Compute $\lim_{n\to\infty}3\frac {{n!}^{\frac 1n}}{n}.$ 
Evaluate $\lim_{n\to\infty}3\frac {{n!}^{\frac 1n}}{n}.$

I tried forcing a riemann sum:
rewrite $\lim_{n\to\infty}3\frac {{n!}^{\frac 1n}}{n}=\lim_{n\to\infty}3(\frac{n!}{n^n})^{\frac 1n}=L.$ Apply $\ln$ on both sides and get:
$$\lim_{n\to\infty}\ln3+ \frac 1n\ln(\frac {n!}{n^n})=\lim_{n\to\infty} \ln3+ \sum_{k=1}^{k=n}\frac1n\ln(\frac kn)=\ln3+\int_0^1\ln(x)dx=\ln(\frac 3e)\to L=\frac 3e.$$
I actually did right I had a typo... Sorry for wasting your time.
 A: 
The OP had mentioned using Riemann Sums to evaluate the limit. So, I thought I would present a way forward, which expresses the limit as a Riemann Sum.  To that end, we proceed.


Note that we can write
$$\begin{align}
\lim_{n\to \infty}\frac{(n!)^{1/n}}{n}&=\lim_{n\to \infty}\left(\frac1n e^{\frac1n \sum_{k=1}^n\log(k)}\right)\\\\
&= e^{\lim_{n\to \infty}\left(\frac1n \sum_{k=1}^n \log(k/n)\right)}\\\\
&=e^{\int_0^1 \log(x)\,dx}\\\\
&=e^{-1}
\end{align}$$
as was to be shown!
It is important to realize that 
$$\int_0^1\log(x)\,dx=\lim_{n\to \infty}\int_{1/n}^1 \log(x)\,dx=\lim_{n\to \infty} \frac1n \sum_{k=1}^n\log(k/n)$$
A: $$
S_n =\ln\left(\frac{n!}{n^n}\right) = \sum_{k=1}^n\ln\left(\frac{k}{n}\right)\frac{1}{n}
$$
$$
\lim_{n\to \infty}S_n = \int_0^1\ln(\xi)d\xi
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\lim_{n \to \infty}{\pars{n!}^{1/n} \over n} & =
\lim_{n \to \infty}{\bracks{\root{2\pi}n^{n + 1/2}\expo{-n}}^{1/n} \over n} =
\lim_{n \to \infty}\bracks{{\pars{2\pi}^{1/\pars{2n}}\ n^{1 + 1/\pars{2n}} \over n}\,\expo{-1}}
\\[5mm] & =
\expo{-1}\lim_{n \to \infty}\exp\pars{\ln\pars{n} \over 2n} =
\expo{-1}
\exp\pars{\lim_{n \to \infty}{\ln\pars{1 + 1/n} \over 2}}
= \bbx{\expo{-1}} \approx 0.3679
\end{align}
A: It is well known  $$\lim _{n\to \infty} \frac {{n!}^{\frac 1n}}{n} =1/e$$
Therefore 
$$\lim_{n\to\infty}3\frac {{n!}^{\frac 1n}}{n}=\lim_{n\to\infty}3\frac {n/e}{n} =3/e$$.
