The limit of $(n!)^{1/n}/n$ as $n\to\infty$ [duplicate]

Put $$a_n = \frac{n!^{1/n}}{n}$$

then $$\log a_n = \frac{1}{n}\left(\log n! - n\log n\right)= \frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}\right)$$

where we have used $\log n! = \log 1 + \log 2 + \ldots + \log n$. The sum above is a Riemann sum for the integral $\int_0^1\log x dx$ so

$$\lim_{n\to\infty} \log a_n = \int_0^1\log x dx = [x\log x - x]_0^1 = -1$$

and it follows that $$\lim_{n\to\infty}a_n = \frac{1}{e}$$

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.

Let $\displaystyle a_n=\frac{n!}{n^n}$.

Then $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}=\lim_{n\to\infty}\frac{1}{(1+\frac{1}{n})^n}=\frac{1}{e}$,

so $\;\;\;\;\displaystyle\lim_{n\to\infty}(a_n)^{\frac{1}{n}}=\lim_{n\to\infty}\frac{(n!)^{\frac{1}{n}}}{n}=\frac{1}{e}$ $\hspace{.2 in}$ (since $\frac{a_{n+1}}{a_n}\rightarrow L\implies(a_n)^{\frac{1}{n}}\rightarrow L)$.