It is evident that sum of the series tends to oo. However, let us find sum up to n terms.
n^(th) term of the series 1*3*5+3*5*7+5*7*9+.... is
(2n-1)(2n+1)(2n+3) and we have to find sum_1^n((2n-1)(2n+1)(2n+3))
As (2n-1)(2n+1)(2n+3)
= (2n+3)(4n^2-1)
= 8n^3+12n^2-2n-3
Hence sum_1^n(8n^3+12n^2-2n-3)
= 8sum_1^n n^3+12sum_1^n n^2-2sum_1^n n-3n
= 8(n^2(n+1)^2)/4+12(n(n+1)(2n+1))/6-2(n(n+1))/2-3n
= 2n^2(n+1)^2+2n(n+1)(2n+1)-n(n+1)-3n
= 2n^2(n^2+2n+1)+(2n^2+2n)(2n+1)-n^2-n-3n
= 2n^4+4n^3+2n^2+4n^3+2n^2+4n^2+2n-n^2-n-3n
= 2n^4+8n^3+7n^2-2n
It is apparent that as n->oo sum of the series tends to oo.