Limit of (n)^(1/n)/n as n approaches infinity

[Dethrone]

Determine $$\lim_{{n}\to{\infty}}\frac{(n!)^{1/n}}{n}$$

 

[Nono713]

Determine $$\lim_{{n}\to{\infty}}\frac{(n!)^{1/n}}{n}$$

Spoiler

By the continuity of the natural logarithm and using Stirling's approximation, we find that:
$$\ln \left [ \lim_{n \to \infty} \frac{(n!)^{\frac{1}{n}}}{n} \right ] = \lim_{n \to \infty} \ln \frac{(n!)^{\frac{1}{n}}}{n} = \lim_{n \to \infty} \left [ \frac{\ln{n!}}{n} - \ln{n} \right ] = \lim_{n \to \infty} \left [ \left ( \ln{n} + O \left ( \frac{\ln{n}}{n} \right ) - 1 \right ) - \ln{n} \right ] = -1$$
Therefore it follows that:
$$\lim_{n \to \infty} \frac{(n!)^{\frac{1}{n}}}{n} = e^{-1} = \frac{1}{e}$$

 

[Dethrone]

Excellent solution Bacterius! Thanks for participating. :D

For anyone who wants to tackle this without Stirling's approximation, I have a few hints:

Spoiler

Use the fact that, assuming $f$ is non-negative and increasing:
$$\sum_{k=1}^{n-1}f(k)\le\int_1^nf(x) \,dx\le\sum_{k=1}^{n}f(k)$$

and apply it to $\ln(x)$ to get $en^ne^{-n}<n!<en^{n+1}e^{-n}$.

 

[Euge]

Here is my method:

Spoiler

If $a_n = \frac{(n!)^{1/n}}{n}$, then

$$a_n = \sqrt[n]{\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\cdots \left(1 - \frac{n-1}{n}\right)} = \exp\left\{\frac{1}{n}\sum_{k = 0}^{n-1}\log\left(1 - \frac{k}{n}\right)\right\},$$

which converges to

$$\exp\left\{\int_0^1 \log(1 - x)\, dx\right\} = \exp\left\{\int_0^1\log x\, dx\right\} = \exp(-1) = \frac{1}{e}.$$

 

[Dethrone]

Hi Euge! Very interesting solution, excellent job! (Cool)

Spoiler

$$\sum_{k}^{n-1}\ln k \le\int_{1}^{n} \ln x\,dx \le \sum_{k=1}^{n}\ln k$$
$$\implies \ln(n-1)! \le n \ln n-n+1 \le \ln n!$$

On the left hand side, it can be shown that $\ln(n-1)! \le n\ln n -n+1 \implies n! \le n^{n+1}e^{-n}e$ and on the right hand side, $n \ln n-n+1 \le \ln n! \implies n^ne^{-n}\le n!$.

$$n^ne^{-n}e \le n! \le n^{n+1}e^{-n}e$$$$\implies\frac{e^{1/n}}{e}\le \frac{(n!)^{1/n}}{n} \le \frac{e^{1/n}n^{1/n}}{e}$$

Furthermore, $\lim_{{n}\to{\infty}}a^{1/n}=1$ and $\lim_{{n}\to{\infty}}n^{1/n}=1$, so by the squeeze theorem,
$$\lim_{{n}\to{\infty}}\frac{(n!)^{1/n}}{n}=\frac{1}{e}$$