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anonymous

  • 5 years ago

determine if the summation to infinity of the alternating sum is absolutely convergent, conditionally convergent or divergent: (-1)^(k+1)/k! (having problems simplifying the absolute convergence test, for some reason i was never taught how to simplify the factorial when it was in a fraction..)

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  1. anonymous
    • 5 years ago
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    Never heard it called the absolute convergence test. Are you talking about the ratio test?

  2. anonymous
    • 5 years ago
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    \[\lim_{n \rightarrow \infty} \left| a(n+1)/a(n) \right|\] ?

  3. anonymous
    • 5 years ago
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    yes haha

  4. anonymous
    • 5 years ago
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    Ah ok. Well what your going to do is you write your function out, replacing "n" with "n+1". Then your going to divide it by your function written with just "n" for your variable.

  5. anonymous
    • 5 years ago
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    But you can bypass that step by multiplying by the reciprocal once you get the hang of it.

  6. anonymous
    • 5 years ago
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    yup got that, i'm having trouble simplifying that

  7. anonymous
    • 5 years ago
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    Your k! when your replacing "k" with "k+1" will end up being "(k+1)!"

  8. anonymous
    • 5 years ago
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    Is that what you meant or as you progress farther into the problem when you have k! over (k+1)!

  9. anonymous
    • 5 years ago
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    further, taking the limit of the already established reciprocal

  10. anonymous
    • 5 years ago
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    multiplaction step

  11. anonymous
    • 5 years ago
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    ok. And what exact thing are you having trouble multiplying?

  12. anonymous
    • 5 years ago
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    finding the limit

  13. anonymous
    • 5 years ago
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    right now i have 1^(k+1)+1/(k+1)! * k!/1^(k+1)

  14. anonymous
    • 5 years ago
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    so whats next? how do i find that limit/simplifiy that fraction

  15. anonymous
    • 5 years ago
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    Ok. there is a property that comes into play when you have n!/(n+1)!. I'm trying to think of how to show it. It reduces it down.

  16. anonymous
    • 5 years ago
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    Let me look in my notes real quick

  17. anonymous
    • 5 years ago
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    ok :)

  18. anonymous
    • 5 years ago
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    cause i'll also need that rule to test the original function for convergence

  19. anonymous
    • 5 years ago
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    Ok here is what happens. Lets say you have n! over (n+1)!. Your result will be (n+1) in the denominator (no factorial)

  20. anonymous
    • 5 years ago
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    The factorials cancel each other in a way.

  21. anonymous
    • 5 years ago
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    ..and you'll keep the constants right?

  22. anonymous
    • 5 years ago
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    Well the only thing affected in what I just told you is your k! and (k+1)! The rest of the problem reduces just like typical multiplication & cancelling.

  23. anonymous
    • 5 years ago
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    So the first term in what you have right now has a (k+1) in the denominator instead of (k+1)!

  24. anonymous
    • 5 years ago
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    so i'll end up taking the limit of 1/k+1 ?

  25. anonymous
    • 5 years ago
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    I believe so

  26. anonymous
    • 5 years ago
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    wonderful :)

  27. anonymous
    • 5 years ago
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    thanks for your help! :)

  28. anonymous
    • 5 years ago
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    Your very welcome

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