Evaluate : underset(n to oo)lim(1)/(n)...

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

Verified by Experts

The correct Answer is:

`(2)/(pi)log2`

Similar Questions

Explore conceptually related problems

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

"tan"^(2)(pi)/(4) "sin"(pi)/(3) "tan"(pi)/(6) "tan"^(2)(pi)/(3) =

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)+(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 [((2n)!)/(n!n^(n))]^((1)/(n))

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

Evaluate: underset(n rarr infty)lim 1/n[sin (pi/(2n)) +sin((2pi)/(2n))+sin ((3pi)/(2n))+......+sin((npi)/(2n))]

Evaluate : underset(x rarr (pi)/(4))lim (1-tan x)/(x-(pi)/(4))

The value of lim_(ntooo)(1)/(n)[sec^2""(pi)/(4n)+sec^2""(2pi)/(4n)+....+sec^2""(npi)/(4n)] is