Greatest Integer Function of Limit of Product Sequence

To find the value of $[L]$, we first express the given sequence $y_n$ in a form suitable for applying Riemann sums.

The given sequence is: $y_n = \frac{1}{n} \left( (n+1)(n+2) \cdots (n+n) \right)^{\frac{1}{n}}$

We can rewrite $\frac{1}{n}$ as $\left( \frac{1}{n^n} \right)^{\frac{1}{n}}$ and bring it inside the $n$-th root: $y_n = \left( \frac{(n+1)(n+2) \cdots (n+n)}{n^n} \right)^{\frac{1}{n}} = \left( \prod_{r=1}^{n} \frac{n+r}{n} \right)^{\frac{1}{n}} = \left( \prod_{r=1}^{n} \left( 1 + \frac{r}{n} \right) \right)^{\frac{1}{n}}$

Taking the natural logarithm on both sides: $\ln(y_n) = \ln \left( \prod_{r=1}^{n} \left( 1 + \frac{r}{n} \right) \right)^{\frac{1}{n}} = \frac{1}{n} \sum_{r=1}^{n} \ln\left( 1 + \frac{r}{n} \right)$

Now, taking the limit as $n \to \infty$: $\lim_{n \to \infty} \ln(y_n) = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \ln\left( 1 + \frac{r}{n} \right)$

Recognizing this limit as a definite integral (Riemann sum with $x = \frac{r}{n}$ and $dx = \frac{1}{n}$): $\ln(L) = \int_{0}^{1} \ln(1+x) \, dx$

To evaluate the integral, we use integration by parts or substitute $u = 1+x$: $\int_{0}^{1} \ln(1+x) \, dx = \left[ (1+x)\ln(1+x) - (1+x) \right]_{0}^{1}$

Evaluating at the limits: $\ln(L) = \left( 2\ln(2) - 2 \right) - \left( 1\ln(1) - 1 \right) = 2\ln(2) - 1 = \ln(4) - \ln(e) = \ln\left(\frac{4}{e}\right)$

Taking the exponential of both sides: $L = \frac{4}{e}$

Since $e \approx 2.71828$, we have: $L = \frac{4}{2.71828\dots} \approx 1.4715$

Thus, the greatest integer value $[L]$ is: $[L] = [1.4715] = 1$