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anonymous

  • 5 years ago

Determine whether the sequence converges or diverges. If it converges, nd the limit n!/2^n

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  1. anonymous
    • 5 years ago
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    You can use the ratio test. \[\lim_{n->\infty}\left|\frac{a_{n+1}}{a_n} \right|=\lim_{n->\infty}\frac{(n+1)!/2^{n+1}}{n!/2^{n}}=\lim_{n->\infty}\frac{(n+1)n!}{2 \times 2^n}\frac{2^n}{n!}\]\[=\lim_{n->\infty}\frac{n}{2} \rightarrow \infty\] The sequence does not converge since the ratio in the limit is not less than 1.

  2. anonymous
    • 5 years ago
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    nice!

  3. anonymous
    • 5 years ago
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    Thanks ;)

  4. anonymous
    • 5 years ago
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    \[\frac{n+1}{2}\]

  5. anonymous
    • 5 years ago
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    A question.. So does it mean that you cant test some values as N... and see whats happening?

  6. anonymous
    • 5 years ago
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    Can you rephrase that? I'm not sure what you mean.

  7. anonymous
    • 5 years ago
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    n!/2^n... Put in some values, as a check.. ..n= 1, 2, 3

  8. anonymous
    • 5 years ago
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    And see if it diverges or converges

  9. anonymous
    • 5 years ago
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    I hope you didnt misunderstand me, it was just a question, if i would in some values, in the funktion , n!/2^n .. could i see right away if it diverges or converges..

  10. anonymous
    • 5 years ago
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    Ah I see. No, because that's not actually testing for ALL n. The tests are derived from the definition of convergence.

  11. anonymous
    • 5 years ago
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    You could get an idea, but it wouldn't prove it. That's all.

  12. anonymous
    • 5 years ago
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    Okey, so u have to, show it for all n. like u did

  13. anonymous
    • 5 years ago
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    Exactly.

  14. anonymous
    • 5 years ago
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    Well, thx.

  15. anonymous
    • 5 years ago
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    No worries.

  16. anonymous
    • 5 years ago
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    Hehe, I thought this was XiaoHong's question.

  17. anonymous
    • 5 years ago
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    So did i ;)

  18. anonymous
    • 5 years ago
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    thanks

  19. anonymous
    • 5 years ago
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    Welcome.

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