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anonymous

  • 5 years ago

series n!/n^n converge or diverge? how do you find out the convergence of ((2n^3+5n^2+3)/(4n^3+n+7))^n?

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  1. anonymous
    • 5 years ago
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    for second one try to apply root test and then factor n^3 out and then take limit.

  2. anonymous
    • 5 years ago
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    ok i understood that. is there any other way i can do this

  3. anonymous
    • 5 years ago
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    existence, you should note that\[\frac{n!}{n^n}-\frac{n}{n}.\frac{n-1}{n}.\frac{n-2}{n}.....\frac{3}{n}\frac{2}{n}\frac{1}{n}\]As n goes to infinity, the left-hand numbers become asymptotically equivalent to 1, while the right-hand numbers approach zero, so you end up with a situation where\[\frac{n!}{n^n} \iff 1.1.1.....0.0.0=0\]where the if and only if sign is meant to mean asymptotic behavior.

  4. anonymous
    • 5 years ago
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    For the last one, you have that\[2n^3\]is asymptotically equivalent to \[2n^3+5n^2+3\](i.e.for n large, they behave the same way) and for \[4n^3+n+7\]you have that this is asymptotically equivalent to \[4n^3\]As a consequence, for large n, you have\[\left( \frac{2n^3+5n^2+3}{4n^3+n+7} \right)^n \approx \left( \frac{2n^3}{4n^3} \right)^n=\left( \frac{1}{2} \right)^n=\frac{1}{2^n}\]so as n tends to infinity, your expression tends to 0 since 1/2^n tends to zero. I've included some information on asymptotic equivalence.

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  5. anonymous
    • 5 years ago
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    thank you ver much

  6. anonymous
    • 5 years ago
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    Thank me by becoming a fan :) Took forever to write out :p And you're welcome.

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