Limit of $\frac{1}{\sqrt[n]{n!}}$ as $n$ approaches infinity So i was trying to evalue this limit:
$$\lim_{n \to \infty}\frac{1}{\sqrt[n]{n!}}, n \in \mathbb{N}$$
This, of course, by common sense is equal to zero (since factorial grows a lot faster). Is there a way to prove this limit without having to tackle with proving function growth rate. I;m not sure how that would be done, but I believe i would have to expand the factorial function to set $\mathbb{R}$ in order to compare, and that's still beyond my abilities. Thanks.
 A: Let $k =\lfloor \frac{n}{2} \rfloor$ that is $n=2k $ or $n=2k+1$. Then $k \geq \frac{n-1}{2}$ and 
$$n! \geq k(k+1) ... n \geq k^k\geq \left(\frac{n-1}{2} \right)^\frac{n-1}{2}$$
Therefore, for $n \geq 3$
$$0 \leq \frac{1}{\sqrt[n]{n!}} \leq \frac{1}{\left(\frac{n-1}{2} \right)^\frac{n-1}{2n}}\leq \frac{1}{\left(\frac{n-1}{2} \right)^\frac{1}{3}}$$
A: If we prove that $n!>\left(\frac{n}{e}\right)^n$, the limit is trivially zero.
Now notice that, by partial summation:
$$ \log\left(n!\right)=\sum_{k=1}^{n}\log k = n\log n-\sum_{k=1}^{n-1}\log\left(1+\frac{1}{k}\right)\geq n\log n-\sum_{k=1}^{n-1}\frac{1}{k}\geq n\left(\log n-1\right)$$
and the inequality is proved.
A: use the series $e^x = 1 + \dfrac{x}{1!} + \cdots + \frac{x^n}{n!} + \cdots$ to get  $$e^n > \dfrac{n^n}{n!}  $$ which gives you $$\left(\frac{1}{n!}\right)^{1/n} < \frac{e}{n} $$ and letting $n \to \infty$ gives you the limit zero.
A: You can use a well-known lower bound for factorials:
$$n! ~~ \geq ~~ \sqrt{2\pi}\ n^{n+1/2}e^{-n} ~~ \geq ~~ n^{n+1/2}e^{-n}  ~~ \geq ~~ n^{n}e^{-n}$$
Now simplify the expression you get when substituting the above into $\sqrt[n]{n!}$:
$$( n^{n}e^{-n})^\frac{1}{n} =  \frac{n}{e} $$
