Question

series (n!)^n / n^4n converges or diverges?

Answer 1

try nth root test. i believe it diverges.

Answer 2

If you just neet to check for convergence or divergence, why not just put the values and check,

\[1 \rightarrow 1!^1/1^4 = 1\]
\[2 \rightarrow 2!^2/2^8 = 4/256 = 1/64\]

Answer 3

matter of fact these terms don't even go to zero, so it clearly diverges.
as an example \[\frac{(10!)^{10}}{10^{40}}\] is huge. the denominator is \[10^{40}\]
while the numerator is about \[3.96\times 10^{65}\]

Answer 4

First tell me, does converging only mean that the terms go on becoming small and small?

Answer 5

If so,
(n+1)!^(n+1)/(n+1)^4(n+1)
=(n+1)^(n+1)(n!)^(n+1) / (n+1)^4(n+1)
\[=(n!)^n (n+1)^{n+1} n! / (n+1)^{(n+1)^{4}}\]

Answer 6

So it has to converge!