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
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