Limit solved by definite integral (Demidovich)

Question

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?

Answer 1

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.

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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: $$ \lim_{n\to +\infty}\frac{\log(a_n)}{n} = \int_{0}^{1}\log(x)\,dx = \color{red}{-1}.$$

Answer 2

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

$$\begin{align} \frac{\sqrt[n]{n!}}{n}&=e^{\frac1n \log(n!)-\log(n)}=e^{-1+\epsilon(n)}\\\\ &\to e^{-1}\,\,\text{as}\,\,n\to \infty \end{align}$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}

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