If `sum_(r=1)^n t_r=n/8(n+1)(n+2)(n+3),` then find `sum_(r=1)^n1/(t_r)dot`

Explanation

`S_(n)=sum_(r=1)^(n)T_(r)=((n+1)(n+2)(n+3))/(8)`
`T_(r)=S_(r)-S_(r-1)=(r(r+1)(r+2)(r+3))/(8)-((r-1)(r+1)(r+2))/(8)`
`=((r+1)(r+2))/(2)`
`(1)/(T_(r))=(2)/(r(r+1)(r+2))=((r_+2)-r)/(r(r+1)(r+2))=((1)/(r(r+1))-(1)/(r(r+1)(r+2)))`
`sum_(r=1)^(n)(1)/(T_(r))=sum_(r=1)^(n)((1)/(r(r+1))-(1)/(r(r+1)(r+2)))`
`=sum_(r=1)^(n){((1)/(r)-(1)/(r+1))-((1)/(r+1)-(1)/(r+2))}`
`=((1)/(1)-(1)/(n+1))-((1)/(2)-(1)/(n+2))`
`=(1)/(2)+(1)/(n+2)-(1)/(n+2)=(n(n+3))/(2(n+1)(n+2))`.