The nth term of the series
Tr=(rth term of the series3,5,7...)/(r^2(r+1)^2)
=(3+(r-1)×2)/(r^2(r+1)^2)
=(2r+1)/(r^2(r+1)^2)
=((r+1)^2-r^2)/(r^2(r+1)^2)
=1/r^2-1/(r+1)^2
Putting r =1,2,3..... up to n and summing we get
Sn=1-1/(n+1)^2
As n tends to infinity 1/(n+1) will tend to zero.
Hence Sn=1 when n tends to infinity.
