$$\lim_{n \to \infty} \left(\frac{(n+1)(n+2)\ldots 3n}{n^{2n}}\right)^{1/n}$$ is equal to

We are asked to find the limit

$$L=\lim_{n \to \infty}\left(\frac{(n+1)(n+2)\ldots 3n}{n^{2n}}\right)^{1/n}.$$

First, notice that the numerator contains every integer beginning with $$n+1$$ and ending with $$3n$$. The total number of factors is

$$3n-(n+1)+1 = 2n.$$

We write each term inside the product as $$n+k$$ where $$k$$ runs from $$1$$ to $$2n$$. Then

$$\frac{(n+1)(n+2)\ldots 3n}{n^{2n}} \;=\; \frac{\displaystyle\prod_{k=1}^{2n}(n+k)}{n^{2n}} \;=\; \frac{\displaystyle\prod_{k=1}^{2n} n\!\left(1+\frac{k}{n}\right)}{n^{2n}} \;=\; \frac{n^{2n}\displaystyle\prod_{k=1}^{2n}\left(1+\frac{k}{n}\right)}{n^{2n}} \;=\; \prod_{k=1}^{2n}\left(1+\frac{k}{n}\right).$$

Hence our limit becomes

$$L=\lim_{n\to\infty}\left[\prod_{k=1}^{2n}\left(1+\frac{k}{n}\right)\right]^{1/n}.$$

To evaluate such limits of products, a standard technique is to take natural logarithms. We set

$$\ln L = \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{2n}\ln\!\left(1+\frac{k}{n}\right).$$

Now we recognize the right-hand side as a Riemann sum. Let

$$t_k=\frac{k}{n},\qquad k=1,2,\ldots,2n.$$

Then $$t_k$$ ranges from $$\frac{1}{n}$$ to $$\frac{2n}{n}=2$$ and the mesh size is $$\Delta t=\frac{1}{n}.$$ Therefore

$$\frac{1}{n}\sum_{k=1}^{2n}\ln\!\left(1+\frac{k}{n}\right) \;=\;\sum_{k=1}^{2n}\ln(1+t_k)\,\Delta t.$$

As $$n\to\infty$$, the sum becomes the definite integral

$$\int_{0}^{2}\ln(1+t)\,dt.$$

We compute this integral by parts. We remember the formula:

$$\int \ln x\,dx = x\ln x - x + C.$$

Replacing $$x$$ with $$1+t$$, we obtain

$$\int \ln(1+t)\,dt = (1+t)\ln(1+t) - t + C.$$

Evaluating from $$t=0$$ to $$t=2$$ gives

$$\int_{0}^{2}\ln(1+t)\,dt =\left[(1+t)\ln(1+t)-t\right]_{0}^{2} =\Big[(3)\ln 3-2\Big]-\Big[(1)\ln 1-0\Big] =3\ln 3-2.$$

Hence

$$\ln L = 3\ln 3 - 2.$$

Exponentiating both sides, we find

$$L = e^{\,3\ln 3 - 2} = e^{3\ln 3}\,e^{-2} = 3^{3}\,e^{-2} = \frac{27}{e^{2}}.$$

Hence, the correct answer is Option D.