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tiffanymak1996 Group Title

is this absolutely convergent or conditionally or divergent?

  • one year ago
  • one year ago

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  1. tiffanymak1996 Group Title
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    \[\sum_{n=1}^{\infty} (-1)^{n}\times e ^{\frac{ 1 }{ n }}\]

    • one year ago
  2. Herp_Derp Group Title
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    Divergent...

    • one year ago
  3. tiffanymak1996 Group Title
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    yes that's the answer given, but i got conditionally throughout the process can u go thru the steps with me?

    • one year ago
  4. Herp_Derp Group Title
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    Well, for some large \(n\), \(e^{1/n}\) is sufficiently close to \(1\) so that we can compare this series to the divergent series:\[\sum_{n=0}^{\infty}(-1)^n\]The larger the \(n\), the closer your series is to this series. Therefore your series diverges. Do you understand why \(\sum_{n=0}^{\infty}(-1)^n\) diverges?

    • one year ago
  5. tiffanymak1996 Group Title
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    but, i thought n=1 under the sigma sign...

    • one year ago
  6. godorovg Group Title
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    No, tiffany they are telling right here, it is diverges

    • one year ago
  7. jeanlouie Group Title
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    we have abs(\[(-1)^{n}\times e ^{1/n}\]) -> 1 when \[n \rightarrow \infty\] but if consider \[a_{n}\] = \[(-1)^{n}\times e ^{1/n}\] , then |\[a_{2n}\] - \[a_{2n+1}\]| > 1 with every n > some N it is diverges

    • one year ago
  8. tiffanymak1996 Group Title
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    yeah, what i meant was i know it is divergent, but when i go thru the steps i got conditionally convergence

    • one year ago
  9. tiffanymak1996 Group Title
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    don't understand this part an = (−1)n×e1/n , then | a2n - a2n+1 | > 1 with every n > some N it is diverges

    • one year ago
  10. jeanlouie Group Title
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    i mean, for example, let N = 1000000, then for every n > N, |a_2n - a_(2n+1)| > 1

    • one year ago
  11. godorovg Group Title
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    Jeanlouie I haven't this type calc for a while since High School, in order for this to be diverges it means the n is closer to 1 or I am lost here?

    • one year ago
  12. tiffanymak1996 Group Title
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    for the ratio test if the lim is <1 then the series converges, if lim >1 or=infinity the series diverges, if lim=1 inconclusive

    • one year ago
  13. tiffanymak1996 Group Title
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    i know that testing the series for absolutely convergent or conditionally or divergent is to try the ratio test with absolute value first then the one without.

    • one year ago
  14. tiffanymak1996 Group Title
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    but when i calculate the one with absolute value the limit =1 so inconclusive; when i test the series for the ratio test without absolute value the limit =-1, so it's convergent; that's how i get conditionally convergent.

    • one year ago
  15. tiffanymak1996 Group Title
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    although i know i'm wrong, cause the answer is divergent:(

    • one year ago
  16. godorovg Group Title
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    what is the value {1} of this when testing ratio =

    • one year ago
  17. tiffanymak1996 Group Title
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    when testing ratio without the absolute value i got -1

    • one year ago
  18. godorovg Group Title
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    How does the value change to -1 I am lost there?

    • one year ago
  19. jeanlouie Group Title
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    1 is only an example here, 0.5 or any other \[\epsilon < -2\] is OK, too. when \[n \rightarrow \infty \], \[a_{2n} \rightarrow 1\] and \[a_{2n+1}\rightarrow -1\] ps:sr for my poor English

    • one year ago
  20. jeanlouie Group Title
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    ϵ< 2, sorry

    • one year ago
  21. tiffanymak1996 Group Title
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    \[\rho = \lim_{n \rightarrow +\infty} \frac{ (-1)^{n+1} \times e ^{\frac{ 1 }{ n }}}{ ((-1)^{n})\times e ^{\frac{ 1 }{ n }}\ }\]

    • one year ago
  22. tiffanymak1996 Group Title
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    ln both top and bottom

    • one year ago
  23. jeanlouie Group Title
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    yeah, ρ = -1, you got it

    • one year ago
  24. tiffanymak1996 Group Title
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    becomes:\[\lim_{n \rightarrow +\infty} -\frac{ \frac{ 1 }{ n+1} }{ \frac{ 1 }{ n } } \]

    • one year ago
  25. jeanlouie Group Title
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    no, top or bottom may be negative, we can't use ln here.

    • one year ago
  26. tiffanymak1996 Group Title
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    (the limit is still there but i'm just lazy: then: -n/(n+1) divide top and bottom with n: -1/(1+1/n) substitute infinity: -1/(1+0) =-1

    • one year ago
  27. godorovg Group Title
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    Tiffany I think you have the right answer.

    • one year ago
  28. tiffanymak1996 Group Title
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    i do? but this is like conditionally ; yet the modal answer is divergent...

    • one year ago
  29. godorovg Group Title
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    Tiffany 1+1n is not -1 it is 2n

    • one year ago
  30. tiffanymak1996 Group Title
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    model, sry

    • one year ago
  31. godorovg Group Title
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    2/n not -1

    • one year ago
  32. godorovg Group Title
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    you see where this is?

    • one year ago
  33. tiffanymak1996 Group Title
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    in -1/(1+1/n), if i substitute n as infinity, then 1/infinity is 0, so -1/(1+0)=-1/1

    • one year ago
  34. tiffanymak1996 Group Title
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    is this correct?

    • one year ago
  35. godorovg Group Title
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    when sub in n, you still have 1+1 which is 2.

    • one year ago
  36. tiffanymak1996 Group Title
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    where does the 1+1 com from?

    • one year ago
  37. godorovg Group Title
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    so if we look at this -1(1+1/n) would give you add the two ones which would 2/n times that by -1 and what is the answer?

    • one year ago
  38. tiffanymak1996 Group Title
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    so sorry, it should be -1/(1+(1/n)) not -1/(1+1/n)

    • one year ago
  39. godorovg Group Title
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    -1/1=-1times +1 =-1/n

    • one year ago
  40. tiffanymak1996 Group Title
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    \[\lim_{n \rightarrow +\infty}\frac{ -1 }{ 1+\frac{ 1 }{ n } }\]

    • one year ago
  41. godorovg Group Title
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    yes..

    • one year ago
  42. tiffanymak1996 Group Title
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    so... \[\lim_{n \rightarrow +\infty} \frac{ -1 }{ 1+\frac{ 1 }{ n } }\]

    • one year ago
  43. godorovg Group Title
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    Tiffany can the bottom be 0? or am I missing something here?

    • one year ago
  44. tiffanymak1996 Group Title
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    the bottom can't be zero, what i did is just divide top and bottom with n

    • one year ago
  45. tiffanymak1996 Group Title
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    and then: \[\frac{ -1 }{ 1+\frac{ 1 }{ \infty } }\]

    • one year ago
  46. tiffanymak1996 Group Title
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    and since \[\frac{ 1 }{ \infty }=0\],

    • one year ago
  47. tiffanymak1996 Group Title
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    \[\frac{ -1 }{ 1+0 }=-1\]

    • one year ago
  48. tiffanymak1996 Group Title
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    and i did this like 5 times and still don't get why it's divergent...

    • one year ago
  49. tiffanymak1996 Group Title
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    T.T

    • one year ago
  50. godorovg Group Title
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    what happens if we -1/1+0=-1 now what if simply the bottom to -1/1=-1 than what?

    • one year ago
  51. tiffanymak1996 Group Title
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    if it's -1 then the series is convergent, not divergent.

    • one year ago
  52. godorovg Group Title
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    yes..

    • one year ago
  53. tiffanymak1996 Group Title
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    well, the model answer IS divergent.

    • one year ago
  54. godorovg Group Title
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    only if stays at one, -1=-1 if you add one you get 0

    • one year ago
  55. tiffanymak1996 Group Title
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    and that's what frustrates me the most.

    • one year ago
  56. tiffanymak1996 Group Title
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    ok...

    • one year ago
  57. godorovg Group Title
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    so I am taking the answer can't be 0 for the final answer?

    • one year ago
  58. tiffanymak1996 Group Title
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    no, it's not that it can't be 0, just that how do u suddenly +1 after the limit?

    • one year ago
  59. godorovg Group Title
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    Because the +1 falls within the limit.

    • one year ago
  60. tiffanymak1996 Group Title
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    oh, ok, thx, i think i get it now. :)

    • one year ago
  61. godorovg Group Title
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    welcome ...

    • one year ago
  62. tiffanymak1996 Group Title
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    sry for bothering you so long...

    • one year ago
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