For each positive integer n , let y n = 1 n ( ( n + 1 ) ( n + 2 ) ... ( n + n ) ) 1 n . If lim n →…

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

yn=1+1n1+2n….1+nn1n
yn=∏r=1n1+rn1n
lnyn=1n∑r=1nln⁡1+rn
⇒  limn→∞⁡ln⁡(yn)=limn→∞⁡∑r=1n1nln⁡1+rn
⇒ lnL=∫01ln⁡1+xdx
⇒ lnL=ln4e
⇒  L=4e
L=1

Asked in: JEE Advanced 2018 (Paper 1)

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