lim 1/n * ( n ! )^1/n when n-->infinitely?

Answer 1

Let your limes be x. Then ln x = ln(1/n) + (1/n)*ln(n!). Using Stirling's formula ln(n!) approaches n*ln(n) - n when n goes to infinity. Therefore you get ln x = - ln n + ln n - 1 = -1. Therefore you limes x = e^-1.
n.b. For Stirling formula see Google.

Answer 2

Lim 1/n * (n!)^1/n = Lim [ (n-1)(n-2)....3*2*1)]^1/n
As n ---> infinity, = 1 as anything to the power 0 is 1 ( 1/infinity =0)

Answer 3

As a math teacher, there is a theorem in math that 1 / infinity = 0, so
lim 1/n * ( n ! )^1/n when n-->infinitely
will become (1/infinity) * (n!)^1/infinity = 0*(n!)^0 = 0*1 = 0.
Understand, to fully understand this, try to look up different theorems on analytical geometry and calculus related math books.

Answer 4

(1/n)[(n!)^(1/n)] as n-> infinity
take logarithms then,
-ln n + (1/n)ln n! = -ln n + (1/n).[n.ln n - n] = -ln n +ln n -1 =-1
then ln {(1/n)[(n!)^(1/n)]} = -1 , as n->infinity
=> (1/n)[(n!)^(1/n)] ->exp (-1) as n-> infinity

I've used Stirling's formula with n.ln n -n for ln n!, it is actually n.ln n - n +1 , but since n->infinity there is no difference as 1/n ->0

Answer 5

Take ln of this problem to get ln(1/n * ( n ! )^1/n) = ln(1/n) + ln(n!)^1/n

1/n*ln(n!) = 1/n* (sum of ln(i) from 1 to n) = average of ln(i) ---> infinity

ln(1/n) --> negative infinity



Hmm...Good question!