If ` sum_(r=1)^(n)T_(r)=(n(n+1)(n+2)(n+3))/(12)` where `T_(r)` denotes the rth term of the series. Find ` lim_(nto oo) sum_(r=1)^(n)(1)/(T_(r))`.

Explanation

We, have, `T_(n)=sum_(r=1)^(n)T_(r)-sum_(r=1)^(n-1)T_(r)`
`=(n(n+1)(n+2)(n+3))/(12)-((n-1)n(n+1)(n+2))/(12)`
`=(n(n+1)(n+2))/(12)[(n+3)-(n-1)]`
`=(n(n+1)(n+2))/(3)(1)/(T_(n))=(3)/(n(n+1)(n-2))`
`:.underset(nto oo)(lim)sum_(r=1)^(n)(1)/(T_(r))=underset(nto oo)(lim)sum_(r=1)^(n)(3)/(r(r+1)(r+2))`
`=3underset(nto oo)(lim)sum_(r=1)^(n)(1)/(r(r+1)(r+2))`
`=3underset(nto oo)(lim)((1)/(1*2*3)+(1)/(2*3*4)+(1)/(3*4*5)+"...."+(1)/(n(+1)(n+2)))`
Maha Shortcut Method
`=3underset(nto oo2)(lim)(1)/(2)((1)/(1*2)-(1)/(n(n+1)(n+2)))`
`(3)/(2)((1)/(2)-0)=(3)/(4)`.