For each positive integer n,let y_n = frac{1} | vedclass

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(D) We have $y_n = \frac{1}{n} \left( \prod_{r=1}^n (n+r) \right)^{\frac{1}{n}} = \left( \prod_{r=1}^n \frac{n+r}{n^n} \right)^{\frac{1}{n}} = \left(

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$1$

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(D) We have $y_n = \frac{1}{n} \left( \prod_{r=1}^n (n+r) \right)^{\frac{1}{n}} = \left( \prod_{r=1}^n \frac{n+r}{n^n} \right)^{\frac{1}{n}} = \left( \prod_{r=1}^n (1 + \frac{r}{n}) \right)^{\frac{1}{n}}$.Taking the natural logarithm on both sides:$\ln(y_n) = \frac{1}{n} \sum_{r=1}^n \ln(1 + \frac{r}{n})$.As $n \rightarrow \infty$,this is a Riemann sum:$\lim_{n \rightarrow \infty} \ln(y_n) = \int_0^1 \ln(1+x) dx$.Using integration by parts:$\int_0^1 \ln(1+x) dx = [x \ln(1+x)]_0^1 - \int_0^1 \frac{x}{1+x} dx = \ln 2 - \int_0^1 (1 - \frac{1}{1+x}) dx = \ln 2 - [x - \ln(1+x)]_0^1 = \ln 2 - (1 - \ln 2) = 2 \ln 2 - 1 = \ln 4 - 1 = \ln(4/e)$.Thus,$\ln(L) = \ln(4/e)$,which implies $L = 4/e$.Since $e \approx 2.718$,$L = 4/2.718 \approx 1.47$.Therefore,$[L] = [1.47] = 1$.

For each positive integer $n$,let $y_n = \frac{1}{n} ((n+1)(n+2) \dots (n+n))^{\frac{1}{n}}$. For $x \in \mathbb{R}$,let $[x]$ be the greatest integer less than or equal to $x$. If $\lim_{n \rightarrow \infty} y_n = L$,then the value of $[L]$ is:

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