and goes

from zero to infinity of pi to 2 3rds

n divided by e to the n times n factorial. What

is this going to be? It seems like it ought to converge.

Hm, let's see. We could try the root test.

The root test tends to work when you have

something with an exponent that has an n in it.

Let's see what happens if we take the limit as n goes to infinity of the nth

root of pi to the two thirds n divided by e of the n times n factorial.

Well, that pi term and the e term work out nicely but

we're left with pie to the two thirds over e, times the nth root of

1, over n factorial. That

is maybe not so easy to do, if you want to get fancy then

we can use Stirling's formula for the

[INAUDIBLE]

of the factorial.

Wow I don't know I'm not really in the mood for that.

I'm not quite sure how to valuate that limit so let's try the ratio test

instead we take ace of n plus one multiplied by ace of n and take the limit.

Zen goes to infinity. Well this one works, and

it works well. We see that row as a limit, is equal to 0.

That means that not only does this series converge, but

the terms in the series get small very, very quickly.

Now, if we've thought about that, we would have said, well, of course that's

true. N factorial grows much, much more quickly

than pi to the 2 3rds n. So, certainly this must converge.