If `f(n)=(1)/(n){(n+1)(n+2)(n+3)...(n+n)}^(1//n)` then `lim_(n to oo)f(n)` equals

Options

A. e
B. `1//e`
C. `2//e`
D. `4//e`

Answer

D

Explanation

Let
`A=lim_(n to oo)=(1)/(n)[(n+1)(n+2)(n+3)...(n+n)]^(1//n)A`
`=lim_(n to oo)[((n+1)/(n))((n+2)/(n))((n+3)/(n))...((n)/(n))]^(1//n)`
`rArr A=lim_(n to oo)[(1+(1)/(n))(1+(2)/(n))(1+(3)/(n))...(1+(1)/(n))]^(1//n)`
`rArr log A= lim_(n to oo) (1)/(n)underset(r=1)overset(n)sum "log"(1+(r)/(n))`
`rArr log A=underset(0)overset(1)int log (1+x)dx=[ x log (1+x)]_(0)^(1)-underset(0)overset(1)int(x)/(x+1)dx`
`rArr log A= log 2-underset(0)overset(1)int (1-(1)/(x+1))dx`
`rArr log A=log2-1+log2= log ((4)/(e)) rarr A=4//e`