anonymous
  • anonymous
series (n!)^n / n^4n converges or diverges?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
try nth root test. i believe it diverges.
anonymous
  • anonymous
If you just neet to check for convergence or divergence, why not just put the values and check, \[1 \rightarrow 1!^1/1^4 = 1\] \[2 \rightarrow 2!^2/2^8 = 4/256 = 1/64\]
anonymous
  • 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 \[3.96\times 10^{65}\]

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anonymous
  • anonymous
First tell me, does converging only mean that the terms go on becoming small and small?
anonymous
  • anonymous
If so, (n+1)!^(n+1)/(n+1)^4(n+1) =(n+1)^(n+1)(n!)^(n+1) / (n+1)^4(n+1) \[=(n!)^n (n+1)^{n+1} n! / (n+1)^{(n+1)^{4}}\]
anonymous
  • anonymous
So it has to converge!

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