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Requiem

  • 2 years ago

Determine convergence or divergence. (-1)^n/(n^3) I know that I want to use the alternating series test. I used (bsubn) as 1/n^3 and (bsubn+1) as 1/(n+1)^3. I know that (bsubn) as n ----> infinity equals 0, but is (bsubn+1) less than or equal to (bsubn)?

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  1. Requiem
    • 2 years ago
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    Or do i need to use a comparison test?

  2. Requiem
    • 2 years ago
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    I know that those two conditions have to be met for it to be convergent...

  3. anonymous
    • 2 years ago
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    is it \[\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}\]

  4. Requiem
    • 2 years ago
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    yes thats it

  5. anonymous
    • 2 years ago
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    it not only converges, but it converges absolutely

  6. anonymous
    • 2 years ago
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    all you need for the alternating series to converge is that the terms go to zero, which thsee certainly do

  7. Requiem
    • 2 years ago
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    oh ok..in my book it says that (bsubn+1) has to be less than or equal to (bsubn) but also that (bsubn) has to equal 0 when the limit is taken...

  8. anonymous
    • 2 years ago
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    but \[\sum_{n=1}^{\infty}\frac{1}{n^3}\] also converges, so this series is not just convergent, it is absolutely convergent as well

  9. Requiem
    • 2 years ago
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    and i wasnt sure if the former was true

  10. anonymous
    • 2 years ago
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    yeah only need the terms go to zero

  11. Requiem
    • 2 years ago
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    ok, thanks satellite

  12. anonymous
    • 2 years ago
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    \[\sum\frac{(-1)^n}{n}\] for example converges, although not absolutely

  13. anonymous
    • 2 years ago
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    hell, even \[\sum\frac{(-1)^n}{\ln(n)}\] converges

  14. Requiem
    • 2 years ago
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    its converges conditionally right?

  15. anonymous
    • 2 years ago
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    the last two i wrote are only conditionally convergent, the one you posted is absolutely convergent, a stronger condition

  16. Requiem
    • 2 years ago
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    ok thanks you for your help! much appreciated

  17. anonymous
    • 2 years ago
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    yw

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