- #1
badtwistoffate
- 81
- 0
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?
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?
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