A calculus problem by Jose Sacramento

Relevant wiki: Riemann Sums

$\begin{aligned} L & = \lim_{n \to \infty} \frac {\sqrt[n]{n(n+1)(n+2)\cdots(n+n)}}n \\ & = \lim_{n \to \infty} \frac {\sqrt[n]{n \cdot n^n \left(1 + \frac 1n \right) \left(1 + \frac 2n \right) \cdots \left(1 + \frac nn \right)}}n \\ & = \lim_{n \to \infty} n^\frac 1n \prod_{k=0}^n \left(1 + \frac kn \right)^\frac 1n \\ & = \lim_{n \to \infty} \exp \left({\color{#3D99F6}\frac {\ln n}n} + \color{#D61F06}{\frac 1n \sum_{k=0 }^n \ln \left(1 + \frac kn \right)} \right) & \small \color{#3D99F6}\text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = \exp \left({\color{#3D99F6}\frac {\frac 1n}1} + \color{#D61F06} {\int_0^1 \ln (1+x) \ dx} \right) & \small \color{#D61F06}\text{By Riemann sums } \\ & = \exp \left({\color{#3D99F6}0} + \color{#D61F06}{(1+x)\ln (1+x) - x\bigg|_0^1}\right) \\ & = \exp \left( \color{#D61F06}{2\ln 2 - 1 - \ln 1 + 0} \right) \\ & = \frac 4e \approx \boxed{1.472} \end{aligned}$

Relevant wiki: Stirling's Formula

We have $\frac{\sqrt[n]{n(n+1)\cdots (n+n)}}{n} \; = \; \frac{1}{n}\left(\frac{n (2n)!}{n!}\right)^{\frac{1}{n}} \; \sim \; \frac{4}{e} (2n)^{\frac{1}{2n}} \hspace{1cm} n \to \infty$ using Stirling's Approximation. Thus the limit is $\frac{4}{e} \approx \boxed{1.471517765}$ . To 3DP, this should be $1.472$ , not $1.471$ .