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anonymous

  • 5 years ago

show or prove that if series |ak| converges then series ak converges. Details: the converse is clearly false. for example series (-1)^n/n converges by AST but series 1/n diverges

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  1. anonymous
    • 5 years ago
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    Assume |ak| converges then assume ak will diverge and show that it doesn't. (this might or might not work but that's what I would try)

  2. anonymous
    • 5 years ago
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    hell

  3. anonymous
    • 5 years ago
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    if series |ak| converges then lim |ak|=0, then lim ak=0 , fine. then what does that show

  4. anonymous
    • 5 years ago
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    i didnt mean to write hell

  5. anonymous
    • 5 years ago
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    that's alright

  6. anonymous
    • 5 years ago
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    If lim ak = 0 that gets us started

  7. anonymous
    • 5 years ago
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    i meant to write hello

  8. anonymous
    • 5 years ago
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    because now it rules out ak diverging for sure and we are left with having to show that it converges

  9. anonymous
    • 5 years ago
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    ok

  10. anonymous
    • 5 years ago
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    so we have shown that lim ak = 0 , and by contradiction assumption we have that series ak diverges

  11. anonymous
    • 5 years ago
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    we may or may not need contradiction, I haven't actually done out the whole proof, I'm just thinking out loud

  12. anonymous
    • 5 years ago
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    because you might be able to use one of the tests for convergence (maybe the alternating series test) to show convergence

  13. anonymous
    • 5 years ago
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    oh , we have two possibilities, either series |ak| = series ak, or it doesnt (by a negative factor)

  14. anonymous
    • 5 years ago
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    if series |ak| = series ak then we have a contradiction. if series |ak| != series ak , then the left side must be an alternating series?

  15. anonymous
    • 5 years ago
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    I may have another approach to the proof to share.

  16. anonymous
    • 5 years ago
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    either |ak| = ak, the sequence, or it doesnt. if they are not equal then ... oh i could be wrong about the alternating series. some series dont alternate like 1/3 + 1/5 - 1/7 - 1/9 ...

  17. anonymous
    • 5 years ago
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    no i was wrong about that, you can have series |ak| not equal to series ak but series ak is not alternating.

  18. anonymous
    • 5 years ago
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    |ak| is either -ak or ak, then we can write this inequality: \[0\le a_k+\left| a_k \right|\le2 \left| a_k \right|\] For instance, I want to show that ak + |ak| is convergent.

  19. anonymous
    • 5 years ago
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    ok where ak is a term in the sequence

  20. anonymous
    • 5 years ago
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    ok so we have series [ ak + |ak| ] < = series 2 |ak|

  21. anonymous
    • 5 years ago
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    because 0 <= [ ak + |ak| ] < = 2 |ak| can you sum both sides ?

  22. anonymous
    • 5 years ago
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    yes we can so we have series ak + series |ak| < = series 2|ak|

  23. anonymous
    • 5 years ago
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    subtract series |ak| from both sides

  24. anonymous
    • 5 years ago
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    so series ak < series |ak| , and if series |ak| diverges than series ak must diverge , contradiction

  25. anonymous
    • 5 years ago
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    First, 2|ak| is clearly a convergent series, since it's just a multiple of one. That implies that ak+|ak| is also a convergent by comparison. Having proved that, consider: \[\sum_{}^{}a_k=\sum_{}^{}(a_k+\left| a_k \right|)-\sum_{}^{}\left| a_k \right|\] Therefore, ak is the difference between two convergent series, and hence it's convergent.

  26. anonymous
    • 5 years ago
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    oh, i think there is a simpler proof. ak <= |ak| , and series ak <= series |ak| , but series |ak| diverges by reductio hypothesis , so series ak must diverge ?

  27. anonymous
    • 5 years ago
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    You want to prove convergence?

  28. anonymous
    • 5 years ago
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    yes but we already assumed ak is convergent, we want to show a contradiction

  29. anonymous
    • 5 years ago
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    oh youre doing a direct proof

  30. anonymous
    • 5 years ago
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    you actually dont need a contradiction here

  31. anonymous
    • 5 years ago
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    I only thought it would help just by taking a quick look at the problem but the way AnwarA posted works better

  32. anonymous
    • 5 years ago
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    I really don't understand, What is it exactly that you want to prove?

  33. anonymous
    • 5 years ago
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    oh

  34. anonymous
    • 5 years ago
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    sorry we were approaching this differently, you proved it

  35. anonymous
    • 5 years ago
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    isnt there a simpler proof, here

  36. anonymous
    • 5 years ago
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    ak < = |ak| , sum both sides series ak <= series |ak| by assumption series |ak| converges... so series ak must converge, oh but ak does not have to be greater than zero, i see ,

  37. anonymous
    • 5 years ago
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    how did you get this

  38. anonymous
    • 5 years ago
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    Get what?

  39. anonymous
    • 5 years ago
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    series ak = series [ (ak + |ak| ) - series |ak |

  40. anonymous
    • 5 years ago
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    so my proof doesnt work ?

  41. anonymous
    • 5 years ago
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    can you tiny url that

  42. anonymous
    • 5 years ago
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    if you copy and paste it you should get it

  43. anonymous
    • 5 years ago
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    It discusses absolute convergence and the first part is the proof AnwarA just posted.

  44. anonymous
    • 5 years ago
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    so you did series ak = series [ ak + |ak| - |ak| ] , and then you split that series on the right

  45. anonymous
    • 5 years ago
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    but theres some condition about splitting terms in series

  46. anonymous
    • 5 years ago
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    series (an + bn ) = series an + series bn , as long as ...

  47. anonymous
    • 5 years ago
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    as long as series an and series bn converge

  48. anonymous
    • 5 years ago
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    sorry you might need to refresh openstudy.com

  49. anonymous
    • 5 years ago
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    Well, it's clear: \[\sum_{}^{}a_k=\sum_{}^{}(a_k+\left| a_k \right|)-\sum_{}^{}\left| a_k \right|=\sum_{}^{}a_k-\sum_{}^{}\left| a_k \right|+\sum_{}^{}\left| a_k \right|=\sum_{}^{}a_k\]

  50. anonymous
    • 5 years ago
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    No i dont you can do that my other question is , which kind of goes with this, prove lim |an| = 0 iff lim an = 0

  51. anonymous
    • 5 years ago
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    suppose you have series [ 2/n - 1/n] = series 2/n - series 1/n ?

  52. anonymous
    • 5 years ago
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    infinity - infinity is not determinate

  53. anonymous
    • 5 years ago
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    im trying to think of a counterexample, where series (an + bn) != series an + series bn

  54. anonymous
    • 5 years ago
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    like an infinite case

  55. anonymous
    • 5 years ago
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    You're confusing me a little bit :).. These two things are totally different, I mean your last three replies.

  56. anonymous
    • 5 years ago
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    well linearity of series works only under certain conditions

  57. anonymous
    • 5 years ago
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    So you're asking if series [ak+an]=series [ak]+series [an]?

  58. anonymous
    • 5 years ago
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    right, you implicitly used that , under what conditions is that true

  59. anonymous
    • 5 years ago
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    Hmm. I don't think there are any conditions.

  60. anonymous
    • 5 years ago
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    ok , how do you prove that lim |an| = 0 iff lim an = 0

  61. anonymous
    • 5 years ago
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    Oh wait, It's true for ak and an convergent series.

  62. anonymous
    • 5 years ago
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    ahhh

  63. anonymous
    • 5 years ago
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    since you might get funny results with divergent series

  64. anonymous
    • 5 years ago
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    ok to recap, we showed that absolute convergence is a stronger condition. since if series |ak| converges then series ak converges. but the converse is not necessarily true, so its called conditional convergence , ie series ak converges but |ak| does not converge

  65. anonymous
    • 5 years ago
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    if we can show series |ak| converges, we get series ak converging for free

  66. anonymous
    • 5 years ago
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    To prove that lim |an| = 0 iff lim an = 0, we have to do it in two direction (since it's iff statement). The first part is to prove that lim |an|=0 if lim an=0 as n approaches infinity.

  67. anonymous
    • 5 years ago
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    Yeah, you're right.

  68. anonymous
    • 5 years ago
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    about which part

  69. anonymous
    • 5 years ago
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    oh the stronger aspect

  70. anonymous
    • 5 years ago
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    Yeah.. Let me finish the first part of the proof :)

  71. anonymous
    • 5 years ago
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    sure, sorry for interrupting. youre the best man

  72. anonymous
    • 5 years ago
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    |an| is either an or -an. If an>=0, that's |an|=an, then: \[\lim_{n \rightarrow \infty}\left| a_n \right|=\lim_{n \rightarrow \infty}a_n=0\]

  73. anonymous
    • 5 years ago
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    ok

  74. anonymous
    • 5 years ago
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    if an <0 then ?

  75. anonymous
    • 5 years ago
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    youre proving that if lim |an| = 0 -> lim an = 0

  76. anonymous
    • 5 years ago
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    If an<0 then |an|=-an, that's: \[\lim_{n \rightarrow \infty}\left| a_n \right|=-\lim_{n \rightarrow \infty}a_n=0\]

  77. anonymous
    • 5 years ago
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    but you can have a mixture of positive and negative

  78. anonymous
    • 5 years ago
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    like alternating , such as a1 >0 , a2 < 0 , etc

  79. anonymous
    • 5 years ago
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    No, I am proving the opposite. That's if I KNOW that lim an=0, then lim |an|=0. I took the two cases when an>=0 or an<0.

  80. anonymous
    • 5 years ago
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    oh

  81. anonymous
    • 5 years ago
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    Do you see it? Try reading it again.

  82. anonymous
    • 5 years ago
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    yes but an can alternate

  83. anonymous
    • 5 years ago
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    An

  84. anonymous
    • 5 years ago
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    Yeah, that makes no difference since you ALREADY KNOW that lim |an| is zero.

  85. anonymous
    • 5 years ago
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    but that*

  86. anonymous
    • 5 years ago
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    i though you said we assume lim an = 0

  87. anonymous
    • 5 years ago
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    and we want to prove lim |an | = 0

  88. anonymous
    • 5 years ago
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    We are not assuming that, the question is. GO and read your question again.

  89. anonymous
    • 5 years ago
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    ok theres too parts

  90. anonymous
    • 5 years ago
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    assuming lim |an| = 0, then show lim an = 0 the other one is assuming lim an = 0 show lim |an | = 0

  91. anonymous
    • 5 years ago
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    Exactly :)

  92. anonymous
    • 5 years ago
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    so which one did you prove

  93. anonymous
    • 5 years ago
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    We have done the second part, you should try the first one.

  94. anonymous
    • 5 years ago
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    but youre mixing up specific ak and a general ak

  95. anonymous
    • 5 years ago
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    What do you mean?

  96. anonymous
    • 5 years ago
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    well for example , take (-1)^n / n^2 , the sequence

  97. anonymous
    • 5 years ago
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    Ok?

  98. anonymous
    • 5 years ago
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    so you said that either |an| = an or | an | = -an, but in this case its neither

  99. anonymous
    • 5 years ago
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    Yeah, this is the definition of absolute value.

  100. anonymous
    • 5 years ago
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    no its neither

  101. anonymous
    • 5 years ago
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    Are you sure?

  102. anonymous
    • 5 years ago
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    here | (-1)^n / n^2 ) | != (-1)^n / n^2, and | (-1)^n / n^2 | != - (-1)^n / n^2

  103. anonymous
    • 5 years ago
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    where != means not equal

  104. anonymous
    • 5 years ago
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    so your statement or proof is incorrect

  105. anonymous
    • 5 years ago
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    | (-1)^n / n^2 ) |= 1 / n^2

  106. anonymous
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    an = (-1)^n / n^2, |an| is not equal to -an , nor an .

  107. anonymous
    • 5 years ago
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    :) I said for an>=0 |an|=an, and for an<0 |an|=-an. This is what absolute value is.

  108. anonymous
    • 5 years ago
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    yes and ?

  109. anonymous
    • 5 years ago
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    we have to show for all lim |an| = 0 , how are you going to put those two things together

  110. anonymous
    • 5 years ago
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    wait so youre splitting up the limit ?

  111. anonymous
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    It's clear!! the limit of the negative part is zero, and so is the positive part. right?

  112. anonymous
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    what do you mean

  113. anonymous
    • 5 years ago
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    oh yes

  114. anonymous
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    the limit of the negative terms is 0

  115. anonymous
    • 5 years ago
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    Got it?

  116. anonymous
    • 5 years ago
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    but how do you show the whole limit is going to zeor

  117. anonymous
    • 5 years ago
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    zero

  118. anonymous
    • 5 years ago
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    you showed that some of the terms go to zero, and some of the terms term go to zero

  119. anonymous
    • 5 years ago
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    Limit is a linear operation, isn't it?

  120. anonymous
    • 5 years ago
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    so the positive terms go to zero, true, and the negative terms go to zero, true , and?

  121. anonymous
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    so you split An into positive and negative terms ?

  122. anonymous
    • 5 years ago
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    You can say so.

  123. anonymous
    • 5 years ago
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    but thats not rigorous

  124. anonymous
    • 5 years ago
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    maybe a contradiction proof

  125. anonymous
    • 5 years ago
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    but graphically it makes sense

  126. anonymous
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    if we graph An = f(n) on the x y plane

  127. anonymous
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    the negative terms are below the x axis

  128. anonymous
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    and they are going to zero

  129. anonymous
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    the positive terms above the x axis are going to zero

  130. anonymous
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    not rigorous? show me what a rigorous proof is.

  131. anonymous
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    err, not deductive

  132. anonymous
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    ok i disproved your proof

  133. anonymous
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    simple counterexample

  134. anonymous
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    are you ready

  135. anonymous
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    Lol, It's a "proof", you can't disprove it. But sure, show me what you got :)

  136. anonymous
    • 5 years ago
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    what if an = (-1)^n / n^2

  137. anonymous
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    no it isnt

  138. anonymous
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    its not a proof

  139. anonymous
    • 5 years ago
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    you wrote that lim |an| = lim an or lim |an| = - lim an , FALSE

  140. anonymous
    • 5 years ago
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    lim |an | = lim an or that lim |an | = lim -an , and i showed you a counterexample

  141. anonymous
    • 5 years ago
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    Yeah?! :)

  142. anonymous
    • 5 years ago
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    thats what you wrote, its a false statement

  143. anonymous
    • 5 years ago
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    Claim: lim |an | = lim an or that lim |an | = lim (-an) False,. let an = (-1)^n / n^2

  144. anonymous
    • 5 years ago
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    lim | (-1)^n / n^2 | is not lim (-1)^n / n^2 and it is not lim - (-1)^n / n^2

  145. anonymous
    • 5 years ago
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    It's lim an when an>=0 and it's lim -an when an<0.

  146. anonymous
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    an = (-1)^n / n^2 lim |an| is not equal to lim an, and lim |an| is not equal to lim (-an)

  147. anonymous
    • 5 years ago
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    and we need to show lim |an| = 0 for all an

  148. anonymous
    • 5 years ago
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    Yeah.

  149. anonymous
    • 5 years ago
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    you didnt show for all an, you showed that if an is always positive, or if an is always negative

  150. anonymous
    • 5 years ago
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    You didn't even bother to understand what I wrote.

  151. anonymous
    • 5 years ago
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    sure i did, relax for a second,

  152. anonymous
    • 5 years ago
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    I am all relaxed.

  153. anonymous
    • 5 years ago
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    you proved that lim |an| goes to zero if an is always positive, and you proved lim |an| goes to zero when an is always negative

  154. anonymous
    • 5 years ago
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    THATS what you proved

  155. anonymous
    • 5 years ago
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    Is it?

  156. anonymous
    • 5 years ago
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    you proved two seperate statements, and thats NOT the conclusion

  157. anonymous
    • 5 years ago
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    yup, go back and think about it

  158. anonymous
    • 5 years ago
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    do you remember your proof? you proved that if an is positive, so and so, if an is negative so and so. thats all. what if an is positive AND negative

  159. anonymous
    • 5 years ago
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    if an is positive lim |an| = 0 , fine. if an is negative lim |an| = 0. great, and? what if an is both positive and negative

  160. anonymous
    • 5 years ago
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    and my counterexample blows your proof to pieces

  161. anonymous
    • 5 years ago
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    an = (-1)^n/n^2, your proof does not handle such a creature

  162. anonymous
    • 5 years ago
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    Haha

  163. anonymous
    • 5 years ago
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    Is this a college calculus course?

  164. anonymous
    • 5 years ago
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    Here is another proof then.

  165. anonymous
    • 5 years ago
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    no, just me thinking

  166. anonymous
    • 5 years ago
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    good thing you didnt freak out and starting calling me names , thanks

  167. anonymous
    • 5 years ago
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    some people cant handle the pressure,

  168. anonymous
    • 5 years ago
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    i like to know all the theorems relevant to a course, like calculus. and i like to see their proofs, if possible

  169. anonymous
    • 5 years ago
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    If lim |an|=0, then -lim|an|=0, and since -|an|<=an<=|an|, lim an=0 by the squeeze theorem.

  170. anonymous
    • 5 years ago
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    ok

  171. anonymous
    • 5 years ago
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    Is that a "proof"?

  172. anonymous
    • 5 years ago
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    the other direction is tougher

  173. anonymous
    • 5 years ago
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    YES , lol

  174. anonymous
    • 5 years ago
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    you done good

  175. anonymous
    • 5 years ago
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    :) I won't call you names other than cartorest (I don't know if that's even a name:P)

  176. anonymous
    • 5 years ago
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    no i meant, you didnt freak out when i challenged your proof

  177. anonymous
    • 5 years ago
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    Haha, why should I freak out? I like that actually.

  178. anonymous
    • 5 years ago
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    :)

  179. anonymous
    • 5 years ago
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    Good luck in your study. I had fun "Studying" with you tonight.

  180. anonymous
    • 5 years ago
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    so , prove if lim an = 0 then lim |an | = 0

  181. anonymous
    • 5 years ago
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    wait

  182. anonymous
    • 5 years ago
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    the entree, the main course,

  183. anonymous
    • 5 years ago
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    Yeah?

  184. anonymous
    • 5 years ago
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    whats the proof , you proved the converse

  185. anonymous
    • 5 years ago
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    I don't have another proof other than what I did before right now.

  186. anonymous
    • 5 years ago
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    you mean the non proof

  187. anonymous
    • 5 years ago
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    ok i have an idea. lets use contradiction, assume lim an = 0 and lim |an| != 0

  188. anonymous
    • 5 years ago
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    i dont know

  189. anonymous
    • 5 years ago
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    well ill post it , thanks for your help. just realize you did not prove that lim |an | = 0 implies that lim an = 0

  190. anonymous
    • 5 years ago
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    i mean, my bad

  191. anonymous
    • 5 years ago
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    I don't know either. I am so sleepy!!

  192. anonymous
    • 5 years ago
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    you proved lim |an| = 0 then lim an = 0 .

  193. anonymous
    • 5 years ago
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    with the sandwich theorem, but the converse we didnt prove yet

  194. anonymous
    • 5 years ago
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    I have to go now, I'll catch up with you tomorrow.

  195. anonymous
    • 5 years ago
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    Bye!! :)

  196. anonymous
    • 5 years ago
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    ok , i just wanted to be clear we are in agreement

  197. anonymous
    • 5 years ago
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    You really are funny.

  198. anonymous
    • 5 years ago
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    well unlike religious views, math people MUST agree on things

  199. anonymous
    • 5 years ago
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    i hope thats a compliment

  200. anonymous
    • 5 years ago
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    It is :)

  201. anonymous
    • 5 years ago
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    alright, maybe we will chat tomorrow , adios

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