Given that $\mathop {\lim }\limits_{n \to \infty } \mathop \sum \limits_{r = 1}^n \frac{{\log \left( {r + n} \right) - \log n}}{n} = 2\left( {\log 2 - \frac{1}{2}} \right),\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^k}}}{\left[ {{{\left( {n + 1} \right)}^k}{{\left( {n + 2} \right)}^k} \ldots {{\left( {n + n} \right)}^k}} \right]^{1/n}}$, is

${\left( {\frac{e}{4}} \right)^k}$

JEE MATHEMATICS MCQ: LIMITS

1 Hard

1 of 1

Application +4 | -1

Given that

\lim\limits_{n \to \infty } \sum \limits_{r = 1}^n \frac{{\log \left( {r + n} \right) - \log n}}{n} = 2\left( {\log 2 - \frac{1}{2}} \right),\lim\limits_{n \to \infty } \frac{1}{{{n^k}}}{\left[ {{{\left( {n + 1} \right)}^k}{{\left( {n + 2} \right)}^k} \ldots {{\left( {n + n} \right)}^k}} \right]^{1/n}}

, is

View answer and explanation

Correct option: C

{\left( {\frac{4}{e}} \right)^k}

Option C (

{\left( {\frac{4}{e}} \right)^k}

) is correct because the limit can be expressed as an integral and evaluated using the given information.