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anonymous

  • 5 years ago

why cant i apply the alternating series test to this series sigma n=1 to infinity (-1)^n-1 sin n

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  1. anonymous
    • 5 years ago
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    Says here, to use alternating it should be of the for -1^(n+1) or -1^n

  2. anonymous
    • 5 years ago
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    is there any other explanation besides that

  3. anonymous
    • 5 years ago
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    No. The point is there are many test and any one of them may be inconclusive and you may have to use another one in conjunction with it. Flexibility is needed on your part.

  4. anonymous
    • 5 years ago
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    so if there was a minus sign n-1 then it would be alternating?

  5. anonymous
    • 5 years ago
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    I am not sure what you are referring to.

  6. anonymous
    • 5 years ago
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    "Says here, to use alternating it should be of the for -1^(n+1) or -1^n 10 minutes ago "

  7. anonymous
    • 5 years ago
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    Yeah. Yours is not like that. Yours has n-1 in the exponent.

  8. anonymous
    • 5 years ago
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    so those are the only two main condition that cause it to have alternative signs

  9. anonymous
    • 5 years ago
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    I think its -1 in front that makes it alternating. Yours is alternating but they don't like the n-1 in the exponent

  10. anonymous
    • 5 years ago
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    when i plugged in values i gott two negatives and then two positive

  11. anonymous
    • 5 years ago
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    Yep, that's what they mean by alternating

  12. anonymous
    • 5 years ago
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    now im kinda of confused then why is it that in this problem we can apply the test if thats what happens

  13. anonymous
    • 5 years ago
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    OK, with the power vested in me, I hereby give you permission to use the alternating series test.

  14. anonymous
    • 5 years ago
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    k

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