i'm supposed to test this series for convergence/divergence
$\displaystyle \sum_{n=1}^{\infty} \frac{(n!)^n}{n^{4n}}$
i guess my real hang-up is with the factorial..but i'm just kind of lost with this one.. please help!
it's divergent because for sufficiently large values of $\displaystyle n$ we have: $\displaystyle n! \geq n(n-1)(n-2)(n-3)(n-4) \geq n^4.$ thus: $\displaystyle \frac{(n!)^n}{n^{4n}} \geq 1.$
this method also proves that for any real number $\displaystyle c$ the series $\displaystyle \sum_{n=1}^{\infty} \frac{(n!)^n}{n^{cn}}$ is divergent.
you can use the root test and make it
$\displaystyle \frac{(n!)^n}{(n^4)^n}$ so that it is
limit (x is approaching infinity) of the nth root of $\displaystyle \frac{(n!)^n}{(n^4)^n}$ and it becomes
limit of $\displaystyle \frac{n!}{n^4}$
and it is $\displaystyle \frac{fast}{slow} $
diverging