If `sum_(r=1)^n a_r=1/6 n(n+1)(n+2)` for all `n gt=1` then `lim_(n rarr oo)sum_(r=1)^n 1/a_r` is

lim_(n rarr oo) sum_(r=0)^(n-1) 1/(n+r) =

If sum_(r=1)^(n) a_(r)=(1)/(6)n(n+1)(n+2) for all nge1 , then lim_(ntooo) sum_(r=1)^(n) (1)/(a_(r)) , is

If sum_(r=1)^(n) a_(r)=(1)/(6)n(n+1)(m+2) for all nge1 , then lim_(ntooo) sum_(r=1)^(n) (1)/(a_(r)) , is

If sum_(r=1)^(n) a_(r) = (n (n+1)(n+2) )/( 6) AA n ge 1 , then Lt_(n to oo) sum_(r=1)^(n) (1)/( a_r)=

If sum_(r=1)^(n)T_(r)=(n(n+1)(n+2)(n+3))/(12) then 4(lim_(x rarr oo)sum_(x-1)^(n)(1)/(T_(r))) is equal to

If sum_(r=1)^(n) T_(r)=(n(n+1)(n+2)(n+3))/(8) , then lim_(ntooo) sum_(r=1)^(n) (1)/(T_(r))=

lim_(n to oo) sum_(r = 1)^(n) 1/n e^(r/n) is :

lim_(n to oo) sum_(r=1)^(n) (1)/(n)e^(r//n) is

lim_(n to oo) sum_(r=1)^(n) (1)/(n)e^(r//n) is