For each positive integer n, let yn=1n((n+1)(n+2)...(n+n))1n . For x∈R, let [x] be the greater integer less than or equal to x. If limn→∞ yn=L, then the value of [L] is___________

yn={(1+1n)(1+2n).....(1+nn)}1nyn=∏r=1n(1+rn)1/nlogyn=∑r=1n1nln(1+rn)⇒limn→∞ log yn=limn→∞∑r=1n1nln(1+rn)⇒logL=∫01 ln(1+x)dx ⇒logL=log4e

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For each positive integer n, let $y_{n} = \frac{1}{n} {((n + 1) (n + 2) ... (n + n))}^{\frac{1}{n}}$ . For $x \in R$ , let $[x]$ be the greater integer less than or equal to x. If $\lim_{n \to \infty} y_{n} = L,$ then the value of $[L]$ is___________

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Detailed Solution

$\begin{array}{l} y_{n} = {(1 + \frac{1}{n}) (1 + \frac{2}{n}) ..... (1 + \frac{n}{n})}^{\frac{1}{n}} \\ y_{n} = \prod_{r = 1}^{n} {(1 + \frac{r}{n})}^{1 / n} \\ \log y_{n} = \sum_{r = 1}^{n} \frac{1}{n} l n (1 + \frac{r}{n}) \\ \Rightarrow \lim_{n \to \infty} {log y}_{n} = \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} l n (1 + \frac{r}{n}) \\ \Rightarrow \log L = \int_{0}^{1} \begin{array}{l} l n (1 + x) d x \end{array} \\ \Rightarrow \log L = \log \frac{4}{e} \end{array}$