1. ## series with n!

i'm supposed to test this series for convergence/divergence

$\sum_{n=1}^{\infty} \frac{(n!)^n}{n^{4n}}$

i guess my real hang-up is with the factorial..but i'm just kind of lost with this one.. please help!

2. Originally Posted by buttonbear
i'm supposed to test this series for convergence/divergence

$\sum_{n=1}^{\infty} \frac{(n!)^n}{n^{4n}}$

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 $n$ we have: $n! \geq n(n-1)(n-2)(n-3)(n-4) \geq n^4.$ thus: $\frac{(n!)^n}{n^{4n}} \geq 1.$

this method also proves that for any real number $c$ the series $\sum_{n=1}^{\infty} \frac{(n!)^n}{n^{cn}}$ is divergent.

3. is this any particular test? or is it just some sort of comparison.. i'm sorry but i need to be able to justify everything i say about these problems

4. Originally Posted by buttonbear
i'm supposed to test this series for convergence/divergence

$\sum_{n=1}^{\infty} \frac{(n!)^n}{n^{4n}}$

i guess my real hang-up is with the factorial..but i'm just kind of lost with this one.. please help!

you can use the root test and make it

$\frac{(n!)^n}{(n^4)^n}$ so that it is
limit (x is approaching infinity) of the nth root of $\frac{(n!)^n}{(n^4)^n}$ and it becomes
limit of $\frac{n!}{n^4}$
and it is $\frac{fast}{slow}$
diverging