Powers versus factorials
[SOUND].
Seen
the convergence of some series that
involved both powers and factorials.
Seen that this series that some angles
from 1 to infinity
of n factorial over n over 2 to the nth
power.
Converges but this series, the sum n it
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 we know that the limit as n approaches
infinity of
n factorial over n over 2 to the nth power
is 0.
And that's also the case if we look at the
limit over here.
The limit of n factorial over n over 3 to
the nth power as n approaches infinity
is infinity.
Not just because this series diverges but
because of the way
in which we showed this series diverge by
applying the ration test.
So 2 is too small.
In that case, the denominator ends up
overpowering the numerator.
And 3 is too big.
In that case, the numerator ends up
overpowering the denominator.
These two forces are balanced.
When two and three are replaced by e.
So if I look at this sequence, the
sequence a sub
n equals n factorial over n over e to the
nth power,
in that case, the limit as n approaches
infinity of the
ratio between the n plus first and the nth
term is 1.
So the limit of the ratio between
subsequent terms is 1.
And that means if you 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 that these facts do not
imply that the
limit of a sub n approaches infinity as 1,
that's not even true.
Right?
But what am I trying to get at here,
right?
What we've seen is that n factorial must
be way
smaller than n over 2 to the nth power
because this
limit was equal to 0.
We've also seen that n factorial must be a
ton bigger than n over 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 analyse this sequence a sub n a
little bit more if we're willing to use
an interval.
Okay.
So, let's evaluate log of n factorial, at
least approximate it.
Log of n factorial, that's log of 1 times
2 times 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 you can approximate
this sum with an integral, this is
approximately the integral from 1 to n.
of log x dx.
I gotta figure out what is this definite
integral.
I need to know an anti 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 anti derivative of log x
and
I can use that anti derivative to evaluate
this integral.
So this integral is, let me write down the
anti derivative, x.
Log x minus x evaluated at n and 1 and
take the difference.
So this is n log n minus n, it'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 n 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.
You know 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 I've got log of n factorial is
approximately
log of n over e to the nth power.
So then,
[LAUGH]
you hope that I just conclude that n
factorial
is approximately n over e to the nth
power.
So how good is this approximation?
Well, we can look at some numerical
evidence, I mean, here, I calculated 10
factorial.
It's about 3 times 10 to the 6th.
And here I've got ten over e to the 10th
power and it's not so far off.
Let's try it with some bigger numbers as
well.
Here's 25 factorial.
It's about 1.55 times 10 to the 25th.
And it's 25 over e to the 25th power.
Well it's 1.23 about times 10 to the 24th.
That's not so terrible.
This is usually called Sterling'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 2 pi n and this usually goes by
the name Stirling's approximation
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