Expert: Paul Klarreich - 12/6/2006

Question
This one I am totally lost on.

Given the sequence {(1/n) n root(n!).  Show that the limit as n --> inf is e^(-1).

Ouch.  Thanks for your help.  I truly appreciate it.  Gotta kick this upcoming final in the butt.

Answer
Questioner:  Gary
Category:  Calculus
 
Subject:  limits (sorta)
Question:  This one I am totally lost on.

Given the sequence {(1/n) n root(n!).  Show that the limit as n --> inf is e^(-1).

Ouch.  Thanks for your help.  I truly appreciate it.  Gotta kick this upcoming final in the butt.
...................................

I assume the term is:
     1
an = --- (n!)^1/n
     n

n! ~~ sqrt(2 pi n) (n/e)^n

So putting that in, we have:
      1
an ~~ --- (2 pi n)^1/n ((n/e)^n)^1/n
      n

      1                n
an ~~ --- (2 pi n)^1/n ---
      n                e

      1
an ~~ --- (2 pi n)^1/n
      e

      1
an ~~ --- (2 pi)^1/n (n)^1/n
      e

Now a constant to the 1/n will approach 1, so we can ignore the (2 pi)^1/n

      1
an ~~ --- (n)^1/n
      e

Now for the  (n)^1/n

Try:
 lim  x^(1/x)
x->inf

This is an indeterminate form, and we should be able to use l'hospitals rule.
                                   ln x
Let  u = ln(x^(1/x)) = (1/x) ln x = -----
                                     x

Now that approaches zero as x --> inf. (Easy to prove.)
Therefore x^(1/x) approaches 1.

So we are up to:
      1             1
an ~~ --- (n)^1/n = ---(1) = 1/e = e^(-1)
      e             e