Show that
limn→∞ln(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!)nn∈N
So we may rewrite xn as:
xn∼ln(2πn)2n+nln(ne)n
Now using the fact that lim(xn+yn)=limxn+limyn :
limn→∞xn=limn→∞ln(2πn)2n+limn→∞nln(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 limn→∞ln(n!)n=∞ is to consider the following property of logarithms: log(n!)=log(n)+log(n−1)+⋯+log(2)>n2log(n2). Now log(n!)n>log(n/2)2. As n→∞, this clearly diverges to +∞.
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