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anonymous

  • 4 years ago

the series ((-1)^n+1)(x-2)^n)/n .... the interval of convergence is 1<x<3, but i need help checking the convergence of the endpoints.

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  1. anonymous
    • 4 years ago
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    another end point one, this time i do it with paper before typing

  2. anonymous
    • 4 years ago
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    \[|x-2|<1\]

  3. anonymous
    • 4 years ago
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    ok got it

  4. anonymous
    • 4 years ago
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    now we try replacing x by 1 btw you have done all the hard work, the rest should be the easy part

  5. anonymous
    • 4 years ago
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    at x=1 i think it is \[\sum_{?}^{?} (-1)^{n+1}(-1)^{n}\div n\] ???

  6. saifoo.khan
    • 4 years ago
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    Whenever free, http://openstudy.com/study#/updates/4f333c8be4b0fc0c1a0b72dd

  7. anonymous
    • 4 years ago
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    ok very smilar to the last one

  8. anonymous
    • 4 years ago
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    but this time we don't do anything stupid \[(-1)^{n+1}(-1)^n=(-1)^{2n+1}=-1\]

  9. anonymous
    • 4 years ago
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    on account of \[2n+1\] is odd.

  10. anonymous
    • 4 years ago
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    so does not converge at x = 1

  11. anonymous
    • 4 years ago
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    what do you want to bet it does converge at x = 3?

  12. anonymous
    • 4 years ago
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    .ah! um a lot? haha.

  13. anonymous
    • 4 years ago
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    yeah well we can just about visualize it because it is the same as the last one. this time you get \[(-1)^{n+1}\times 1=(-1)^{n+1}\]so we are in good shape

  14. anonymous
    • 4 years ago
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    thank you! another question...

  15. anonymous
    • 4 years ago
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    this one alternates in other words, whereas the other one does not

  16. anonymous
    • 4 years ago
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    keep posting

  17. anonymous
    • 4 years ago
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    you got zarkon watching so if i mess up he will step in wish he had stepped in to the last one so i wouldn't have made a fool out of myself

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spraguer (Moderator)
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