<s>[INST] Find $\lim\limits_{n\to+\infty}\frac{\sqrt[n]{n!}}{n}$ I tried using Stirling's approximation and d'Alambert's ratio test but can't get the limit. Could someone show how to evaluate this limit?
 [/INST] This is an alternate solution for those who don't know stirling approximation yet like me
$$
\lim _{ n\longrightarrow +\infty  }{ \frac { \sqrt [ n ]{ n! }  }{ n }  } \quad =\quad \lim _{ n\longrightarrow +\infty  }{ \frac { \sqrt [ n ]{ \prod _{ k=0 }^{ n-1 }{ (n-k) }  }  }{ n }  } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty  }{ \frac { \sqrt [ n ]{ \prod _{ k=0 }^{ n-1 }{ n(1-\frac { k }{ n } ) }  }  }{ n }  } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty  }{ \frac { \sqrt [ n ]{ { n }^{ n }\prod _{ k=0 }^{ n-1 }{ (1-\frac { k }{ n } ) }  }  }{ n }  } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty  }{ \frac { n\sqrt [ n ]{ \prod _{ k=0 }^{ n-1 }{ (1-\frac { k }{ n } ) }  }  }{ n }  } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty  }{ { \left( \prod _{ k=0 }^{ n-1 }{ \left( 1-\frac { k }{ n }  \right)  }  \right)  }^{ \frac { 1 }{ n }  } } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty  }{ { e }^{ \frac { 1 }{ n } \sum _{ k=0 }^{ n-1 }{ \ln { \left( 1-\frac { k }{ n }  \right)  }  }  } } 
$$
Now we will evaluate that infinite sum:
$$
\underset { n\longrightarrow +\infty  }{ lim } \frac { 1 }{ n } \sum _{ k=0 }^{ n-1 }{ \ln { \left( 1-\frac { k }{ n }  \right)  }  } \quad =\underset { n\longrightarrow +\infty  }{ lim } \quad \frac { 1-0 }{ n } \sum _{ k=0 }^{ n-1 }{ \ln { \left( 1-\frac { k }{ n }  \right)  }  } \\ \qquad \qquad \qquad \qquad \qquad =\quad \int _{ 0 }^{ 1 }{ ln(1-x)\quad dx } \\ \qquad \qquad \qquad \qquad \qquad =\quad { \left[ x\ln { (1-x) }  \right]  }_{ 0 }^{ 1 }+\int _{ 0 }^{ 1 }{ \frac { x }{ 1-x } dx } \\ \qquad \qquad \qquad \qquad \qquad =\quad { \left[ x\ln { (1-x) }  \right]  }_{ 0 }^{ 1 }+{ \left[ -\ln { (1-x)-x }  \right]  }_{ 0 }^{ 1 }\\ \qquad \qquad \qquad \qquad \qquad =\quad { \left[ (x-1)\ln { (1-x)-x }  \right]  }_{ 0 }^{ 1 }\quad =\quad -1
$$
So your limit is: ${e}^{-1}$
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