`lim(n->oo){(1^m+2^m+3^m+....+n^m)/n^(m+1)}` equals

lim_(n to oo){(1^(m)+2^(m)+3^(m)+...+ n^(m))/(n^(m+1))} equals

lim_(n rarr oo) {(1^(m)+2^(m)+3^(m)+…….+n^(m))/(n^(m+1))} equals

Evaluate: lim_(nrarroo)(1^(m)+2^(m)+3^(m)+...+n^(m))/(n^(m+1))(mgt-1)

lim_( equals )(n rarr oo){(1^(m)+2^(m)+3^(m)+...+n^(m))/(n^(m+1))}

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

Evaluate : underset(nrarrinfty)lim((1^m+2^m+...+n^m)/n^(m+1))

Given that lim_(n to oo)sum_(r=1)^(n)(log(n^(2)+r^(2))-2logn)/(n)=log2+(pi)/(2)-2 , then lim_(n to oo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+r^(2))^(m)......(n^(2))^(m)]^(1//n) is equal to

2{(m-n)/(m+n)+1/3((m-n)/(m+n))^(3)+1/5((m-n)/(m+n))^(5) +..} is equals to