limit [(n!)^1/n]/n as n->infinity.....answer is 1/e; how ??

Question

 question is of jee exams; of calulus.

it can be solved by puiseux series.

but finding a better method.

Answer 1

An alternative could be to use Stirling's approximation.

 

n! ∼ √(2πn) (n/e)ⁿ as n → ∞

 

Then we get the following

 

lim_n→∞ (n!)^1/n / n

 

= lim_n→∞ π^1/(2n)(2n)^1/(2n) (1/e)

 

= 1·1·1/e

 

= 1/e

 

Note that I used lim_x→∞ x^1/x = 1, which is straight forward to prove by log-transforming and applying L'Hopital's rule.