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
Best Answer
$$\left(\frac{\left(n!\right)}{n}\right)^{\frac{1}{n}}=\left((n-1)!\right)^{\frac{1}{n}}$$
$$ \geq \sqrt[n]{\left((n-1)\times 1\right) \times \left((n-2)\times 2\right) \times \cdots \times \left( \left(n-\left\lceil \frac{n}{2} \right\rceil + 1 \right)\times \left( \left\lceil \frac{n}{2} \right\rceil - 1 \right) \right) } $$
$$ \geq \left( n-1 \right) ^ {\frac{\left\lfloor \frac{n}{2} \right\rfloor-1}{n}} \geq \left( n-1 \right)^{1/3} \text{ for } n\geq \ 17.$$
Therefore,
$$\left(\frac{\left(n!\right)}{n}\right)^{\frac{1}{n}} \to\infty.$$