Convergence Tests for Sin(2n)/n^2 and (-3)^n/n!: Ratio Test vs Comparison Test

[badtwistoffate]

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 don't 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?

 

[Pyrrhus]

On the 1st one: what about the comparison test to its absolute value series?

On the 2nd one: Consider the Ratio Test or d'Alambert's ratio.