How to compute $\lim_{n\rightarrow\infty}\frac1n\left\{(2n+1)(2n+2)\cdots(2n+n)\right\}^{1/n}$

As there are $n$ terms as the multipliers,

$$\displaystyle f(n)=\frac1n\Big\{(2n+1)(2n+2)\cdots(2n+n)\Big\}^{1/n}=\left(\prod_{1\le r\le n}\frac{2n+r}n\right)^{\frac1n}$$

hence $$\ln f(n)=\frac1n\sum_{1\le r\le n}\ln\left(2+\frac rn\right)$$ Using $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n g\left(\frac rn\right)=\int_0^1g(x)dx,$$

$$\lim_{n\to\infty}\ln f(n)=\int_0^1\ln(x+2)dx$$

Note that \begin{eqnarray} \int\ln(x+2)dx&=&x\ln(x+2)-\int\frac x{x+2}dx\\ &=& x\ln(x+2)-\int\frac{x+2-2}{x+2}dx\\ &=& x\ln(x+2)-\int\ dx+2\int\frac1{x+2}dx\\ &=& x\ln(x+2)-x+2\ln(x+2)\\ &=&(x+2)\ln(x+2)-x \end{eqnarray} hence $$\int_0^1\ln(x+2)dx=3\ln3-1-\{2\ln2-0\}=\ln (3^3)-\ln e-\ln(2^2)=\ln \frac{27}{4e}$$

Hint (for this multiple choice question): $2 \le f(n) \le 3$.

$$\frac{1}{n}\left[\frac{(3n)!}{(2n)!}\right]^{\frac{1}{n}}\sim\frac{1}{n}\left[\frac{\left(\frac{3n}{e}\right)^{3n}\sqrt{6\pi n}}{\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}}\right]^{\frac{1}{n}}=\frac{1}{n}\left[\left(\frac{n}{e}\right)^n\frac{27^n}{4^n}\sqrt{\frac{3}{2}}\right]^{\frac{1}{n}}\to \frac{27}{4e}$$