Limit solved by definite integral (Demidovich) I was solving this limit from the Demidovich's book of exercises:
$$\lim_{n\to\infty} \frac{\sqrt[n]{\vphantom{\Large a}\, n!\,}}{n}$$
and I managed to get it to this state but then I got stuck:
$$e^{\frac{1}{n}(\log(n) + \log(n - 1) + ... + \log(1)) - \log(n)}$$
where $\log x$ is a natural logarithm of x. Can you please provide any hint?
 A: Apply the ratio test to the sequence $\{a_n\}_{n\geq 1}$ given by:
$$ a_n = \frac{n!}{n^n}. $$
We have:
$$ \frac{a_{n+1}}{a_n} = \frac{1}{\left(1+\frac{1}{n}\right)^n} $$
hence $\lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=\frac{1}{e}$ implies $\lim_{n\to +\infty}\sqrt[n]{a_n}=\frac{1}{e}$, no integrals needed.

If you want to use an integral at all costs, notice that:
$$ \log(a_n) = \sum_{k=1}^{n}\log\left(\frac{k}{n}\right) $$
hence, by Riemann sums:
$$ \lim_{n\to +\infty}\frac{\log(a_n)}{n} = \int_{0}^{1}\log(x)\,dx = \color{red}{-1}.$$
A: Here is a brute force approach.
$$\frac{\sqrt[n]{n!}}{n}=e^{\frac1n \log(n!)-\log(n)}$$
Now, note that the term $\log(n!)$ can be written as
$$\begin{align}
\log(n!)&=\sum_{k=1}^n \log(k)\\\\
&=n\left(\frac1n \sum_{k=1}^n\log(k/n)+\log(n)\right) \tag 1
\end{align}$$
Note that the sum in $(1)$ is the Riemann sum for the logarithm function.  Therefore,
$$\lim_{n\to \infty}\left(\frac1n \sum_{k=1}^n\log(k/n)\right)
=\int_0^1 \log(x)\,dx=-1 \tag 2$$
From $(2)$ we have
$$\frac1n \sum_{k=1}^n\log(k/n)=-1+\epsilon(n)$$
where $\lim_{n\to \infty}\epsilon(n)=0$.  Then, we can write
$$\begin{align}
\frac{\sqrt[n]{n!}}{n}&=e^{\frac1n \log(n!)-\log(n)}=e^{-1+\epsilon(n)}\\\\
&\to e^{-1}\,\,\text{as}\,\,n\to \infty
\end{align}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
You can use the Stirling Asymptotic Expression for 'large' $n$:
\begin{align}
\color{#f00}{\lim_{n \to \infty}{\root[n]{n!} \over n}} & =
\lim_{n \to \infty}{\pars{\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/\pars{2n}}\expo{-1}}
= \color{#f00}{{1 \over \expo{}}} \approx 0.3679
\end{align}

Note that
$$
\lim_{n \to \infty}{\ln\pars{n} \over 2n} =
\lim_{n \to \infty}{\ln\pars{n + 1} - \ln\pars{n} \over \pars{2n + 2} - 2n} = 0
\quad\imp\quad\lim_{n \to \infty}n^{1/\pars{2n}} = 1
$$
