Evaluate :
`underset(n to oo) lim [(n!)/(n^(n))]^((1)/(n))`

Verified by Experts

The correct Answer is:

`(1)/(e )`

Similar Questions

Explore conceptually related problems

Evaluate : underset(n to oo)lim [((2n)!)/(n!n^(n))]^((1)/(n))

Evaluate : underset(n to oo)lim(n)/((n!)^((1)/(n)))

Evaluate : underset(n to oo) lim[(1)/(n)+(1)/(n+1)+(1)/(n+2)+…+(1)/(4n)]

Evaluate : underset(n to oo) lim[(1)/(n)+(1)/(n+1)+(1)/(n+2)+…+(1)/(3n)]

Evaluate : underset(n to oo) lim[(1+(1)/(n))(1+(2)/(n))(1+(3)/(n))…(1+(n)/(n))]^((1)/(n))

Evaluate : underset(n to oo) lim[((1+1^(2)/n^(2)))(1+(2^(2))/(n^(2)))(1+(3^(2))/(n^(2)))…(1+(n^(2))/(n^(2)))]^((1)/(n))

Evaluate : underset(n to oo)lim (1)/(n)[sin (pi/(2n))+sin((2pi)/(2n))+sin((3pi)/(2n))+…+ sin((npi)/(2n))]

Evaluate : underset(n to oo)lim(1)/(n)["tan"(pi)/(4n)+"tan"(2pi)/(4n)+"tan"(3pi)/(4n)+…+ "tan"(npi)/(4n)]

The value of underset(n to oo)lim((n!)^((1)/(n)))/(n) is -

Evaluate (with the help of definite integral) : underset(n to oo)lim[(1)/(sqrt(n))+(1)/(sqrt(2n))+(1)/(sqrt(3n))+…+(1)/(n)]