Limits, factorials, exponents

$y=\lim_{n\to\infty} \left( \dfrac{n!}{n^n} \right)^{\dfrac{1}{n}}$

$\ln{y}= \lim_{n\to\infty} \dfrac{1}{n} \ln{\left(\dfrac{1 \times 2 \times \dots \times n}{n \times n \times \dots \times n}\right)}$

$\ln{y}=\lim_{n\to\infty} \dfrac{1}{n} \left(\ln{\dfrac{1}{n}} +\ln{\dfrac{2}{n}} + \dots + \ln{\dfrac{n}{n}}\right)$

$\ln{y}= \lim_{n\to\infty} \dfrac{1}{n} \sum_{k=1}^{n} \ln{\dfrac{k}{n}}$

We know that such an infinite summation can be converted into an integral by the fundamental definition of integration as a limit of an infinite sum:

$\lim_{n\to\infty} \dfrac{1}{n} \sum_{k=1}^{n} f\left(\dfrac{k}{n}\right) = \int_{0}^{1} f(x) dx$

$x=\dfrac{k}{n} \Rightarrow dx=\dfrac{1}{n}$

Hence,

$\begin{aligned} \ln{y} &= \int_{0}^{1} \ln{x} dx \\ &= \left[ x \ln{x}-x \right]_{0}^{1} \\ &= -1 \end{aligned}$

Note that substituting the lower limit requires us to evaluate the following:

$\lim_{x\to 0} x \ln{x} = 0$

$\Longrightarrow y = \boxed{e^{-1}}$

This can be done with Stolz-Cesaro or Stirling's but Stirling's is much more elegant.

From Stirling's formula we know that $n! \sim \sqrt{2\pi n} n^n e^{-n}$ So the limit can be written as $\lim_{n \to \infty} \left( \frac{\sqrt{2\pi n} n^n }{n^n e^n}\right)^{1/n}$ And simplifying down we get $\lim_{n\to \infty}e^{-1} (2 \pi)^{1/2n}n^{1/2n}$ The product on the right will go to $1$ which is not difficult to show. Thus the answer is $\boxed{e^{-1}}$

Well, first of all, nice question!

Let ${\frac{n!}{{n}^{n}}}^{\frac{1}{n}} = y$

$\frac{1}{n}(ln(\frac{n!}{{n}^{n}})) = ln y$

${e}^{\frac{1}{n}(ln(\frac{n!}{{n}^{n}}))} = y$

Now we need to find ${lim}_{n\to\infty} y$

We can find the power till infinity and then evaluate with e.

Hence, we need to find

${lim}_{n\to\infty} \frac{1}{n}(ln(\frac{n!}{{n}^{n}}))$

$= {lim}_{n\to\infty} \frac{(ln(n!) - ln({n}^{n})}{n}$

Using L'Hopital's rule,

$= {lim}_{n\to\infty} (\frac{d(ln(n!))}{dn} - (ln(n) +1))$ [ $ln({n}^{n}) = nln(n)$ , and then we can use product rule to evaluate its derivative]

$= {lim}_{n\to\infty} (\frac{d(ln(n!))}{dn} - {lim}_{n\to\infty} (ln n) -1$

We know that

${lim}_{n\to\infty} (ln n) = \infty$

and

${lim}_{n\to\infty} (\frac{d(ln(n!))}{dn} = \infty$ (because $ln(n!) = ln(n) + ln(n-1) + ln(n-2)..... ln(2) + ln(1)$ and so when we take its derivative and then take the limit to infinity, the limit will go to $\infty$ )*

As a result,

$= \infty - \infty -1$

$= -1$

Now, don't forget that this is the power of e and the whole limit will be -

$= \boxed{{e}^{-1}}$