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anonymous

  • 5 years ago

sigma (k=1) to infinity (-1)^(k+1) sqrt(k)/k+1 . I tried to solve it using the ratio test, but got stuck in the end, and I can't use root test since the whole thing is not to the power of 1/k. Which way can I use?

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  1. anonymous
    • 5 years ago
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    \[\frac{(k+1)\sqrt(k+1)}{k^{3/2} + 2 \sqrt(k)}\] that's where I got stuck after using the ratio test

  2. anonymous
    • 5 years ago
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    \[\sum_{k=1}^{\infty}(-1)^{k+1} \frac{\sqrt(k)}{k+1}\] that's the question

  3. anonymous
    • 5 years ago
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    oh wow, alas, visitors lol ^_^

  4. anonymous
    • 5 years ago
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    It doesn't say what you are doing? test for convergence?

  5. anonymous
    • 5 years ago
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    lol, yes ^_^ but I'm not sure which way to choose, I tried ratio test

  6. anonymous
    • 5 years ago
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    umm it looks like you could maybe use divergence test

  7. anonymous
    • 5 years ago
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    kth term divergence test? ^_^

  8. anonymous
    • 5 years ago
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    so from where I stopped , I can use the k-th term divergence test? right?

  9. anonymous
    • 5 years ago
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    hold on

  10. anonymous
    • 5 years ago
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    kk ^_^

  11. anonymous
    • 5 years ago
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    ok with divergence test you get = 0 so that is inconclusive

  12. anonymous
    • 5 years ago
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    so i'll try a ratio

  13. anonymous
    • 5 years ago
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    no, so that means it's convergent!

  14. anonymous
    • 5 years ago
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    but, the book says it's conditionally convergent . That's where I'm confused :(

  15. anonymous
    • 5 years ago
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    oh I see now! The series is increasing, but it's limit tends to zero, so that's why it's conditonally convergent right? ^_^

  16. anonymous
    • 5 years ago
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    okay so when you get 0 from the divergence test, that tells you to try another test basically. And if you can get it to converge with another test (that is not absolute convergence) then its conditionally convergent

  17. anonymous
    • 5 years ago
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    um, that's a good guess. But I don't think so

  18. anonymous
    • 5 years ago
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    so with the ratio test I got 1

  19. anonymous
    • 5 years ago
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    so again, that means nothing

  20. anonymous
    • 5 years ago
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    um so what other tests are there

  21. anonymous
    • 5 years ago
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    AST, p-series, root test, geometric series

  22. anonymous
    • 5 years ago
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    isn't my logic right lol? the series must be decreasing to be absolutely convergent, but it's increasing instead

  23. anonymous
    • 5 years ago
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    I'm not sure enough to say yes or no to that. See what happens when you do more conditionally convergent problems

  24. anonymous
    • 5 years ago
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    what about intergral test

  25. anonymous
    • 5 years ago
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    OH I think I got it

  26. anonymous
    • 5 years ago
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    One sec

  27. anonymous
    • 5 years ago
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    why integral test? we don't have any exponents lol

  28. anonymous
    • 5 years ago
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    nvm. My final guess is that its similar to the alternating harmonic series, which is absolutely convergent.

  29. anonymous
    • 5 years ago
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    but it's conditionally convergent lol

  30. anonymous
    • 5 years ago
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    \[\sqrt{k}\div x = 1/(k^1/2)\]

  31. anonymous
    • 5 years ago
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    K^(1/2)/K = 1/(k^(1/2))

  32. anonymous
    • 5 years ago
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    yeah idk this one is a feather

  33. anonymous
    • 5 years ago
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    >_< watch your language

  34. anonymous
    • 5 years ago
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    ..serious?

  35. anonymous
    • 5 years ago
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    lol

  36. anonymous
    • 5 years ago
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    for series to converge they must be decreasing, but this one is increasing so that's why it's conditionally convergent ^_^

  37. anonymous
    • 5 years ago
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    right? .-.

  38. anonymous
    • 5 years ago
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    You can use the alternating series test.

  39. anonymous
    • 5 years ago
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    I did, but it didn't work >_<

  40. anonymous
    • 5 years ago
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    I ended up with 1 , which means no conclusion, the limit must be equal to 0

  41. anonymous
    • 5 years ago
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    You just need to show that sqrt(k)/(k+1) goes to zero as k goes to infinity (easy) and that sqrt(k)/(k+1) is monotonic decreasing. You can do this by taking the derivative and showing that the derivative is negative for all x greater than 1. \[f'(x)=\frac{1-x}{2\sqrt{x}(x+1)^2}=0 \rightarrow x=1\]partitions the interval. You take a test point x=4, say (anything in this interval) and show that \[f'(4)<0\]. It will be negative for all x in this interval (i.e x >= 1). By the Mean Value Theorem, you can choose any interval in [1, infinity), [a,b], say, and show that since f'(x)<0 on [1,infty), it is the case that\[f'(c)=\frac{f(b)-f(a)}{b-a}<0 \rightarrow f(b)<f(a)\]i.e. monotonic decreasing.

  42. anonymous
    • 5 years ago
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    but the limit is equal to = 1 loki

  43. anonymous
    • 5 years ago
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    \[\lim_{k \rightarrow \infty}\frac{k^{1/2}}{k+1}=\lim_{k \rightarrow \infty}\frac{1/k^{1/2}}{1+1/k}=0\]

  44. anonymous
    • 5 years ago
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    \[\lim_{k \rightarrow \infty} \frac{\sqrt(k)}{k+1} = \lim_{k \rightarrow \infty} \frac{|k|}{k} = 1\]

  45. anonymous
    • 5 years ago
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    No, it goes to zero...

  46. anonymous
    • 5 years ago
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    how come? .-.

  47. anonymous
    • 5 years ago
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    look above

  48. anonymous
    • 5 years ago
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    didn't I compute it right?

  49. anonymous
    • 5 years ago
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    I did lol

  50. anonymous
    • 5 years ago
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    I assume your numerator is\[\sqrt{k}\]

  51. anonymous
    • 5 years ago
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    there are alot of ways ^_^ you take the biggest power above, and the biggest power bellow then compute :)

  52. anonymous
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=lim+n+goes+to+infty+sqrt%28n%29%2F%281%2Bn%29

  53. anonymous
    • 5 years ago
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    yep, which will be |k|........oh wait! LOL

  54. anonymous
    • 5 years ago
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    >_< silly silly mistake, k^1/2 - K = 1/k = 0 you're right, I'm sorry ^_^"

  55. anonymous
    • 5 years ago
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    so it's limit = 0 and it's decreasing using the ratio test/finding the derivative, so by AST it's convergent?

  56. anonymous
    • 5 years ago
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    Yes

  57. anonymous
    • 5 years ago
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    but the book says it's conditionally convergent?

  58. anonymous
    • 5 years ago
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    that's the confusing part

  59. anonymous
    • 5 years ago
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    Well, the alternating series test is dependent upon the series having that (-)^k term in there. If it didn't, the sum wouldn't be killed off enough at every second step.

  60. anonymous
    • 5 years ago
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    Absolute convergence would have sqrt(k)/(k+1) as your series, not (-)^k etc...different.

  61. anonymous
    • 5 years ago
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    Actually, wait...

  62. anonymous
    • 5 years ago
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    hmm, the theorem says the following: if sigma (ak) is convergent, but sigma |ak| is divergent, then sigma|ak| is conditionally convergent. but using AST, both are convergent, which means that ak is absolutely convergent

  63. anonymous
    • 5 years ago
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    or wait! LOL you can use limit comparison test >_< where : \[ak = \frac{\sqrt(k)}{k+1}\] and\[bk = \frac{1}{\sqrt(k)}\]

  64. anonymous
    • 5 years ago
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    You can't use alternating series test on something that isn't alternating.

  65. anonymous
    • 5 years ago
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    |ak| isn't alternating.

  66. anonymous
    • 5 years ago
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    lol yep! so we can use limit comparison test ! ^_^

  67. anonymous
    • 5 years ago
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    then we must find limit of ak/bk

  68. anonymous
    • 5 years ago
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    right? :)

  69. anonymous
    • 5 years ago
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    I think you're being crazy. Use the alternating series test.

  70. anonymous
    • 5 years ago
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    Truuuuuuuuuuuusssssssssssssssttttttttttttttt mmmmmmmmmmmmmmmeeeeeeeeeeeeeeeeeeeee!!!!!!!!!!!!!!!! :)

  71. anonymous
    • 5 years ago
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    >_< lol, but AST gives absolute convergence, there's something wrong

  72. anonymous
    • 5 years ago
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    So are you now saying you want to test for convergence of |ak|?

  73. anonymous
    • 5 years ago
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    when I find the limit of ak/bk = 1 > 0, so it diverges hmmmm

  74. anonymous
    • 5 years ago
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    the question wants you to determine whether the series is absolutely convergent, conditionally convergent or divergent.

  75. anonymous
    • 5 years ago
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    wait limit comparison test won't work since it says if ak converges, then bk must converge and the vice versa, since they are a couple<- as pointed out by my professor lol

  76. anonymous
    • 5 years ago
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    WAIT! I GOT IT

  77. anonymous
    • 5 years ago
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    P -SERIES LOL

  78. anonymous
    • 5 years ago
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    Okay, if it's not abs. convergent, but convergent otherwise, it's conditionally convergent. If it's not convergent for either, it's divergent. We've shown it's convergent as is, now we show that it is divergent for |ak|...it will therefore be conditionally convergent.

  79. anonymous
    • 5 years ago
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    You can use the integral test for the absolute case.

  80. anonymous
    • 5 years ago
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    remember when I said that ak = swrt(k)/k+1 and bk = 1/sqrt(k)? bk is divergent as p-series, 1/2 < 1 so it's divergent, but ak is convergent, so it's conditionally convergent?

  81. anonymous
    • 5 years ago
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    why integral test? we use the integral test if we face exponentials and etc

  82. anonymous
    • 5 years ago
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    nah, it won't work, it's either both converge, or both diverge

  83. anonymous
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    have mercy T_T

  84. anonymous
    • 5 years ago
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    hold on, can't we use root test? ^_^

  85. anonymous
    • 5 years ago
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    You can use limit comparison, with what you have. I got limit =1. It's greater than zero and finite, and since 1/sqrt(k) is divergent, |ak| is divergent too.

  86. anonymous
    • 5 years ago
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    but ak is convergent, so that's why it's conditonally convergent!

  87. anonymous
    • 5 years ago
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    LOKI, you're a genius LOL

  88. anonymous
    • 5 years ago
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    You're finished. 1) Limit comparison using 1/sqrt{k} to show |ak| divergent 2) Alternating series to show ak is convergent 3) Therefore, the series is conditionally convergent.

  89. anonymous
    • 5 years ago
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    o_o why use both ways?

  90. anonymous
    • 5 years ago
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    you were right just now >_<

  91. anonymous
    • 5 years ago
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    abt the limit comparison test lol

  92. anonymous
    • 5 years ago
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    You have to use both ways. Don't type anything until I've finished...

  93. anonymous
    • 5 years ago
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    okay, I'll behave ._.

  94. anonymous
    • 5 years ago
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    Absolute convergence is a stronger deal than conditional convergence. If a series is absolutely convergent, it is conditionally convergent. So abs. conv. is a *sufficient* condition to show convergence. If a series is not absolutely convergent, it doesn't mean it is necessary divergent. It may still be convergent. If it is, and does not have absolute convergence, you have *conditional convergence*. So people attack these problems first by looking for absolute convergence...why? Because the tests are easy and abs. convergence guarantees the other convergence. If it isn't abs. convergent, you STILL have to check for convergence.

  95. anonymous
    • 5 years ago
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    You can write now :D

  96. anonymous
    • 5 years ago
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    so, you're saying that absolute convergence = conditional convergence?

  97. anonymous
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    = convergence?

  98. anonymous
    • 5 years ago
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    yes

  99. anonymous
    • 5 years ago
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    but conditional convergence does not have to equal absolute convergence

  100. anonymous
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    It doesn't go both ways.

  101. anonymous
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    that doesn't make sense lol, then why did they put the theorem? why did they say if ak diverges and |ak| converges, then |ak| is conditionally convergent?

  102. anonymous
    • 5 years ago
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    there must be a reason, since they also said if both converge then it's "absolutely" convergent.

  103. anonymous
    • 5 years ago
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    I've just checked my old lecture notes, and it says: Definition (Conditional Convergence): A series \[\sum_{}{}u_n\]which is convergent, but not absolutely convergent, is called *conditionally convergent*.

  104. anonymous
    • 5 years ago
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    when is it not absolutely convergent?

  105. anonymous
    • 5 years ago
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    When you show that\[\sum_{}{}|a_n|\]is divergent.

  106. anonymous
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    but we've showed that both are convergent?

  107. anonymous
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    No, we haven't. It's NOT absolutely convergent using the limit comparison test taking a test series of 1/sqrt{k} (which is divergent).

  108. anonymous
    • 5 years ago
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    O_O okay okay, but that is bk and not |ak|

  109. anonymous
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    What is bk? I have the following:\[a_k=(-)^{k+1}\frac{\sqrt{k}}{k+1}\]and \[|a_k|=\frac{\sqrt{k}}{k+1}\]

  110. anonymous
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    we've let |ak| = sqrt(k)/k+1 and bk = 1/sqrt(k)

  111. anonymous
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    so? .-. now we find the limit of ak/bk right?

  112. anonymous
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    whic is 0, but since bk isdivergent as p-series 1/2 < 1 then bk is divergent, then what?

  113. anonymous
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    whic is 0, but since bk isdivergent as p-series 1/2 < 1 then bk is divergent, then what?

  114. anonymous
    • 5 years ago
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    Yes, and so, since \[\sum_{k=1}^{\infty}\frac{1}{\sqrt{k}}\rightarrow \infty\]and also, since\[\lim_{k \rightarrow \infty}\frac{1/\sqrt{k}}{\left( \frac{\sqrt{k}}{1+k} \right)}=\lim_{k \rightarrow \infty} \frac{k+1}{k} =1\], and as the limit of the ratio is greater than zero and finite, you have by the limit comparison test that the series is divergent.

  115. anonymous
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    yes, but what abt the conditional convergence thing?

  116. anonymous
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    Okay...the proof I just did shows that it is NOT ABSOLUTELY CONVERGENT. So now we ask, is it CONDITIONALLY CONVERGENT? Use the alternating series test to show that \[\sum_{}{}a_k\]is CONVERGENT. Since it's NOT ABSOLUTELY CONVERGENT BUT CONVERGENT in the NON-ABSOLUTE it is CONDITIONALLY CONVERGENT

  117. anonymous
    • 5 years ago
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    O_O.....o.k

  118. anonymous
    • 5 years ago
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    phew

  119. anonymous
    • 5 years ago
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    I think you've misinterpreted a definition as a theorem.

  120. anonymous
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    thank you for being patient with me o_o lol, I can be stubborn sometimes ^_^"

  121. anonymous
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    So you're fine with the process?

  122. anonymous
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    yes, I'll just go over it again, thank you ^_^

  123. anonymous
    • 5 years ago
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    OK. I'm drawing a summary.

  124. anonymous
    • 5 years ago
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    alright~

  125. anonymous
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  126. anonymous
    • 5 years ago
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    thaaaaaaaaank yoouuuuu looookiiiiiii~

  127. anonymous
    • 5 years ago
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    Did that clear it up?

  128. anonymous
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    yep~ :) and I'm showing that to my professor lol

  129. anonymous
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    go get some sleep now ^_^

  130. anonymous
    • 5 years ago
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    oh...well, I hope he agrees!

  131. anonymous
    • 5 years ago
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    It's 5:30pm here - I'm more inclined to get dinner.

  132. anonymous
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    he will ~ :)

  133. anonymous
    • 5 years ago
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    lol, go eat then :) and bonne apetite if that's how you spell it lol

  134. anonymous
    • 5 years ago
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    I assume the spelling's right :)

  135. anonymous
    • 5 years ago
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    you and your assumptions lol, take care :)

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