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anonymous

  • one year ago

Limits of series. I am going to type the equation using equation editor.

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  1. anonymous
    • one year ago
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    I am trying to find the following limit. \[\lim_{n \rightarrow \inf} \frac{ 2(n!) }{ (-6)^n }\]

  2. SolomonZelman
    • one year ago
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    Factorial will grow much much bigger than any exponential growth (after some point).

  3. anonymous
    • one year ago
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    Is it because the factorial is multiplied by 2?

  4. anonymous
    • one year ago
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    I was under the assumption that the (-6)^n would grow faster than the factorial.

  5. SolomonZelman
    • one year ago
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    two does not do anything at all.... whether it has a two or not factorial is much bigger. Wrong assumption! The opposite is true

  6. anonymous
    • one year ago
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    Then I guess the limit diverges to both negative and positive infinity.

  7. SolomonZelman
    • one year ago
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    Yes:) positive and negative infinity, because it alternates.

  8. anonymous
    • one year ago
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    Alright, Thanks. I guess my understand of factorials are a bit off.

  9. SolomonZelman
    • one year ago
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    We can play around and maybe apply Stirling's approximation: (Note, that the factorial is greater than its stirlings approximation) \(\large\color{black}{ \displaystyle n!\approx\frac{\sqrt{2\pi n}{~}n^n}{e^n} }\) \(\large\color{black}{ \displaystyle \frac{n!}{6^n}\approx\frac{\sqrt{2\pi n}{~}n^n}{6^ne^n} }\) \(\large\color{black}{ \displaystyle \frac{n!}{6^n}\approx\sqrt{2\pi n}\frac{{~}n^n}{(6e)^n} }\)

  10. some.random.cool.kid
    • one year ago
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    thx i learned something new just by watching lol

  11. SolomonZelman
    • one year ago
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    You would then agree that the numerator is greater, right?

  12. anonymous
    • one year ago
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    I do agree that the numerator is greater.

  13. SolomonZelman
    • one year ago
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    Good:) And I understand that you were applying the Diverghence test to \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{2(n!)}{(-6)^n}}\)

  14. SolomonZelman
    • one year ago
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    (not necessarily n=1, but that series)

  15. anonymous
    • one year ago
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    yep

  16. SolomonZelman
    • one year ago
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    Ok, so if it was the other way around, then you would apply: \(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty } ~ \frac{x^n}{n!}=e^x}\) So for example: \(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty } ~ \frac{(-6)^n}{n!}=e^{-6}= 1/e^6}\)

  17. SolomonZelman
    • one year ago
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    Anyway, good luck....

  18. anonymous
    • one year ago
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    Thanks for the help. It really made since. I am now making forward progress on my assignments.

  19. SolomonZelman
    • one year ago
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    Good to hear that:)

  20. SolomonZelman
    • one year ago
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    be well

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