limit question (n! )^1/n / ( n )

[peblez]

Limit as n --> infinity of ( n ! ) ^ 1/n / ( n )

This is how i solved it but i'm not sure if this is correct

first i ln top and bottom.

ln ( n ! ) ^1/n / ( ln n ) then i do e^ everything

e^ [( ln (n! ) / n ) - ( ln n ) ]

lim as n --> infinity ln ( n!) / n - ln n

ln ( n !) / n = 0 b/c n goes to infinity faster than the top
so u end up with -infinity

e^- infinity = 0

thus the limit is 0. Is this correct?

 

[galactus]

Actually, this is a famous limit and it converges to 1/e

That is because it is related to Striling's formula. Which says that n! is asymptotic to \(\displaystyle n^{n}e^{-n}\sqrt{2{\pi}n}\)

So, \(\displaystyle (n!)^{\frac{1}{n}}--->\frac{n}{e}\)

Therefore, we divide both sides by n and get your limit which converges to 1/e.

 

[peblez]

Is there another way to do it without prior knowledge of Striling's formula? Because we were never taught this formula in our calc class. Can you tell me why my method does not work?

 

[peblez]

Hi. I think this time i have the correct method. Can you verify?

lim n --> infinity ( n !) ^1/n / (n) = 1 / [(n / (n!) ^1/n ] Using e^x = Sigma x^n / n!
I then take the nth power, and also take the nth root so i get

[ 1 ^ n / ( n ^ n / (n! ) ) ] ^1/n

[ 1^n / e^n ] ^1/n = 1/ e

 

[daon]

Where did your 'Sigma' go?