Find `lim_(n->oo) S_n` ; if `S_n=1/(2n)+1/(sqrt(4n^2-1))+1/(sqrt(4n^2-4))+......+1/(sqrt(3n^2+2n-1))`.

Evaluate: ("lim")_(nvecoo)(1/(sqrt(4n^2-1))+1/(sqrt(4n^2-2^2))++1/(sqrt(3n^2)))

Evaluate: ("lim")_(n rarr oo)(1/(sqrt(4n^2-1))+1/(sqrt(4n^2-2^2))++1/(sqrt(3n^2)))

Evaluate: ("lim")_(n rarr oo)(1/(sqrt(4n^2-1))+1/(sqrt(4n^2-2^2))++1/(sqrt(3n^2)))

If quad S_(n)=(1)/(2n)+(1)/(sqrt(4n^(2)-1))+(1)/(sqrt(4n^(2)-4))+...+(1)/(sqrt(3n_(2)^(2)+2n-1)),n in N then lim_(n rarr oo)S_(n) is equal to (pi)/(2)(b)2(c)1(d)(pi)/(6),n in N

lim_(n rarr oo)(1)/(sqrt(n^(2)))+(1)/(sqrt(n^(2)+1))+(1)/(sqrt(n^(2) +2))+...(.1)/(sqrt(n^(2)+2n))=

Definite integration as the limit of a sum : lim_(ntooo)[(1)/(n)+(1)/(sqrt(n^(2)+n))+(1)/(sqrt(n^(2)+2n))+.......+(1)/(sqrt(n^(2)+(n-1)n))]=.........

Evaluate: lim_(n rarr oo)((1)/(sqrt(4n^(2)-1))+(1)/(sqrt(4n^(2)-2^(2)))+...+(1)/(sqrt(3n^(2))))

lim_(n->oo) ((sqrt(n^2+n)-1)/n)^(2sqrt(n^2+n)-1)

{:(" "Lt),(n rarr oo):} ((1)/(sqrt(4n^(2)-1))+(1)/(sqrt(4n^(2)-2^(2)))+....+(1)/(sqrt(3n^(2))))=

The value of lim_(nto oo)(1/(sqrt(n^(2)))+1/(sqrt(n^(2)+1))+…..+1/(sqrt(n^(2)+2n))) is