Finding the limit of $\frac {n}{\sqrt[n]{n!}}$ I'm trying to find 
$$\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} .$$ 
I tried couple of methods: Stolz, Squeeze, D'Alambert
Thanks!
Edit: I can't use Stirling.
 A: Let $\displaystyle{a_n=\frac{n^n}{n!}}$.  Then the power series $\displaystyle{\sum_{n=1}^\infty a_n x^n}$ has radius of convergence $R$ satisfying $\displaystyle{\frac{1}{R}=\lim_{n\to \infty} \sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}}$, provided these limits exist.  The first limit is what you're looking for, and the second limit is $\displaystyle{\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n}$.
Added: I just happened upon a good reference for the equality of limits above, which gives a more general result which is proved directly without reference to power series. Theorem 3.37 of Rudin's Principles of mathematical analysis, 3rd Ed., says:

For any sequence $\{c_n\}$ of positive numbers,
  $$\liminf_{n\to\infty}\frac{c_{n+1}}{c_n}\leq\liminf_{n\to\infty}\sqrt[n]{c_n},$$
  $$\limsup_{n\to\infty}\sqrt[n]{c_n}\leq\limsup_{n\to\infty}\frac{c_{n+1}}{c_n}.$$

In the present context, this shows that $$\liminf_{n\to\infty}\left(1+\frac{1}{n}\right)^n\leq\liminf_{n\to\infty}\frac{n}{\sqrt[n]{n!}}\leq\limsup_{n\to\infty}\frac{n}{\sqrt[n]{n!}}\leq\limsup_{n\to\infty}\left(1+\frac{1}{n}\right)^n.$$
Assuming you know what $\displaystyle{\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n}$ is, this shows both that the limit in question exists (in case you didn't already know by other means) and what it is.

From the comments: User9176 has pointed out that the case of the theorem above where $\displaystyle{\lim_{n\to\infty}\frac{c_{n+1}}{c_n}}$ exists follows from the Stolz–Cesàro theorem applied to finding the limit of $\displaystyle{\frac{\ln(c_n)}{n}}$.  Explicitly, 
$$\lim_{n\to\infty}\ln(\sqrt[n]{c_n})=\lim_{n\to\infty}\frac{\ln(c_n)}{n}=\lim_{n\to\infty}\frac{\ln(c_{n+1})-\ln(c_n)}{(n+1)-n}=\lim_{n\to\infty}\ln\left(\frac{c_{n+1}}{c_n}\right),$$ 
provided the latter limit exists, where the second equality is by the Stolz–Cesàro theorem.
A: If you take the log, it is:
$$\frac{1}{n}\sum_{k=1}^n \log\left(\frac{k}{n}\right)$$
Which is a Riemann sum for $\int_{0}^1 \log x$.
The indefinite integral is $F(x)=x\log x-x$ and $\lim_{x\to 0} x\log x -x =0$, and $F(1)=-1$. 
You have to deal with the fact that this integral is an improper integral, but it "just works."
A: This is going to be a bit difficult (since apparently lots of things aren't allowed). Here's how I would do it (this is far from a complete solution but just a couple of hints):
I hope you know that $e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n}$ (this is often taken as the definition of $e$).
You can show easily that the sequence $c_{k} = \left(1 + \frac{1}{k}\right)^k$ is monotonically increasing and that the sequence $d_{k} = \left(1 + \frac{1}{k}\right)^{k+1}$ is monotonically decreasing. This gives the squeezing
$$\displaystyle
\left(1 + \frac{1}{k}\right)^k = c_k \lt e \lt d_k = \left(1 + \frac{1}{k}\right)^{k+1}.$$
By taking the products $c_{1} c_{2} \cdots c_{n}$ and $d_{1} d_{2} \cdots d_{n}$ you can then show
$$\displaystyle
\frac{(n+1)^n}{n!} \lt e^n \lt \frac{(n+1)^{n+1}}{n!}
$$
using a few manipulations.
Now extract roots on both sides of the last inequalities and you're there.
A: Let $[x]$ denote the largest integer not exceeding $x.$  For $n\geq 1$ we have $$\log n! =\int_1^{n+1}\log [x]\; dx<\int_1^{n+1}\log x\; dx=-n+(n+1)\log (n+1)$$ and $$\log n!=\int_1^n \log (1+[x]) \;dx\geq \int_1^n\log x \;dx=1-n+n\log n.$$ So $$1/n\leq  1+\log ( (n!^{1/n}/n)<(1+1/n)\log (n+1)-\log n=\log (1+1/n)+(1/n)\log (n+1).$$ Since $(1/n)\log (n+1)\to 0$ as $n\to \infty$ we have $$\lim_{n\to \infty}\log (n!^{1/n}/n)=-1.$$
A: By applying Cauchy-d'Alembert criterion we get that:
$$\lim_{n\to\infty} \frac{n}{n!^{\frac{1}{n}}}=\lim_{n\to\infty}\left(\frac{n^n}{n!}\right)^{\frac{1}{n}}
\\ \lim_{n\to\infty} \frac{n}{n!^{\frac{1}{n}}} = \lim_{n\to\infty} \frac{(n+1)^{(n+1)}}{(n+1)!}\cdot \frac{n!}{n^n} 
\\ \lim_{n\to\infty} \frac{n}{n!^{\frac{1}{n}}}= \lim_{n\to\infty} \frac{(n+1)^n}{n^n} 
\\ \lim_{n\to\infty} \frac{n}{n!^{\frac{1}{n}}}=\lim_{n\to\infty} {\left(1+\frac{1}{n}\right)^{n}}=e. $$
Q.E.D.
A: what's wrong with just logging the expression? 
$$
\varphi
(n) = \frac{n}{n!^{\frac{1}{n}}}\\
L \varphi(n) = \log \varphi(n) = \log n - \frac{\log n!}{n} = \log n -\sum_{k=1}^{n}\frac{\log k}{n} \\
\sim \log n -\frac{n \log n -n + 1 }{n} = \log n - \log n +1 + \frac{1}{n}= 1 + o(1)
$$
Hence $\lim_{n \to \infty} \varphi(n) =e^1 = e$
EDIT: to make things sharper, here's the approximation using Euler-Maclaurin formula: $\sum_{k=1}^{n} \log k = \int_{1}^{n}\log x dx + O(\log n) = n \log n -n +1 +O(\log n ).$ Obviously $\lim_{n \to \infty} \frac{\log n }{n} = 0$, hence the statement above holds: 
$$
\frac{n \log n -n -\frac{1}{2} \log n +1}{n} = \log n -1 +o(1)
$$
and the result holds because $e^{o(1)} = 1$. 
A: If $f(n)=\frac{n}{\sqrt[n]{n!}}$ and $g(n) = f(n)^n$ then 
$$g(n) = \frac{n^n}{n!}$$
and taking the ratio of terms, removing the factorials and using $\frac{n+1}{n} = 1+\frac{1}{n}$,
$$ \frac{g(n+1)}{g(n)} = \left(1 + \frac{1}{n}\right)^n $$
You may recognise this as having a limit of $e$.  It implies 
$$\lim_{n \to \infty} \frac{g(n+1)}{g(n)} \frac{1}{e} = 1$$
and so multiplying a string of these together
$$\lim_{n \to \infty} \frac{g(n)}{e^n h(n)} = 1$$ 
for some function $h(n)$ which grows more slowly than $e^n$ or decays more gently than $e^{-n}$, [not that it matters, but $h(n)$ is about $1/\sqrt{2 \pi n}$] so taking the $n$-th root
$$\lim_{n \to \infty} \frac{f(n)}{e} = \lim_{n \to \infty} h(n)^{1/n} = 1$$
and so $\lim_{n \to \infty} f(n) = e$.
A: $$\lim\frac{n}{\sqrt[^n]{n!}}=\lim \sqrt[n]{\frac{n^n}{n!}}=\lim \sqrt[n]{\frac{n}{1}\frac{n}{2}\dots\frac{n}{n}}=\lim \exp\bigg({\cfrac{\ln\frac{n}{1}+\ln\frac{n}{2}+\dots+\ln\frac{n}{n}}{n}}\bigg)=$$$$=\lim \exp\bigg({\frac{1}{n}\sum_{k=1}^n\ln\frac{n}{k}}\bigg)=\exp\bigg(\lim{\frac{1}{n}\sum_{k=1}^n\ln\frac{n}{k}}\bigg)=$$$$=\exp\bigg(\int_0^1(-\ln x)dx\bigg)=\exp(1)=e \ \ \ \ \square$$

By the way, you can calculate $\int_0^1(-\ln x)dx$ easily by integrating by parts.
A: One other method is to also derive an assymptote for the factorial function on the fly. Here's a short proof.
Take,
$$\ln(n!)=\sum_{i=1}^{n} ln(i) \sim \int_{1}^{n}\ln(t) dt=n\ln(n)-n+1$$
Exponentiation each side, we get,
$$n!\sim n^ne^{-n}e=e\left(\frac{n}{e}\right)^n \sim \left(\frac{n}{e}\right)^n$$
Substituting this in the question, we get
$$\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}=\lim_{n\to\infty}\frac{n}{\sqrt[n]{\left(\frac{n}{e}\right)^n}}=e$$ 

Edit, by the way you have used the Stirling's approximation, indirectly, here's the Stirling's approximation used for practical purposes;
$$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$
In the above case, we won't need to worry about the $\sqrt{2\pi n}$ thing. As $(n/e)^n$ is the dominating factor here.
A: $$
\begin{align*}
\lim_{n\to +\infty}\frac{n}{\sqrt[n]{n}}
&=\lim_{n\to \infty}\sqrt[n]{\frac{n^n}{n!}}\\
&=\lim_{n\to \infty}\frac{n}{n+1}\sqrt[n]{\prod_{k=1}^n(\frac{k+1}{k})^k}\\
&=\lim_{n\to \infty} \frac{n}{n+1}\lim_{n\to \infty}(\frac{n+1}{n})^n\\
&=e
\end{align*}
$$
We only need to know that, if $a_n>0$ and
$$
\lim_{n\to \infty} a_n=a
$$
exists, then
$$
\lim_{n\to \infty} \sqrt[n]{a_1a_2\cdots a_n}=\lim_{n\to \infty}a_n=a
$$
In fact, for any $\epsilon>0$, there exists $N$ such that if $n>N$, then
$|a_n-a|<\epsilon$(Assume that $a>0$). Then we can observe that
$$
\sqrt[n]{a_1\cdots a_N}(a-\epsilon)^{n-N}\le \sqrt[n]{a_1a_2\cdots a_n}\le \sqrt[n]{a_1\cdots a_N}(a+\epsilon)^{n-N}
$$
Let $n\to \infty$, we can get that
$$
a-\epsilon\le \lim_{n\to \infty}\sqrt[n]{a_1a_2\cdots a_n}\le a+\epsilon
$$
Thus,
$$
\lim_{n\to \infty} \sqrt[n]{a_1a_2\cdots a_n}=a
$$
