What will be this limit? I'm having a hard time finding this.... What will be this limit?
\begin{equation}
\lim_{\ n\to\infty}\ \left(\frac{\left(n!\right)}{n}\right)^{\frac{1}{n}}=P\
\end{equation}
I tried it like this:
its in the form (infinity)^(0), so taking natural log on both sides, we can convert it to (inf/inf) or (0/0) form, so that we can apply L'Hôpital's rule. Eventually it ended up like:
\begin{equation}   
 \lim_{n\to\infty}\ \left(\frac{\log_{e}\left(\int_{0}^{\infty}e^{-t}t^{\left(n-1\right)}dt\right)}{n}\right)\ =\ln\left(P\right)
  \end{equation}
So it's in the form (inf/inf) and we can apply L'Hôpital's rule. It results in:
\begin{equation}
\lim_{n\to\infty}\left(\text{Digamma}\left(n\right)\right)=\ \ln\left(P\right),\text{ which tends to} +\infty
\end{equation}
So it will tend to   (+infinity)
But in my book, the limit is given as (1/e)
I don't know how it is 1/e
 A: Note that we have
$$n!\ge \left(\frac{n}{2}\right)^{n/2}$$
Therefore,
$$\left(\frac{n!}{n}\right)^{1/n}\ge \frac{\sqrt{n/2}}{n^{1/n}}$$
Can you finish now?


Now, let's examine the limit
$$\lim_{n\to \infty}\frac{\left(n!\right)^{1/n}}{n}$$
Applying Stirling's Approximation we find that as $n\to \infty$
$$\frac{(n!)^{1/n}}{n}\sim \frac{(2\pi n)^{1/2n}(n/e)}{n}\to \frac1e$$
And we are done.
A: $$\left(\frac{\left(n!\right)}{n}\right)^{\frac{1}{n}}=\left((n-1)!\right)^{\frac{1}{n}}$$
$$ \geq \sqrt[n]{\left((n-1)\times 1\right) \times \left((n-2)\times 2\right) \times  \cdots \times \left( \left(n-\left\lceil \frac{n}{2} \right\rceil + 1 \right)\times \left( \left\lceil \frac{n}{2} \right\rceil - 1 \right) \right) } $$
$$ \geq \left( n-1 \right) ^ {\frac{\left\lfloor \frac{n}{2} \right\rfloor-1}{n}} \geq \left( n-1 \right)^{1/3} \text{ for } n\geq \ 17.$$
Therefore,
$$\left(\frac{\left(n!\right)}{n}\right)^{\frac{1}{n}} \to\infty.$$
A: Try and use Sterling's approximation which is:
$$n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n$$
so you get:
$$\left(\frac{n!}n\right)^{1/n}\sim\left(\frac{\sqrt{2\pi n}\left(\frac ne\right)^n}n\right)^{1/n}=\frac ne\left(\sqrt{\frac{2\pi}{n}}\right)^{1/n}=\frac{n^{1-\frac1{2n}}}{e}(2\pi)^{\frac1{2n}}$$
which you can see as you increase $n$ is asymptotic to:
$$\frac ne$$
which diverges. This is the same result that many calculators came to so I believe there is a mistake in the book
A: As Yiogoros S. Smyrlis said, i assume you mean
$$\lim_{n \to \infty}  \frac{(n!)^{\frac{1}{n}}}{n}$$
So, first, set it equal to a variable, say it will be x.
$$x=\lim_{n \to \infty}  \frac{(n!)^{\frac{1}{n}}}{n}$$
$$\log(x) = \frac{1}{n}\log n! - n\log n$$
$$=\frac{1}{n} \sum_{q=1} ^{n} \log\bigg(\frac{q}{n}\bigg)$$
Where the sum is a Riemann sum for $\int_{0} ^{1}\log xdx$, can you work it out from here?
