For each positive integer n, let y_n=1n((n+1)(n+2)...(n+n))^1n. If limn→∞⁡y_n=L, then the value of L (where x is the greatest integer less than or equal to x) is ____

Correct answer: 1. y_n=1+1^(n1+2n….1+nn1n) y_n=∏r=1n1+r^n1n lny_n=1/n∑r=1nln⁡1+r/n ⇒ limn→∞⁡ln⁡(y_n)=limn→∞⁡∑r=1n1/nln⁡1+r/n ⇒ lnL=∫01ln⁡1+xdx ⇒ lnL=ln4/e ⇒ L=4/e L=1

Q29 / 54 · JEE Advanced 2018 (Paper 1) · Mathematics

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

Correct answer: 1

Solution

y_{n} = {\{(1 + \frac{1}{n}) (1 + \frac{2}{n}) \dots . (1 + \frac{n}{n})\}}^{\frac{1}{n}}

y_{n} = \prod_{r = 1}^{n} {(1 + \frac{r}{n})}^{\frac{1}{n}}

\ln (y_{n}) = \frac{1}{n} \sum_{r = 1}^{n} \ln (1 + \frac{r}{n})

\Rightarrow \lim_{n \to \infty} \ln (y_{n}) = \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} \ln (1 + \frac{r}{n})

\Rightarrow \ln (L) = \int_{0}^{1} \ln (1 + x) d x

\Rightarrow \ln (L) = \ln (\frac{4}{e})

\Rightarrow L = \frac{4}{e}

[L] = 1

Hence, the correct answer is 1.

For each positive integer n, let yn=1n((n+1)(n+2)...(n+n))1n. If limn→∞⁡yn=L, then the value of L (where x is the greatest integer less than or equal to x) is ____

Solution

Relevant Questions

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