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Loser66

  • one year ago

\(\sum_{n =1}^\infty (1/n) \) converges. How about \(\sum_{n=1}^\infty -(1/n)\) ? converges also? Please, help

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  1. Loser66
    • one year ago
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    @imqwerty

  2. idku
    • one year ago
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    diverges!

  3. Loser66
    • one year ago
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    why?

  4. idku
    • one year ago
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    sum from 1 to ∞ of 1/n Hormonic series

  5. idku
    • one year ago
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    Converges means sum is defined

  6. idku
    • one year ago
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    diverges means the sum doesn't exist.

  7. idku
    • one year ago
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    In case you mixed up terms

  8. idku
    • one year ago
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    \(\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...\) this is at least: \(\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\ge\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+...\) \(\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\ge\frac{1}{1}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...\)

  9. Loser66
    • one year ago
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    ok, I think I mess up something. Sorry about that .

  10. ParthKohli
    • one year ago
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    Harmonic series diverges. For the second, if you mean\[\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} = 1-\frac{1}{2} + \frac{1}3 - \frac{1}4 + \cdots\]The second sum is in fact convergent and equal to \(\ln 2\)!

  11. idku
    • one year ago
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    alternating series test. Converges

  12. Loser66
    • one year ago
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    Let me post the original one.

  13. Loser66
    • one year ago
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    \(\sum_{n=1}^\infty \dfrac{(-1)^n i^{n(n+1)}}{n}\) converge or diverge?

  14. idku
    • one year ago
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    \(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\) it meets conditions: \(\large \displaystyle \lim_{k\to \infty} (a_n)=0\) \(\large \displaystyle|a_n|>|a_{n+1}\) for all n. Thus, converges.

  15. idku
    • one year ago
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    i mean in the limit "n -> "

  16. idku
    • one year ago
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    what does i do in your series?

  17. idku
    • one year ago
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    is that *i* to indicate imaginary value, or what>?

  18. ParthKohli
    • one year ago
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    it does

  19. Loser66
    • one year ago
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    I lost the net......:(

  20. Loser66
    • one year ago
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    Please, leave your guide here, I will take it if I can see what you guys write.

  21. idku
    • one year ago
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    \(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n-1)}}{n}\) I haven't done series with imaginary numbers mixed in...... my guess is: \(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n-1)}}{n}\le\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=\rm converges\)

  22. ParthKohli
    • one year ago
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    \[n=1\Rightarrow i^{n(n+1)}=-1\]\[n=2 \Rightarrow i^{n(n+1)}=-1\]\[n=3 \Rightarrow i^{n(n+1)}=1\]\[n=4 \Rightarrow i^{n(n+1)}=1\]And this continues.

  23. imqwerty
    • one year ago
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    \[a_{n}=\frac{ -1 }{ n }\] \[\lim_{x \rightarrow \infty}\frac{ -1 }{ n } =0\] so because the limit is approaching a certain value we will say that it converges

  24. ParthKohli
    • one year ago
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    Nah, it diverges @imqwerty

  25. ParthKohli
    • one year ago
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    Even though the limit of the sequence exists and is equal to zero, that is not sufficient to say that the series should converge

  26. imqwerty
    • one year ago
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    but whenever we take the limit and if it approaches certain value then it converges right?

  27. ParthKohli
    • one year ago
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    nope...

  28. idku
    • one year ago
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    \(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n-1)}}{n}\) \(\large \displaystyle\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n}{n+1}\right)\)

  29. idku
    • one year ago
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    is that right?

  30. ParthKohli
    • one year ago
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    first of all, for convergence, the limit has to be zero and no other value. but that is not sufficient for convergence.

  31. imqwerty
    • one year ago
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    we will also consider the limit of partial sums right?

  32. idku
    • one year ago
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    yes, true dirverges test only shows divergence if lim \(\ne\)0

  33. idku
    • one year ago
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    divergence test ... i meant

  34. ParthKohli
    • one year ago
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    yes

  35. idku
    • one year ago
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    I was thinking it is an equivalent representation: \(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n}{n+1}\right)\)

  36. Empty
    • one year ago
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    The divergence test is not called the convergence test for the reason that it can only show you if it diverges or not.

  37. idku
    • one year ago
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    yes.

  38. idku
    • one year ago
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    But is this reasonable? (What I said above)

  39. imqwerty
    • one year ago
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    why cant we say that a series diverges or converges by jst calculating the limit ?

  40. Empty
    • one year ago
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    The intuitive reason for what happens when you use the divergence test is this: You have an infinite sum. If the 'last' term isn't 0, then you know it diverges since infinity never ends.

  41. idku
    • one year ago
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    Because when the sequence an approaches 0, that can be a divergent series \(a_n|) eg Harmonic series

  42. idku
    • one year ago
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    If I am right, \(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n}{n+1}\right)\) then the series converges

  43. Empty
    • one year ago
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    you can have your 'last' term equal to 0 in an infinite sum, but you already went through an infinite number of terms prior to that, so those could still possibly diverge.

  44. idku
    • one year ago
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    and that would in fact be: ln(2)+ [ ln(2)-1 ] = 2ln(2) -1

  45. idku
    • one year ago
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    for my representation, if it is right

  46. idku
    • one year ago
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    I checked the signs, so should be fine.

  47. idku
    • one year ago
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    i is imaginary

  48. idku
    • one year ago
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    so, as parth said the i's yield: -1 -1 1 1

  49. Empty
    • one year ago
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    I don't understand I thought the problem was \[\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\]

  50. idku
    • one year ago
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    \(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}\) this is the prob

  51. idku
    • one year ago
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    and then I suggest that just as: \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}=1/1-1/2-1/3+1/4+1/5-1/6-1/7+1/8...\) so is: \(\displaystyle\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^{n}}{n+1}\right)=1/1-1/2-1/3+1/4+1/5-1/6-1/7+1/8...\)

  52. Empty
    • one year ago
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    I was looking at the wrong problem no wonder haha

  53. idku
    • one year ago
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    oh no not that

  54. idku
    • one year ago
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    i was wrong, because my terms cancel, I have to skeep over 1

  55. idku
    • one year ago
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    \(\displaystyle\large\sum_{n=0}^{\infty}\left(\frac{(-1)^{n}}{2n+1}+\frac{(-1)^{n+1}}{2n+2}\right)\)

  56. idku
    • one year ago
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    this should be better: = +1/1 - 1/2 - 1/3 + 1/4 +1/5 - 1/6 - 1/7 + 1/8 +1/9 - 1/10 - 1/11 + 1/12 etc.... but that cycle is essentially just a combo of two altern. series. \(\pi/4\) + -ln(2)/2 \(\left[\pi-2\ln2\right]/2\)

  57. Empty
    • one year ago
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    Actually we can probably evaluate it but to see if this series converges or diverges I think we can use a bit of a trick, look at the real and imaginary parts separately: \[\Re(S) = \frac{S+S^*}{2}\]\[\Im(S) = \frac{S-S^*}{i2}\] These are both going to be completely real series themselves, and you can use the alternating series test to check for convergence on both: \[\Re(S) = \sum_{n=1}^\infty \frac{(-1)^n}{n} \cos\left( \frac{\pi}{2} n(n+1) \right) \] \[\Im(S) = \sum_{n=1}^\infty \frac{(-1)^n}{n} \sin\left( \frac{\pi}{2} n(n+1) \right) \] Inside you have \(\frac{\pi}{2}\) so every 4 values of n you'll repeat so you could separate it out like this into 4 terms, find out what the cosine and sine terms are doing, they'll only give you 0, 1, or -1 in some funny order at worst, but we can see that the inside will repeat every 4 because of this: \[\cos[ \frac{\pi}{2}n(n+1)]\] \[n \to n+4\] \[\cos [ \frac{\pi}{2}(n+4)(n+4+1)] = \cos[ \frac{\pi}{2}n(n+1)+\frac{\pi}{2}(4n+4n+4)]=\cos[ \frac{\pi}{2}n(n+1)]\] really it's just case \(\frac{\pi}{2}(4n+4n+4)\) is a multiple of \(2 \pi\). Maybe this ends up sounding more complicated than just solving this damn thing I don't know haha. But it's just 4 terms haha.

  58. idku
    • one year ago
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    Loser66, how are you so far?

  59. Loser66
    • one year ago
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    :) To me, I do something like let \(a_n = 1/n\) \(b_n = (-1)^n i^{n(n+1)}\) Now, just calculate \(b_n\) Note: \(i^2 =-1\\n(n+1) is~~even\) Hence, for n = 4k, \(b_n = (-1)^4k i^{4k (4k +1) }= 1* ((i^2)^2)^{k(4k+1)}=1\) Hence \(b_n =1, \sum_{n=1}^\infty a_n b_n = \sum_{n=1}^\infty \dfrac{1}{n}\). You say, is it divergent or convergent?

  60. Loser66
    • one year ago
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    for n= 4k+1, similar argument \[b_n = (-1)^{4k+1}i^{(4k+1)(4k+2)}= -1* (i^2)^{(4k+1)(2k+1)}= -1*-1=1\] Hence \[\sum_{n=1}^\infty a_n b_n = \sum_{n=1}^\infty \dfrac{1}{n}\] You say, is it divergent or convergent?

  61. idku
    • one year ago
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    I hope you aren't factoring all that imaginary component out of the series......

  62. idku
    • one year ago
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    also, it is not (-1)^(n(n+1)), but (i)^(n(n+1))

  63. idku
    • one year ago
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    so it is not always positive

  64. idku
    • one year ago
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    -1, -1, 1, 1.....

  65. Loser66
    • one year ago
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    Hey, all natural number can be expressed as one of the forms : 4k, 4k+1, 4k+2 , 4k+3.

  66. Loser66
    • one year ago
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    So as n

  67. Loser66
    • one year ago
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    Why do we have to do that? because that is the way we turn imaginary i to real since i^2 =-1

  68. Loser66
    • one year ago
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    and the circle to get real number from i is 4. I was taught that. :)

  69. Loser66
    • one year ago
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    \(\sum_{n=1}^\infty a_n = 0~~ if~~ n \cong 0~~~ mod4\) \(\sum_{n=1}^\infty a_n = -1~~ if~~ n \cong 1~~~ mod4\) \(\sum_{n=1}^\infty a_n = 0~~ if~~ n \cong2~~~ mod4\) \(\sum_{n=1}^\infty a_n = 1~~ if~~ n \cong 3~~~ mod4\) That shows the sum is divergent.

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