calculus-Determine whether the sequence is divergent or convergent

Question

darkigloo · Answer

$$\lim_{n \rightarrow \infty}\frac{ 4n! }{(-6)^n }$$

satellite73 · Answer

not a sum, right, just a sequence?

darkigloo · Answer

yes

satellite73 · Answer

goes bouncing all over the place
trying to think of a good explation, but $n!$ is way larger than $6^n$ and also it alternates in sign

anonymous · Answer

in the limit, $n!\sim e^{-n} n^{n+1/2}$ so clearly the numerator dominates the denominator and the terms diverge to grow arbitrarily large

anonymous · Answer

but yes, since it alternates in sign as well, it doesn't diverge in any particular direction (i.e. $\pm\infty$) but instead oscillates ever more greatly in amplitude

satellite73 · Answer

how you are supposed to know that $n!$ is larger than $e^n$ is not clear to me unless you know other stuff

darkigloo · Answer

so it diverges?

satellite73 · Answer

definitely

satellite73 · Answer

try with only a small number, like 10 and 11 say

darkigloo · Answer

what would that show?

satellite73 · Answer

here it is with n = 100 http://www.wolframalpha.com/input/?i=%284*100!%29%2F%28-6%29^100

satellite73 · Answer

just gives you an idea of the relative size of $n!$ vs $e^n$ or $6^n$

satellite73 · Answer

how you are supposed to justify it is not clear to me
it is not like you can use l'hopital or anything
you sort of just know it 
maybe there is a better explanation in the book, although @oldrin.bataku wrote one

darkigloo · Answer

ok thank you