DOES THE SEQUENCE CONVERGE OR DIVERGE?

7n!
----
3^n

Question

DOES THE SEQUENCE CONVERGE OR DIVERGE?

7n!
----
3^n

anonymous · Answer

As n-> infinity

anonymous · Answer

diverge

anonymous · Answer

as n -> infinity, the sequence converge only when it reachs 0, diverge when it goes infinity

anonymous · Answer

?

anonymous · Answer

diverges because the limit as n goes to infinity is infinite 
you can prove that $n!> 3^n$ for $n>6$ or $n>7$ one of those two

anonymous · Answer

How to simplify the fraction though to show this? That factorial is really throwing me off

anonymous · Answer

As satellite73 showed, n!>3n for n>7. For example, 
3^7 = 2187 and 7! = 5040
From n = 7, if we increase n by 1 3^n increase 3 times but n! increase 8, 9, 10... times. Thus n! is much greater than 3^n as n -> infinity.

There is no way to simply n! and 3^n. However, we can use the rule when n -> infinity. If the numerator increases faster than the denominator, the fraction approaches infinity. On the other hand, if the numerator increases slower than the denominator, the fraction approaches 0.

anonymous · Answer

By the way, satellite included the coefficient 7, so his prove only needs n = 6. I ignored the coefficient 7, so I started n =7. In both cases, the sequence approaches infinity.