The ratio test is awesome.
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.
Now what can I do?
Well, I've got an n plus one factorial in
numerator, and an n factorial in
the denominator, so this n factorial
cancels everything
except for the n plus one term here.
So this is the limit as n approaches
infinity of just n plus one
times n to the n.
Divided by, so that n factorial's gone
now, n plus one to the n plus oneth power.
Well I've got an n plus one on the
numerator, and the power of n plus one in
the denominator, so I can use this to
change
this n plus one in just an nth power.
So this is the limit as n approaches
infinity of n to the n over n
plus one to the n.
And if you like, I can just rewrite this a
bit too or analyze this limit.
If you really love l'Hôpital's rule, you
could just apply l'Hôpital's rule.
I don't really like l'Hôpital'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?
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.
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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