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.
Answer
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.
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