We are given $y_n=\frac{1}{n}\left((n+1)(n+2)\cdots(2n)\right)^{1/n}.$ We need to find $L=\lim_{n\to\infty} y_n,$ and then compute $[L]$.

1. Rewrite the product

Notice that $(n+1)(n+2)\cdots(2n)=\frac{(2n)!}{n!}.$ So, $y_n=\frac{1}{n}\left(\frac{(2n)!}{n!}\right)^{1/n}.$

2. Take logarithms

Let us study $\ln y_n$: $\ln y_n= -\ln n + \frac{1}{n}\sum_{k=1}^n \ln(n+k).$ Now write $\ln(n+k)=\ln n+\ln\left(1+\frac{k}{n}\right).$ Thus, \begin{align*} \ln y_n &=-\ln n+\frac{1}{n}\sum_{k=1}^n \left[\ln n+\ln\left(1+\frac{k}{n}\right)\right] \ &=-\ln n+\frac{1}{n}(n\ln n)+\frac{1}{n}\sum_{k=1}^n \ln\left(1+\frac{k}{n}\right) \ &=\frac{1}{n}\sum_{k=1}^n \ln\left(1+\frac{k}{n}\right). \end{align*}

As $n\to\infty$, this is a Riemann sum: $\ln L=\int_0^1 \ln(1+x)\,dx.$

3. Evaluate the integral

We compute $\int \ln(1+x)\,dx=(1+x)\ln(1+x)-(1+x)+C.$ Therefore, \begin{align*} \int_0^1 \ln(1+x),dx &=\left[(1+x)\ln(1+x)-(1+x)\right]_0^1 \ &=(2\ln 2-2)-(0-1) \ &=2\ln 2-1. \end{align*} Hence, $\ln L=2\ln 2-1.$ So, $L=e^{2\ln 2-1}=\frac{4}{e}.$

4. Find the greatest integer

Now, $\frac{4}{e}\approx \frac{4}{2.718}\approx 1.47.$ Thus, $[L]=\left[\frac{4}{e}\right]=1.$

5. Compare with stored answer

Our derived answer is $1$, which matches the stored correct answer.