$$\lim_{n\to \infty} \frac{1}{n}\cdot \big((m+1)(m+2) \ldots(m+n)\big)^{\frac{1}{n}}$$
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 $\lim\limits_{n\to \infty}\frac {m+n}{n}$ $\Rightarrow lim_{n\to∞}(1+\frac{m}{n})=1$
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 $e^{-1}$.
