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anonymous

  • 4 years ago

i need to find the interval of convergence of the series (from 1 to infinity) of ((-1^n+1)(x-2)^n)/n2^n and then check for the convergence at the endpoints. i got the interval but i am having trouble with checking the endpoints. help ):

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  1. anonymous
    • 4 years ago
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    \[\sum_1^{\infty}\frac{(-1)^{n+1}(x-2)^n}{n2^n}\]

  2. anonymous
    • 4 years ago
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    i am going to make a guess that the radius is 2, so that you will have \[|x-2|<2\] is that what you got?

  3. anonymous
    • 4 years ago
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    YES. and then i added 2 to both sides. so 0<x<4 ?

  4. anonymous
    • 4 years ago
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    that is what i got but i kind of guessed. ok so now we have to check at x = 0 and x = 4 right?

  5. anonymous
    • 4 years ago
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    yerp....

  6. anonymous
    • 4 years ago
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    maybe i am off by a minus sign, not sure, but this is an alternating series so surely converges as the terms go to zero, so we can include 0

  7. anonymous
    • 4 years ago
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    wait. how did you do that. how does it cancel off?

  8. anonymous
    • 4 years ago
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    i cancled the \[2^n\]

  9. anonymous
    • 4 years ago
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    oooh lol sorry

  10. anonymous
    • 4 years ago
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    my mistake i meant \[\sum\frac{(-1)^n}{n}\] oops

  11. anonymous
    • 4 years ago
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    i canceled wrong...

  12. anonymous
    • 4 years ago
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    no actually my algebra is terrible, can you explain the cancelling please?

  13. anonymous
    • 4 years ago
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    but conclusion is right. converges because you have an alternating series wehre terms go to zero

  14. anonymous
    • 4 years ago
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    you have in the numerator \[(-2)^n\] and in the denominator you have \[2^n\] but \[(-2)^n=(-1)^n2^n\] so you can cancel a \[2^n\]top and bottom

  15. anonymous
    • 4 years ago
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    kick the minus sign in to the first part to make it \[(-1)^n\] instead of \[(-1)^{n+1}\]

  16. anonymous
    • 4 years ago
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    that just leaves you with \[\sum\frac{(-1)^n}{n}\]

  17. anonymous
    • 4 years ago
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    ohh okay. and then by alternating series test it converges. got it! what about at x=4?

  18. anonymous
    • 4 years ago
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    well i didn't do that yet, so lets try it

  19. anonymous
    • 4 years ago
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    haha okay.

  20. anonymous
    • 4 years ago
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    i think this one is a problem because it will not alternate

  21. anonymous
    • 4 years ago
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    i think it will alternate? and it diverges because 1/n is harmonic.

  22. anonymous
    • 4 years ago
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    thank you so much though! do you know how to differentiate and integrate the series ones too?

  23. anonymous
    • 4 years ago
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    wait wait!!

  24. anonymous
    • 4 years ago
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    hold on a second i think i might have made a big big mistake!!

  25. anonymous
    • 4 years ago
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    why? i think it is correct...

  26. anonymous
    • 4 years ago
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    let me be more careful with zero

  27. anonymous
    • 4 years ago
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    whoa did i make a mistake!! lets go slow

  28. anonymous
    • 4 years ago
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    oh oops i think the x=4 is convergent too.

  29. anonymous
    • 4 years ago
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    lets look at the numerator when x = 0

  30. anonymous
    • 4 years ago
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    \[(-1)^{n+1}(-2)^n\] \[(-1)^{n+1}(-1)^n2^n\] \[(-1)^{2n+1}2^n\]

  31. anonymous
    • 4 years ago
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    now we can certainly cancel a \[2^n\] with the denominator

  32. anonymous
    • 4 years ago
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    get \[\sum\frac{(-1)^{2n+1}}{n}\]

  33. anonymous
    • 4 years ago
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    ohh i seee...

  34. anonymous
    • 4 years ago
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    ratio test?

  35. anonymous
    • 4 years ago
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    but \[2n+1\] is odd, and so the numerator is -1

  36. anonymous
    • 4 years ago
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    in other words you get \[\sum\frac{-1}{n}\] which diverges for sure

  37. anonymous
    • 4 years ago
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    ohhh okay okay. sounds good. um can you help me with another one?

  38. anonymous
    • 4 years ago
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    sorry about that , it was a big mistake on my part. i just thought to kick over the minus sign, but i forgot to add the exponents

  39. anonymous
    • 4 years ago
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    we did not do it with 4 yet, wanna try it?

  40. anonymous
    • 4 years ago
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    4 is going to converge, because this time the numerator is \[(-1)^{n+1}2^n\]

  41. anonymous
    • 4 years ago
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    cancel the \[2^n\] get \[\sum \frac{(-1)^{n+1}}{n}\]

  42. anonymous
    • 4 years ago
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    and this one alternates, so it converges for x = 4 but not for x = 0

  43. anonymous
    • 4 years ago
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    oh okay i get it!!

  44. anonymous
    • 4 years ago
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    yeah well i made a mistake, sorry it took so long

  45. anonymous
    • 4 years ago
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    you were so much help regardless. i barely know what i am doing. can i ask you for help on another one?

  46. anonymous
    • 4 years ago
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    sure but post in a new thread and i will come look it is much easier than scrolling down

  47. anonymous
    • 4 years ago
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    thanks !

  48. anonymous
    • 4 years ago
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    yw

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