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?
 A: Without Stirling:
Let $u_n=\frac{\sqrt[n]{n!}}{n}$. We have that 
$$\ln(u_n)=\frac{1}{n}\sum_{k=1}^n\ln\left(\frac{k}{n}\right)$$
and thus
$$\lim_{n\to\infty }\ln(u_n)=\int_0^1 \ln(x)\mathrm{d}x=-1.$$
Since $x\mapsto e^x$ is continuous at $x=-1$, $$\lim_{n\to\infty }u_n=e^{-1}=\frac{1}{e}.$$
A: 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: ${e}^{-1}$
A: Any version of Stirling's formula, including approximate ones that only give the order of magnitude without the right constant or power of $n$, will show that limit is $1/e$.  The dominant piece of the formula is $(n/e)^n$ and any corrections to that die out when taking $n$-th roots. 
A: Use equivalents:
$$\frac{\sqrt[n]{n!}}n\sim_{\infty}\frac{\bigl(\sqrt{2\pi n}\bigr)^{\tfrac 1n}}{n}\cdot\frac n{\mathrm{e}}=\frac 1{\mathrm{e}}\bigl({2\pi n}\bigr)^{\tfrac 1{2n}}$$
Now $\;\ln\bigl({2\pi n}\bigr)^{\tfrac 1{2n}}=\dfrac{\ln\pi+\ln 2n}{2n}\xrightarrow[n\to\infty]{}0$, hence
$$\frac{\sqrt[n]{n!}}n\sim_{\infty}\frac 1{\mathrm{e}}. $$
A: Stirling's Formula is 
$$n!=\sqrt{2\pi\,n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac1n\right)\right)$$
From this it is easy to see that 
$$(n!)^{1/n}=(2\pi\,n)^{1/2n}\frac{n}{e}\left(1+O\left(\frac{1}{n^2}\right)\right)$$
whereupon dividing by $n$ and letting $n\to \infty$ yields 
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{(n!)^{1/n}}{n}=e^{-1}}$$
A: $$\frac { \sqrt [ n ]{ n! }  }{ n } =\sqrt [ n ]{ \frac { n! }{ { n }^{ n } }  } =\sqrt [ n ]{ { x }_{ n } } \\ x_{ n }=\frac { n! }{ { n }^{ n } } \\$$ 
Hence $\forall n\epsilon \quad N,{ x }_{ n }>0$ ,this formula is true 

$$\lim _{ n\rightarrow \infty  }{ \sqrt [ n ]{ { x }_{ n } } =\lim _{ n\rightarrow \infty  }{ \frac { { x }_{ n } }{ { x }_{ n-1 } }  }  } $$

$$ \lim _{ n\rightarrow \infty  }{ \frac { { x }_{ n-1 } }{ { x }_{ n } } = } \lim _{ n\rightarrow \infty  }{ { \left( 1+\frac { 1 }{ n-1 }  \right)  }^{ n-1 } } =\frac { 1 }{ e } $$
