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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0try nth root test. i believe it diverges.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you just neet to check for convergence or divergence, why not just put the values and check, \[1 \rightarrow 1!^1/1^4 = 1\] \[2 \rightarrow 2!^2/2^8 = 4/256 = 1/64\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0matter 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 \[3.96\times 10^{65}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First tell me, does converging only mean that the terms go on becoming small and small?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If so, (n+1)!^(n+1)/(n+1)^4(n+1) =(n+1)^(n+1)(n!)^(n+1) / (n+1)^4(n+1) \[=(n!)^n (n+1)^{n+1} n! / (n+1)^{(n+1)^{4}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So it has to converge!
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