TITLE: The limit of $(n!)^{1/n}/n$ as $n\to\infty$
QUESTION [13 upvotes]: (Proof necessary) 
$$\lim_{n \to \infty}  \frac{(n!)^{\frac{1}{n}}}{n}$$
I don't have an answer yet, but I know it exists, and is less than $1$.
Edit. Winther's answer is the most correct I don't understand how he is jumping from (log(n!) - nlog( n )) to it equal to the Sum from k=1 to n of log(k/n). Don't presume, it's wrong, I need to go, and I'll keep looking at it when I get back
Any help is appreciated

REPLY [7 votes]: Here is a simple elementary proof I found, but first of all, some lemmas:

This one could easily be proven by induction: $\displaystyle \prod_{k=1}^{n}\left(1+\frac{1}{k} \right) = n+1$
You can try to prove this inequality yourself since it's not difficult: $\displaystyle \left (1+\frac{1}{k} \right )^k\leq e \leq \left (1+\frac{1}{k}  \right)^{k+1}\\$
This inequality is the one I'm going to use though because it gives a much tighter bound on our sequence and that's just more fun, though you could use the second inequality without change in proof. I could give you the proof if needed:$\displaystyle \left (1+\frac{1}{k} \right )^k\leq e \leq \left (1+\frac{1}{k}  \right)^{k+1/2}\\$

Ok, so here is the proof:


We first write $n^n/n!$ in a better way: $\displaystyle \frac{n^n}{n!}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k} \right)^n \cdot \prod_{i=1}^{n-1}\prod_{k=1}^{i}  \left(1+\frac{1}{k} \right)^{-1}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k} \right)^n\cdot \prod_{i=1}^{n-1}  \left(1+\frac{1}{i} \right)^{-(n-i)}=\prod_{i=1}^{n-1}  \left(1+\frac{1}{i} \right)^{i}$
Now, we use our inequalities to bound our sequence. First, an upper bound: $\displaystyle \prod_{i=1}^{n-1}  \left(1+\frac{1}{i} \right)^{i} \leq \prod_{i=1}^{n-1}e^{1}=e^{n-1}$
Then, a lower bound: $\displaystyle \prod_{i=1}^{n-1}  \left(1+\frac{1}{i} \right)^{i} \geq \prod_{i=1}^{n-1}\left (e^{1} \cdot  \left(1+\frac{1}{i}\right)^{-1/2} \right )=\frac{e^{n-1}}{\sqrt[2]{n}}$
Now, since $\frac{n!^{1/n}}{n}=(\frac{n^n}{n!})^{-1/n}$, we get: $\displaystyle e^{\frac{1}{n}-1}  \leq \frac{(n!)^{1/n}}{n} \leq e^{\frac{1}{n}-1} \cdot \sqrt[2n]{n}$
Finally, by the squeeze theorem, we get $\displaystyle e^{0-1}  \leq \lim_{n \to \infty} \frac{(n!)^{1/n}}{n} \leq e^{0-1} \cdot 1$
Hence,  $\displaystyle \lim_{n \to \infty} \frac{(n!)^{1/n}}{n}=e^{-1}$


I know there are simpler proofs, but this one is elementary and I feel like it gives you the direct intuition as to why it's true.