help determining where the series here is abs. convergent. its Sin(2n)/n^2, i thought about the ratio test but it gets nasty, is there a easier way? nm, i think it is convergence if I use the comparison test? Sound right? what about (-3)^n/n!, i used the ratio test, and got 0, which means it abs. convergent since 0<1, but i dont think i did it right . I did: Lim n-> infinity : (-3^(n+1)/(n+1)!)(n!/-3^n)= -3 lim n!/n+1!.... does that look right?