n! = l - 2 - . . ( n - l ) . n n n .

But this is not a very good bound. One might try averaging the terms

between 1,... ,n and estimating n! by (n/2)n = 2 _ n n n . However, a

little computation (with a computer, say) would show that this is not

very good. This leads, however, to trying

a~nnn

for some number

a 1. What a should we choose? Let's take logarithms. Clearly,

ln[a~nnn] = n(lnn - In a).

To estimate Inn! = In 1 + In2 -f • • • + Inn we approximate a sum by

an integral:

pn

n

pn+1

I ln(x) dx Y j l n j / ln(x) dx.

By doing the integral (a standard exercise in integration by parts) we

get

n ( l n n - 1 ) 4 - 1 Inn!

(n + 1) ln(n + 1) - n

= n(lnn - 1) -f ln(n) + (n + 1) In 1 + -

L n

The error term en = Inn + (n + 1) ln[l -f (1/^)] satisfies

0, n — oc

n

(why?), so our best guess for a satifies In a = 1, i.e., a = e. We now

have the approximation

It is a good approximation in the sense that

Inn!

hm

—T

r = 1.

n-^oo

ln[nne_nJ

We now ask the question: does the limit

n!

lim

n^oo

nne n