$$\lim_{n\to \infty} \frac{1}{n}\cdot \big((m+1)(m+2) \ldots(m+n)\big)^{\frac{1}{n}}$$
where $m$ is a fixed positive integer.
Here is my attempt:
According to Cauchy's theorem of limit if $\lim\limits_{n\to \infty}a_n=l$ then $\lim{(a_1a_2 \ldots a_n)}^{\frac{1}{n}}=l$
hence $\lim\limits_{n\to \infty}\frac {m+n}{n}$ $\Rightarrow \lim\limits_{n\to\infty}\left(1+\frac{m}{n}\right)=1$
I'm 90 percent clear that my solution is correct. If not then please give me the right solution.
We have
$$\frac {1}{n}[{(m+1)(m+2)............(m+n)}]^\frac{1}{n}=\left[\frac{{(m+1)(m+2)............(m+n)}}{n^n}\right]^\frac{1}{n}=a_n^\frac1n$$
and by ratio-root criteria
$$\frac{a_{n+1}}{a_n}=\frac{{(m+1)(m+2)............(m+n+1)}}{(n+1)^{n+1}}\frac{n^n}{{(m+1)(m+2)............(m+n)}}=$$
$$=\frac{m+n+1}{n+1}\frac{1}{\left(1+\frac1n\right)^n}\to\frac1e \implies a_n^\frac1n\to \frac1e$$
Another way is to take logs. For $n\to \infty$ we get a Riemann integral:
$$\ln S=\lim_{n\to \infty}\frac{1}{n} \sum_{k=1}^{n} \log(\tfrac{m+k}{n})\\ = \int_0^1 \ln(x) dx = -1$$
So required limit is $e^{-1}$.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\lim_{n\to \infty}{1 \over n} \bracks{\rule{0pt}{5mm}\pars{m + 1}\pars{m + 2} \ldots\pars{m + n}}^{\,1/n}} \\[5mm] = & \ \lim_{n\to \infty}{1 \over n} \bracks{\rule{0pt}{5mm}\Gamma\pars{m + 1 + n} \over \Gamma\pars{m + 1}}^{\,1/n} = \lim_{n\to \infty}\ {1 \over n} \bracks{\rule{0pt}{5mm}\pars{n + m}!}^{\,1/n} \\[5mm] = & \ \lim_{n\to \infty}\ {1 \over n} \bracks{\rule{0pt}{5mm}\root{2\pi}\pars{n + m}^{n + m + 1/2}\,\,\, \expo{-n - m}}^{\,1/n} \\[5mm] = & \ \lim_{n\to \infty}\ {1 \over n} \bracks{\rule{0pt}{5mm}n^{n + m + 1/2}\,\,\, \pars{1 + {m \over n}}^{n}\,\,\,\expo{-n - m}}^{\,1/n} \\[5mm] = & \ \lim_{n\to \infty}\ {1 \over n} \bracks{\rule{0pt}{5mm}n^{1 + m/n + 1/\pars{2n}}\,\,\,\,\expo{m/n} \expo{-1 - m/n}} \\[5mm] = & \ \bbx{\color{#44f}{\large{1 \over \expo{}}}} \approx 0.3679\\ & \end{align} $\ds{m}$ must satisfy $\ds{\quad \lim_{n \to \infty}\Gamma^{1/n}\pars{m + 1} = 1}$.
