Find the sum of the series `1/(3^2+1)+1/(4^2+2)+1/(5^2+3)+1/(6^2+4)+oo`

Explanation

`T_(r)=1/(r^(2)+(r-2))=1/((r+2)(r-1))`, where n=3,4,5,..
`=1/3[1/(r-1)-1/(r+2)]`
`=1/3[V(r )-V(r+3)],` where `V(r )=1/(r-1)`
`therefore` Sum of n terms of the series,
`sum_(r=3)^(n)T_(r )=sum_(r=3)^(n)1/3[V( r)-V(r+3)]`
`=1/3[V(3)+V(4)+V(5)-V(n+1)-V(n+2)-V(n+3)]`
`=1/3[1/2+1/3+1/4-1/n-1/(n+1)-1/(n+2)]`
`thereforesum(r=3)^(oo)T_(r)=1/3[1/2+1/3+1/4]=13/36`