Tuesday, February 19, 2019

calculus - Show that limntoinftyfracln(n!)n=+infty




Show that
limnln(n!)n=+




The only way i've been able to show that is using Stirling's approximation:
n!2πn(ne)n



Let:
{xn=ln(n!)nnN



So we may rewrite xn as:
xnln(2πn)2n+nln(ne)n




Now using the fact that lim(xn+yn)=limxn+limyn :
limnxn=limnln(2πn)2n+limnnln(ne)n=0+



I'm looking for another way to show this, since Stirling's approximation has not been introduced at the point where i took the exercise from yet.


Answer



Another way to show limnln(n!)n=

is to consider the following property of logarithms: log(n!)=log(n)+log(n1)++log(2)>n2log(n2).
Now log(n!)n>log(n/2)2.
As n, this clearly diverges to +.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection f:AB and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...