Determine if $$\sum^{\infty}_{n=1} \dfrac{(n!)^n}{n^{4n}}$$ convergent or divergent.
I see powers of $n$ and so my first thought is to use root test.
I can rewrite the series as follows:
$$\sum^{\infty}_{n=1} \dfrac{(n!)^n}{n^{4n}} = \sum^{\infty}_{n=1} (\dfrac{n!}{n^{4}})^n$$
Applying the ratio test, we consider $\lim_{n\to\infty} (|a_n|)^{1/n}$, but since $a_n$ is always positive, we can neglect that. So we have:
$$\lim_{n\to\infty}((\dfrac{n!}{n^{4}})^n)^{1/n} =\lim_{n\to\infty}\dfrac{n!}{n^{4}} $$
How do I evaluate this limit? I think since $n!$ grows without bound compared to $n^4$, that the limit approaches $\infty$, but how can I show this? Is there an easier test?
This is similar to another question found in here, but I simply want to know how to evaluate the limit in this case.
For large values of $n$,
$\frac{n!}{n^4} = \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)!}{n^4} = \frac{n^4(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n})(1-\frac{4}{n})(n-5)!}{n^4} $
$= (1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n})(1-\frac{4}{n})(n-5)! $
which clearly diverges as $n \to \infty$.