For each positive integer n, let ` y_(n) = 1/n ((n+1)(n+2)…(n+n))^(1/n)`. For ` x in R`, let [x] be the greatest integer less than or equal to x. If ` lim_( n to infty) y_(n) = L`, then the value of [L] is …………….. .

For each positive integer n, let ` y_(n) = 1/n ((n+1)(n+2)…(n+n))^(1

1.	For each positive integer n, let ` y_(n) = 1/n ((n+1)(n+2)…(n+n))^(1/n)`. For ` x in R`, let [x] be the greatest integer less than or equal to x. If ` lim_( n to infty) y_(n) = L`, then the value of [L] is …………….. .
Answer» Correct Answer - `(1)` We have, ` y_(n)=1/n [( n+ 1)(n+1)(n+2)...(n+n)]^(1//n) and underset( n to infty) lim y_(n) = L ` ` rArr L = underset( n to infty) lim 1/n [(n+1)(n+2)(n+3)...(n+n)]^(1//n) ` ` rArr L = underset( n to infty) lim [( 1+1/n)(1+2/n)(1+3/n)...(1+n/n)]^(1/n)` `rArr log L = underset( n to infty) lim 1/n [log (1+1/n)+ log (1+2/n) ...log(1+n/n)]` ` rArr log L = underset( n to infty) lim 1/n underset( r =1) overset( n) sum log (1+ r/n)` `rArr log L = int_(0)^(1) underset"II" 1 xx underset("I")log(1+x) dx ` ` rArr log L = (x* log (1+ x))_(0)^(1) -int_(0)^(1) [d/(dx)(log(1+x))int dx]dx` [by using integration by parts] `rArr log L = [ x log (1+x)]_(0)^(1) - int _(0)^(1) x/(1+x) dx ` ` rArr log L = log 2 - int_(0)^(1) ((x+1)/(x+1) -1/(x+1)) dx` ` rArr log L = log 2 - [x]_(0)^(1) + [log(x+1)]_(0)^(1)` ` rArr log L = log 2 -1 + log 2 -0` ` rArr log L = log 4 - log e = log e = log 4/e rArr L = 4/e rArr [L] = [4/e] = 1`