## anonymous 5 years ago series (n!)^n / n^4n converges or diverges?

1. anonymous

try nth root test. i believe it diverges.

2. anonymous

If you just neet to check for convergence or divergence, why not just put the values and check, $1 \rightarrow 1!^1/1^4 = 1$ $2 \rightarrow 2!^2/2^8 = 4/256 = 1/64$

3. anonymous

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 $3.96\times 10^{65}$

4. anonymous

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

5. anonymous

If so, (n+1)!^(n+1)/(n+1)^4(n+1) =(n+1)^(n+1)(n!)^(n+1) / (n+1)^4(n+1) $=(n!)^n (n+1)^{n+1} n! / (n+1)^{(n+1)^{4}}$

6. anonymous

So it has to converge!