How Do I Prove That the Limit of \( n!^{n^{-n}} \) Equals 1?

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SUMMARY

The limit of \( n!^{n^{-n}} \) as \( n \) approaches infinity equals 1. The discussion outlines the process of letting \( y = (n!)^{(n^{-n})} \) and applying the natural logarithm to both sides, resulting in \( \ln y = \frac{\ln(n!)}{n^n} \). By utilizing L'Hôpital's rule or recognizing that \( n^n \) grows faster than \( \ln(n!) \), the limit \( L \) can be determined. Finally, substituting back gives \( y = e^L \), confirming that the limit is indeed 1.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with factorial notation and properties
  • Knowledge of L'Hôpital's rule
  • Basic concepts of logarithms and exponential functions
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  • Study the application of L'Hôpital's rule in various limit problems
  • Explore Stirling's approximation for factorials
  • Learn about the growth rates of functions, particularly \( n! \) versus \( n^n \)
  • Investigate the properties of logarithmic functions in calculus
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How do I show that [tex]\mathop {\lim }\limits_{n \to \infty } \left( {n!} \right)^{\left( {n^{ - n} } \right)} = 1[/tex] ?

(The n^-n forces the value to decrease faster than n! increases, I believe. But how to work out that?)
 
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1)Firstly we'll let the expression (n!)^(n^-n)=y, then taking ln on both sides give,
lny=ln(n!)/n^n.
2) We'll then find the limit of the expression at the RHS using Le Hopital's rule or by observation that n^n increases faster than ln(n!). =)
3) After finding the limit, L, all we have to do is substitute back the value of y, which is e^L
 
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